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Vincent Selhorst-Jones

Vincent Selhorst-Jones

Intermediate Value Theorem and Polynomial Division

Slide Duration:

Table of Contents

I. Introduction
Introduction to Precalculus

10m 3s

Intro
0:00
Title of the Course
0:06
Different Names for the Course
0:07
Precalculus
0:12
Math Analysis
0:14
Trigonometry
0:16
Algebra III
0:20
Geometry II
0:24
College Algebra
0:30
Same Concepts
0:36
How do the Lessons Work?
0:54
Introducing Concepts
0:56
Apply Concepts
1:04
Go through Examples
1:25
Who is this Course For?
1:38
Those Who Need eExtra Help with Class Work
1:52
Those Working on Material but not in Formal Class at School
1:54
Those Who Want a Refresher
2:00
Try to Watch the Whole Lesson
2:20
Understanding is So Important
3:56
What to Watch First
5:26
Lesson #2: Sets, Elements, and Numbers
5:30
Lesson #7: Idea of a Function
5:33
Lesson #6: Word Problems
6:04
What to Watch First, cont.
6:46
Lesson #2: Sets, Elements and Numbers
6:56
Lesson #3: Variables, Equations, and Algebra
6:58
Lesson #4: Coordinate Systems
7:00
Lesson #5: Midpoint, Distance, the Pythagorean Theorem and Slope
7:02
Lesson #6: Word Problems
7:10
Lesson #7: Idea of a Function
7:12
Lesson #8: Graphs
7:14
Graphing Calculator Appendix
7:40
What to Watch Last
8:46
Let's get Started!
9:48
Sets, Elements, & Numbers

45m 11s

Intro
0:00
Introduction
0:05
Sets and Elements
1:19
Set
1:20
Element
1:23
Name a Set
2:20
Order The Elements Appear In Has No Effect on the Set
2:55
Describing/ Defining Sets
3:28
Directly Say All the Elements
3:36
Clearly Describing All the Members of the Set
3:55
Describing the Quality (or Qualities) Each member Of the Set Has In Common
4:32
Symbols: 'Element of' and 'Subset of'
6:01
Symbol is ∈
6:03
Subset Symbol is ⊂
6:35
Empty Set
8:07
Symbol is ∅
8:20
Since It's Empty, It is a Subset of All Sets
8:44
Union and Intersection
9:54
Union Symbol is ∪
10:08
Intersection Symbol is ∩
10:18
Sets Can Be Weird Stuff
12:26
Can Have Elements in a Set
12:50
We Can Have Infinite Sets
13:09
Example
13:22
Consider a Set Where We Take a Word and Then Repeat It An Ever Increasing Number of Times
14:08
This Set Has Infinitely Many Distinct Elements
14:40
Numbers as Sets
16:03
Natural Numbers ℕ
16:16
Including 0 and the Negatives ℤ
18:13
Rational Numbers ℚ
19:27
Can Express Rational Numbers with Decimal Expansions
22:05
Irrational Numbers
23:37
Real Numbers ℝ: Put the Rational and Irrational Numbers Together
25:15
Interval Notation and the Real Numbers
26:45
Include the End Numbers
27:06
Exclude the End Numbers
27:33
Example
28:28
Interval Notation: Infinity
29:09
Use -∞ or ∞ to Show an Interval Going on Forever in One Direction or the Other
29:14
Always Use Parentheses
29:50
Examples
30:27
Example 1
31:23
Example 2
35:26
Example 3
38:02
Example 4
42:21
Variables, Equations, & Algebra

35m 31s

Intro
0:00
What is a Variable?
0:05
A Variable is a Placeholder for a Number
0:11
Affects the Output of a Function or a Dependent Variable
0:24
Naming Variables
1:51
Useful to Use Symbols
2:21
What is a Constant?
4:14
A Constant is a Fixed, Unchanging Number
4:28
We Might Refer to a Symbol Representing a Number as a Constant
4:51
What is a Coefficient?
5:33
A Coefficient is a Multiplicative Factor on a Variable
5:37
Not All Coefficients are Constants
5:51
Expressions and Equations
6:42
An Expression is a String of Mathematical Symbols That Make Sense Used Together
7:05
An Equation is a Statement That Two Expression Have the Same Value
8:20
The Idea of Algebra
8:51
Equality
8:59
If Two Things Are the Same *Equal), Then We Can Do the Exact Same Operation to Both and the Results Will Be the Same
9:41
Always Do The Exact Same Thing to Both Sides
12:22
Solving Equations
13:23
When You Are Asked to Solve an Equation, You Are Being Asked to Solve for Something
13:33
Look For What Values Makes the Equation True
13:38
Isolate the Variable by Doing Algebra
14:37
Order of Operations
16:02
Why Certain Operations are Grouped
17:01
When You Don't Have to Worry About Order
17:39
Distributive Property
18:15
It Allows Multiplication to Act Over Addition in Parentheses
18:23
We Can Use the Distributive Property in Reverse to Combine Like Terms
19:05
Substitution
20:03
Use Information From One Equation in Another Equation
20:07
Put Your Substitution in Parentheses
20:44
Example 1
23:17
Example 2
25:49
Example 3
28:11
Example 4
30:02
Coordinate Systems

35m 2s

Intro
0:00
Inherent Order in ℝ
0:05
Real Numbers Come with an Inherent Order
0:11
Positive Numbers
0:21
Negative Numbers
0:58
'Less Than' and 'Greater Than'
2:04
Tip To Help You Remember the Signs
2:56
Inequality
4:06
Less Than or Equal and Greater Than or Equal
4:51
One Dimension: The Number Line
5:36
Graphically Represent ℝ on a Number Line
5:43
Note on Infinities
5:57
With the Number Line, We Can Directly See the Order We Put on ℝ
6:35
Ordered Pairs
7:22
Example
7:34
Allows Us to Talk About Two Numbers at the Same Time
9:41
Ordered Pairs of Real Numbers Cannot be Put Into an Order Like we Did with ℝ
10:41
Two Dimensions: The Plane
13:13
We Can Represent Ordered Pairs with the Plane
13:24
Intersection is known as the Origin
14:31
Plotting the Point
14:32
Plane = Coordinate Plane = Cartesian Plane = ℝ²
17:46
The Plane and Quadrants
18:50
Quadrant I
19:04
Quadrant II
19:21
Quadrant III
20:04
Quadrant IV
20:20
Three Dimensions: Space
21:02
Create Ordered Triplets
21:09
Visually Represent This
21:19
Three-Dimension = Space = ℝ³
21:47
Higher Dimensions
22:24
If We Have n Dimensions, We Call It n-Dimensional Space or ℝ to the nth Power
22:31
We Can Represent Places In This n-Dimensional Space As Ordered Groupings of n Numbers
22:41
Hard to Visualize Higher Dimensional Spaces
23:18
Example 1
25:07
Example 2
26:10
Example 3
28:58
Example 4
31:05
Midpoints, Distance, the Pythagorean Theorem, & Slope

48m 43s

Intro
0:00
Introduction
0:07
Midpoint: One Dimension
2:09
Example of Something More Complex
2:31
Use the Idea of a Middle
3:28
Find the Midpoint of Arbitrary Values a and b
4:17
How They're Equivalent
5:05
Official Midpoint Formula
5:46
Midpoint: Two Dimensions
6:19
The Midpoint Must Occur at the Horizontal Middle and the Vertical Middle
6:38
Arbitrary Pair of Points Example
7:25
Distance: One Dimension
9:26
Absolute Value
10:54
Idea of Forcing Positive
11:06
Distance: One Dimension, Formula
11:47
Distance Between Arbitrary a and b
11:48
Absolute Value Helps When the Distance is Negative
12:41
Distance Formula
12:58
The Pythagorean Theorem
13:24
a²+b²=c²
13:50
Distance: Two Dimensions
14:59
Break Into Horizontal and Vertical Parts and then Use the Pythagorean Theorem
15:16
Distance Between Arbitrary Points (x₁,y₁) and (x₂,y₂)
16:21
Slope
19:30
Slope is the Rate of Change
19:41
m = rise over run
21:27
Slope Between Arbitrary Points (x₁,y₁) and (x₂,y₂)
22:31
Interpreting Slope
24:12
Positive Slope and Negative Slope
25:40
m=1, m=0, m=-1
26:48
Example 1
28:25
Example 2
31:42
Example 3
36:40
Example 4
42:48
Word Problems

56m 31s

Intro
0:00
Introduction
0:05
What is a Word Problem?
0:45
Describes Any Problem That Primarily Gets Its Ideas Across With Words Instead of Math Symbols
0:48
Requires Us to Think
1:32
Why Are They So Hard?
2:11
Reason 1: No Simple Formula to Solve Them
2:16
Reason 2: Harder to Teach Word Problems
2:47
You Can Learn How to Do Them!
3:51
Grades
7:57
'But I'm Never Going to Use This In Real Life'
9:46
Solving Word Problems
12:58
First: Understand the Problem
13:37
Second: What Are You Looking For?
14:33
Third: Set Up Relationships
16:21
Fourth: Solve It!
17:48
Summary of Method
19:04
Examples on Things Other Than Math
20:21
Math-Specific Method: What You Need Now
25:30
Understand What the Problem is Talking About
25:37
Set Up and Name Any Variables You Need to Know
25:56
Set Up Equations Connecting Those Variables to the Information in the Problem Statement
26:02
Use the Equations to Solve for an Answer
26:14
Tip
26:58
Draw Pictures
27:22
Breaking Into Pieces
28:28
Try Out Hypothetical Numbers
29:52
Student Logic
31:27
Jump In!
32:40
Example 1
34:03
Example 2
39:15
Example 3
44:22
Example 4
50:24
II. Functions
Idea of a Function

39m 54s

Intro
0:00
Introduction
0:04
What is a Function?
1:06
A Visual Example and Non-Example
1:30
Function Notation
3:47
f(x)
4:05
Express What Sets the Function Acts On
5:45
Metaphors for a Function
6:17
Transformation
6:28
Map
7:17
Machine
8:56
Same Input Always Gives Same Output
10:01
If We Put the Same Input Into a Function, It Will Always Produce the Same Output
10:11
Example of Something That is Not a Function
11:10
A Non-Numerical Example
12:10
The Functions We Will Use
15:05
Unless Told Otherwise, We Will Assume Every Function Takes in Real Numbers and Outputs Real Numbers
15:11
Usually Told the Rule of a Given Function
15:27
How To Use a Function
16:18
Apply the Rule to Whatever Our Input Value Is
16:28
Make Sure to Wrap Your Substitutions in Parentheses
17:09
Functions and Tables
17:36
Table of Values, Sometimes Called a T-Table
17:46
Example
17:56
Domain: What Goes In
18:55
The Domain is the Set of all Inputs That the Function Can Accept
18:56
Example
19:40
Range: What Comes Out
21:27
The Range is the Set of All Possible Outputs a Function Can Assign
21:34
Example
21:49
Another Example Would Be Our Initial Function From Earlier in This Lesson
22:29
Example 1
23:45
Example 2
25:22
Example 3
27:27
Example 4
29:23
Example 5
33:33
Graphs

58m 26s

Intro
0:00
Introduction
0:04
How to Interpret Graphs
1:17
Input / Independent Variable
1:47
Output / Dependent Variable
2:00
Graph as Input ⇒ Output
2:23
One Way to Think of a Graph: See What Happened to Various Inputs
2:25
Example
2:47
Graph as Location of Solution
4:20
A Way to See Solutions
4:36
Example
5:20
Which Way Should We Interpret?
7:13
Easiest to Think In Terms of How Inputs Are Mapped to Outputs
7:20
Sometimes It's Easier to Think In Terms of Solutions
8:39
Pay Attention to Axes
9:50
Axes Tell Where the Graph Is and What Scale It Has
10:09
Often, The Axes Will Be Square
10:14
Example
12:06
Arrows or No Arrows?
16:07
Will Not Use Arrows at the End of Our Graphs
17:13
Graph Stops Because It Hits the Edge of the Graphing Axes, Not Because the Function Stops
17:18
How to Graph
19:47
Plot Points
20:07
Connect with Curves
21:09
If You Connect with Straight Lines
21:44
Graphs of Functions are Smooth
22:21
More Points ⇒ More Accurate
23:38
Vertical Line Test
27:44
If a Vertical Line Could Intersect More Than One Point On a Graph, It Can Not Be the Graph of a Function
28:41
Every Point on a Graph Tells Us Where the x-Value Below is Mapped
30:07
Domain in Graphs
31:37
The Domain is the Set of All Inputs That a Function Can Accept
31:44
Be Aware That Our Function Probably Continues Past the Edge of Our 'Viewing Window'
33:19
Range in Graphs
33:53
Graphing Calculators: Check the Appendix!
36:55
Example 1
38:37
Example 2
45:19
Example 3
50:41
Example 4
53:28
Example 5
55:50
Properties of Functions

48m 49s

Intro
0:00
Introduction
0:05
Increasing Decreasing Constant
0:43
Looking at a Specific Graph
1:15
Increasing Interval
2:39
Constant Function
4:15
Decreasing Interval
5:10
Find Intervals by Looking at the Graph
5:32
Intervals Show x-values; Write in Parentheses
6:39
Maximum and Minimums
8:48
Relative (Local) Max/Min
10:20
Formal Definition of Relative Maximum
12:44
Formal Definition of Relative Minimum
13:05
Max/Min, More Terms
14:18
Definition of Extrema
15:01
Average Rate of Change
16:11
Drawing a Line for the Average Rate
16:48
Using the Slope of the Secant Line
17:36
Slope in Function Notation
18:45
Zeros/Roots/x-intercepts
19:45
What Zeros in a Function Mean
20:25
Even Functions
22:30
Odd Functions
24:36
Even/Odd Functions and Graphs
26:28
Example of an Even Function
27:12
Example of an Odd Function
28:03
Example 1
29:35
Example 2
33:07
Example 3
40:32
Example 4
42:34
Function Petting Zoo

29m 20s

Intro
0:00
Introduction
0:04
Don't Forget that Axes Matter!
1:44
The Constant Function
2:40
The Identity Function
3:44
The Square Function
4:40
The Cube Function
5:44
The Square Root Function
6:51
The Reciprocal Function
8:11
The Absolute Value Function
10:19
The Trigonometric Functions
11:56
f(x)=sin(x)
12:12
f(x)=cos(x)
12:24
Alternate Axes
12:40
The Exponential and Logarithmic Functions
13:35
Exponential Functions
13:44
Logarithmic Functions
14:24
Alternating Axes
15:17
Transformations and Compositions
16:08
Example 1
17:52
Example 2
18:33
Example 3
20:24
Example 4
26:07
Transformation of Functions

48m 35s

Intro
0:00
Introduction
0:04
Vertical Shift
1:12
Graphical Example
1:21
A Further Explanation
2:16
Vertical Stretch/Shrink
3:34
Graph Shrinks
3:46
Graph Stretches
3:51
A Further Explanation
5:07
Horizontal Shift
6:49
Moving the Graph to the Right
7:28
Moving the Graph to the Left
8:12
A Further Explanation
8:19
Understanding Movement on the x-axis
8:38
Horizontal Stretch/Shrink
12:59
Shrinking the Graph
13:40
Stretching the Graph
13:48
A Further Explanation
13:55
Understanding Stretches from the x-axis
14:12
Vertical Flip (aka Mirror)
16:55
Example Graph
17:07
Multiplying the Vertical Component by -1
17:18
Horizontal Flip (aka Mirror)
18:43
Example Graph
19:01
Multiplying the Horizontal Component by -1
19:54
Summary of Transformations
22:11
Stacking Transformations
24:46
Order Matters
25:20
Transformation Example
25:52
Example 1
29:21
Example 2
34:44
Example 3
38:10
Example 4
43:46
Composite Functions

33m 24s

Intro
0:00
Introduction
0:04
Arithmetic Combinations
0:40
Basic Operations
1:20
Definition of the Four Arithmetic Combinations
1:40
Composite Functions
2:53
The Function as a Machine
3:32
Function Compositions as Multiple Machines
3:59
Notation for Composite Functions
4:46
Two Formats
6:02
Another Visual Interpretation
7:17
How to Use Composite Functions
8:21
Example of on Function acting on Another
9:17
Example 1
11:03
Example 2
15:27
Example 3
21:11
Example 4
27:06
Piecewise Functions

51m 42s

Intro
0:00
Introduction
0:04
Analogies to a Piecewise Function
1:16
Different Potatoes
1:41
Factory Production
2:27
Notations for Piecewise Functions
3:39
Notation Examples from Analogies
6:11
Example of a Piecewise (with Table)
7:24
Example of a Non-Numerical Piecewise
11:35
Graphing Piecewise Functions
14:15
Graphing Piecewise Functions, Example
16:26
Continuous Functions
16:57
Statements of Continuity
19:30
Example of Continuous and Non-Continuous Graphs
20:05
Interesting Functions: the Step Function
22:00
Notation for the Step Function
22:40
How the Step Function Works
22:56
Graph of the Step Function
25:30
Example 1
26:22
Example 2
28:49
Example 3
36:50
Example 4
46:11
Inverse Functions

49m 37s

Intro
0:00
Introduction
0:04
Analogy by picture
1:10
How to Denote the inverse
1:40
What Comes out of the Inverse
1:52
Requirement for Reversing
2:02
The Basketball Factory
2:12
The Importance of Information
2:45
One-to-One
4:04
Requirement for Reversibility
4:21
When a Function has an Inverse
4:43
One-to-One
5:13
Not One-to-One
5:50
Not a Function
6:19
Horizontal Line Test
7:01
How to the test Works
7:12
One-to-One
8:12
Not One-to-One
8:45
Definition: Inverse Function
9:12
Formal Definition
9:21
Caution to Students
10:02
Domain and Range
11:12
Finding the Range of the Function Inverse
11:56
Finding the Domain of the Function Inverse
12:11
Inverse of an Inverse
13:09
Its just x!
13:26
Proof
14:03
Graphical Interpretation
17:07
Horizontal Line Test
17:20
Graph of the Inverse
18:04
Swapping Inputs and Outputs to Draw Inverses
19:02
How to Find the Inverse
21:03
What We Are Looking For
21:21
Reversing the Function
21:38
A Method to Find Inverses
22:33
Check Function is One-to-One
23:04
Swap f(x) for y
23:25
Interchange x and y
23:41
Solve for y
24:12
Replace y with the inverse
24:40
Some Comments
25:01
Keeping Step 2 and 3 Straight
25:44
Switching to Inverse
26:12
Checking Inverses
28:52
How to Check an Inverse
29:06
Quick Example of How to Check
29:56
Example 1
31:48
Example 2
34:56
Example 3
39:29
Example 4
46:19
Variation Direct and Inverse

28m 49s

Intro
0:00
Introduction
0:06
Direct Variation
1:14
Same Direction
1:21
Common Example: Groceries
1:56
Different Ways to Say that Two Things Vary Directly
2:28
Basic Equation for Direct Variation
2:55
Inverse Variation
3:40
Opposite Direction
3:50
Common Example: Gravity
4:53
Different Ways to Say that Two Things Vary Indirectly
5:48
Basic Equation for Indirect Variation
6:33
Joint Variation
7:27
Equation for Joint Variation
7:53
Explanation of the Constant
8:48
Combined Variation
9:35
Gas Law as a Combination
9:44
Single Constant
10:33
Example 1
10:49
Example 2
13:34
Example 3
15:39
Example 4
19:48
III. Polynomials
Intro to Polynomials

38m 41s

Intro
0:00
Introduction
0:04
Definition of a Polynomial
1:04
Starting Integer
2:06
Structure of a Polynomial
2:49
The a Constants
3:34
Polynomial Function
5:13
Polynomial Equation
5:23
Polynomials with Different Variables
5:36
Degree
6:23
Informal Definition
6:31
Find the Largest Exponent Variable
6:44
Quick Examples
7:36
Special Names for Polynomials
8:59
Based on the Degree
9:23
Based on the Number of Terms
10:12
Distributive Property (aka 'FOIL')
11:37
Basic Distributive Property
12:21
Distributing Two Binomials
12:55
Longer Parentheses
15:12
Reverse: Factoring
17:26
Long-Term Behavior of Polynomials
17:48
Examples
18:13
Controlling Term--Term with the Largest Exponent
19:33
Positive and Negative Coefficients on the Controlling Term
20:21
Leading Coefficient Test
22:07
Even Degree, Positive Coefficient
22:13
Even Degree, Negative Coefficient
22:39
Odd Degree, Positive Coefficient
23:09
Odd Degree, Negative Coefficient
23:27
Example 1
25:11
Example 2
27:16
Example 3
31:16
Example 4
34:41
Roots (Zeros) of Polynomials

41m 7s

Intro
0:00
Introduction
0:05
Roots in Graphs
1:17
The x-intercepts
1:33
How to Remember What 'Roots' Are
1:50
Naïve Attempts
2:31
Isolating Variables
2:45
Failures of Isolating Variables
3:30
Missing Solutions
4:59
Factoring: How to Find Roots
6:28
How Factoring Works
6:36
Why Factoring Works
7:20
Steps to Finding Polynomial Roots
9:21
Factoring: How to Find Roots CAUTION
10:08
Factoring is Not Easy
11:32
Factoring Quadratics
13:08
Quadratic Trinomials
13:21
Form of Factored Binomials
13:38
Factoring Examples
14:40
Factoring Quadratics, Check Your Work
16:58
Factoring Higher Degree Polynomials
18:19
Factoring a Cubic
18:32
Factoring a Quadratic
19:04
Factoring: Roots Imply Factors
19:54
Where a Root is, A Factor Is
20:01
How to Use Known Roots to Make Factoring Easier
20:35
Not all Polynomials Can be Factored
22:30
Irreducible Polynomials
23:27
Complex Numbers Help
23:55
Max Number of Roots/Factors
24:57
Limit to Number of Roots Equal to the Degree
25:18
Why there is a Limit
25:25
Max Number of Peaks/Valleys
26:39
Shape Information from Degree
26:46
Example Graph
26:54
Max, But Not Required
28:00
Example 1
28:37
Example 2
31:21
Example 3
36:12
Example 4
38:40
Completing the Square and the Quadratic Formula

39m 43s

Intro
0:00
Introduction
0:05
Square Roots and Equations
0:51
Taking the Square Root to Find the Value of x
0:55
Getting the Positive and Negative Answers
1:05
Completing the Square: Motivation
2:04
Polynomials that are Easy to Solve
2:20
Making Complex Polynomials Easy to Solve
3:03
Steps to Completing the Square
4:30
Completing the Square: Method
7:22
Move C over
7:35
Divide by A
7:44
Find r
7:59
Add to Both Sides to Complete the Square
8:49
Solving Quadratics with Ease
9:56
The Quadratic Formula
11:38
Derivation
11:43
Final Form
12:23
Follow Format to Use Formula
13:38
How Many Roots?
14:53
The Discriminant
15:47
What the Discriminant Tells Us: How Many Roots
15:58
How the Discriminant Works
16:30
Example 1: Complete the Square
18:24
Example 2: Solve the Quadratic
22:00
Example 3: Solve for Zeroes
25:28
Example 4: Using the Quadratic Formula
30:52
Properties of Quadratic Functions

45m 34s

Intro
0:00
Introduction
0:05
Parabolas
0:35
Examples of Different Parabolas
1:06
Axis of Symmetry and Vertex
1:28
Drawing an Axis of Symmetry
1:51
Placing the Vertex
2:28
Looking at the Axis of Symmetry and Vertex for other Parabolas
3:09
Transformations
4:18
Reviewing Transformation Rules
6:28
Note the Different Horizontal Shift Form
7:45
An Alternate Form to Quadratics
8:54
The Constants: k, h, a
9:05
Transformations Formed
10:01
Analyzing Different Parabolas
10:10
Switching Forms by Completing the Square
11:43
Vertex of a Parabola
16:30
Vertex at (h, k)
16:47
Vertex in Terms of a, b, and c Coefficients
17:28
Minimum/Maximum at Vertex
18:19
When a is Positive
18:25
When a is Negative
18:52
Axis of Symmetry
19:54
Incredibly Minor Note on Grammar
20:52
Example 1
21:48
Example 2
26:35
Example 3
28:55
Example 4
31:40
Intermediate Value Theorem and Polynomial Division

46m 8s

Intro
0:00
Introduction
0:05
Reminder: Roots Imply Factors
1:32
The Intermediate Value Theorem
3:41
The Basis: U between a and b
4:11
U is on the Function
4:52
Intermediate Value Theorem, Proof Sketch
5:51
If Not True, the Graph Would Have to Jump
5:58
But Graph is Defined as Continuous
6:43
Finding Roots with the Intermediate Value Theorem
7:01
Picking a and b to be of Different Signs
7:10
Must Be at Least One Root
7:46
Dividing a Polynomial
8:16
Using Roots and Division to Factor
8:38
Long Division Refresher
9:08
The Division Algorithm
12:18
How It Works to Divide Polynomials
12:37
The Parts of the Equation
13:24
Rewriting the Equation
14:47
Polynomial Long Division
16:20
Polynomial Long Division In Action
16:29
One Step at a Time
20:51
Synthetic Division
22:46
Setup
23:11
Synthetic Division, Example
24:44
Which Method Should We Use
26:39
Advantages of Synthetic Method
26:49
Advantages of Long Division
27:13
Example 1
29:24
Example 2
31:27
Example 3
36:22
Example 4
40:55
Complex Numbers

45m 36s

Intro
0:00
Introduction
0:04
A Wacky Idea
1:02
The Definition of the Imaginary Number
1:22
How it Helps Solve Equations
2:20
Square Roots and Imaginary Numbers
3:15
Complex Numbers
5:00
Real Part and Imaginary Part
5:20
When Two Complex Numbers are Equal
6:10
Addition and Subtraction
6:40
Deal with Real and Imaginary Parts Separately
7:36
Two Quick Examples
7:54
Multiplication
9:07
FOIL Expansion
9:14
Note What Happens to the Square of the Imaginary Number
9:41
Two Quick Examples
10:22
Division
11:27
Complex Conjugates
13:37
Getting Rid of i
14:08
How to Denote the Conjugate
14:48
Division through Complex Conjugates
16:11
Multiply by the Conjugate of the Denominator
16:28
Example
17:46
Factoring So-Called 'Irreducible' Quadratics
19:24
Revisiting the Quadratic Formula
20:12
Conjugate Pairs
20:37
But Are the Complex Numbers 'Real'?
21:27
What Makes a Number Legitimate
25:38
Where Complex Numbers are Used
27:20
Still, We Won't See Much of C
29:05
Example 1
30:30
Example 2
33:15
Example 3
38:12
Example 4
42:07
Fundamental Theorem of Algebra

19m 9s

Intro
0:00
Introduction
0:05
Idea: Hidden Roots
1:16
Roots in Complex Form
1:42
All Polynomials Have Roots
2:08
Fundamental Theorem of Algebra
2:21
Where Are All the Imaginary Roots, Then?
3:17
All Roots are Complex
3:45
Real Numbers are a Subset of Complex Numbers
3:59
The n Roots Theorem
5:01
For Any Polynomial, Its Degree is Equal to the Number of Roots
5:11
Equivalent Statement
5:24
Comments: Multiplicity
6:29
Non-Distinct Roots
6:59
Denoting Multiplicity
7:20
Comments: Complex Numbers Necessary
7:41
Comments: Complex Coefficients Allowed
8:55
Comments: Existence Theorem
9:59
Proof Sketch of n Roots Theorem
10:45
First Root
11:36
Second Root
13:23
Continuation to Find all Roots
16:00
IV. Rational Functions
Rational Functions and Vertical Asymptotes

33m 22s

Intro
0:00
Introduction
0:05
Definition of a Rational Function
1:20
Examples of Rational Functions
2:30
Why They are Called 'Rational'
2:47
Domain of a Rational Function
3:15
Undefined at Denominator Zeros
3:25
Otherwise all Reals
4:16
Investigating a Fundamental Function
4:50
The Domain of the Function
5:04
What Occurs at the Zeroes of the Denominator
5:20
Idea of a Vertical Asymptote
6:23
What's Going On?
6:58
Approaching x=0 from the left
7:32
Approaching x=0 from the right
8:34
Dividing by Very Small Numbers Results in Very Large Numbers
9:31
Definition of a Vertical Asymptote
10:05
Vertical Asymptotes and Graphs
11:15
Drawing Asymptotes by Using a Dashed Line
11:27
The Graph Can Never Touch Its Undefined Point
12:00
Not All Zeros Give Asymptotes
13:02
Special Cases: When Numerator and Denominator Go to Zero at the Same Time
14:58
Cancel out Common Factors
15:49
How to Find Vertical Asymptotes
16:10
Figure out What Values Are Not in the Domain of x
16:24
Determine if the Numerator and Denominator Share Common Factors and Cancel
16:45
Find Denominator Roots
17:33
Note if Asymptote Approaches Negative or Positive Infinity
18:06
Example 1
18:57
Example 2
21:26
Example 3
23:04
Example 4
30:01
Horizontal Asymptotes

34m 16s

Intro
0:00
Introduction
0:05
Investigating a Fundamental Function
0:53
What Happens as x Grows Large
1:00
Different View
1:12
Idea of a Horizontal Asymptote
1:36
What's Going On?
2:24
What Happens as x Grows to a Large Negative Number
2:49
What Happens as x Grows to a Large Number
3:30
Dividing by Very Large Numbers Results in Very Small Numbers
3:52
Example Function
4:41
Definition of a Vertical Asymptote
8:09
Expanding the Idea
9:03
What's Going On?
9:48
What Happens to the Function in the Long Run?
9:51
Rewriting the Function
10:13
Definition of a Slant Asymptote
12:09
Symbolical Definition
12:30
Informal Definition
12:45
Beyond Slant Asymptotes
13:03
Not Going Beyond Slant Asymptotes
14:39
Horizontal/Slant Asymptotes and Graphs
15:43
How to Find Horizontal and Slant Asymptotes
16:52
How to Find Horizontal Asymptotes
17:12
Expand the Given Polynomials
17:18
Compare the Degrees of the Numerator and Denominator
17:40
How to Find Slant Asymptotes
20:05
Slant Asymptotes Exist When n+m=1
20:08
Use Polynomial Division
20:24
Example 1
24:32
Example 2
25:53
Example 3
26:55
Example 4
29:22
Graphing Asymptotes in a Nutshell

49m 7s

Intro
0:00
Introduction
0:05
A Process for Graphing
1:22
1. Factor Numerator and Denominator
1:50
2. Find Domain
2:53
3. Simplifying the Function
3:59
4. Find Vertical Asymptotes
4:59
5. Find Horizontal/Slant Asymptotes
5:24
6. Find Intercepts
7:35
7. Draw Graph (Find Points as Necessary)
9:21
Draw Graph Example
11:21
Vertical Asymptote
11:41
Horizontal Asymptote
11:50
Other Graphing
12:16
Test Intervals
15:08
Example 1
17:57
Example 2
23:01
Example 3
29:02
Example 4
33:37
Partial Fractions

44m 56s

Intro
0:00
Introduction: Idea
0:04
Introduction: Prerequisites and Uses
1:57
Proper vs. Improper Polynomial Fractions
3:11
Possible Things in the Denominator
4:38
Linear Factors
6:16
Example of Linear Factors
7:03
Multiple Linear Factors
7:48
Irreducible Quadratic Factors
8:25
Example of Quadratic Factors
9:26
Multiple Quadratic Factors
9:49
Mixing Factor Types
10:28
Figuring Out the Numerator
11:10
How to Solve for the Constants
11:30
Quick Example
11:40
Example 1
14:29
Example 2
18:35
Example 3
20:33
Example 4
28:51
V. Exponential & Logarithmic Functions
Understanding Exponents

35m 17s

Intro
0:00
Introduction
0:05
Fundamental Idea
1:46
Expanding the Idea
2:28
Multiplication of the Same Base
2:40
Exponents acting on Exponents
3:45
Different Bases with the Same Exponent
4:31
To the Zero
5:35
To the First
5:45
Fundamental Rule with the Zero Power
6:35
To the Negative
7:45
Any Number to a Negative Power
8:14
A Fraction to a Negative Power
9:58
Division with Exponential Terms
10:41
To the Fraction
11:33
Square Root
11:58
Any Root
12:59
Summary of Rules
14:38
To the Irrational
17:21
Example 1
20:34
Example 2
23:42
Example 3
27:44
Example 4
31:44
Example 5
33:15
Exponential Functions

47m 4s

Intro
0:00
Introduction
0:05
Definition of an Exponential Function
0:48
Definition of the Base
1:02
Restrictions on the Base
1:16
Computing Exponential Functions
2:29
Harder Computations
3:10
When to Use a Calculator
3:21
Graphing Exponential Functions: a>1
6:02
Three Examples
6:13
What to Notice on the Graph
7:44
A Story
8:27
Story Diagram
9:15
Increasing Exponentials
11:29
Story Morals
14:40
Application: Compound Interest
15:15
Compounding Year after Year
16:01
Function for Compounding Interest
16:51
A Special Number: e
20:55
Expression for e
21:28
Where e stabilizes
21:55
Application: Continuously Compounded Interest
24:07
Equation for Continuous Compounding
24:22
Exponential Decay 0<a<1
25:50
Three Examples
26:11
Why they 'lose' value
26:54
Example 1
27:47
Example 2
33:11
Example 3
36:34
Example 4
41:28
Introduction to Logarithms

40m 31s

Intro
0:00
Introduction
0:04
Definition of a Logarithm, Base 2
0:51
Log 2 Defined
0:55
Examples
2:28
Definition of a Logarithm, General
3:23
Examples of Logarithms
5:15
Problems with Unusual Bases
7:38
Shorthand Notation: ln and log
9:44
base e as ln
10:01
base 10 as log
10:34
Calculating Logarithms
11:01
using a calculator
11:34
issues with other bases
11:58
Graphs of Logarithms
13:21
Three Examples
13:29
Slow Growth
15:19
Logarithms as Inverse of Exponentiation
16:02
Using Base 2
16:05
General Case
17:10
Looking More Closely at Logarithm Graphs
19:16
The Domain of Logarithms
20:41
Thinking about Logs like Inverses
21:08
The Alternate
24:00
Example 1
25:59
Example 2
30:03
Example 3
32:49
Example 4
37:34
Properties of Logarithms

42m 33s

Intro
0:00
Introduction
0:04
Basic Properties
1:12
Inverse--log(exp)
1:43
A Key Idea
2:44
What We Get through Exponentiation
3:18
B Always Exists
4:50
Inverse--exp(log)
5:53
Logarithm of a Power
7:44
Logarithm of a Product
10:07
Logarithm of a Quotient
13:48
Caution! There Is No Rule for loga(M+N)
16:12
Summary of Properties
17:42
Change of Base--Motivation
20:17
No Calculator Button
20:59
A Specific Example
21:45
Simplifying
23:45
Change of Base--Formula
24:14
Example 1
25:47
Example 2
29:08
Example 3
31:14
Example 4
34:13
Solving Exponential and Logarithmic Equations

34m 10s

Intro
0:00
Introduction
0:05
One to One Property
1:09
Exponential
1:26
Logarithmic
1:44
Specific Considerations
2:02
One-to-One Property
3:30
Solving by One-to-One
4:11
Inverse Property
6:09
Solving by Inverses
7:25
Dealing with Equations
7:50
Example of Taking an Exponent or Logarithm of an Equation
9:07
A Useful Property
11:57
Bring Down Exponents
12:01
Try to Simplify
13:20
Extraneous Solutions
13:45
Example 1
16:37
Example 2
19:39
Example 3
21:37
Example 4
26:45
Example 5
29:37
Application of Exponential and Logarithmic Functions

48m 46s

Intro
0:00
Introduction
0:06
Applications of Exponential Functions
1:07
A Secret!
2:17
Natural Exponential Growth Model
3:07
Figure out r
3:34
A Secret!--Why Does It Work?
4:44
e to the r Morphs
4:57
Example
5:06
Applications of Logarithmic Functions
8:32
Examples
8:43
What Logarithms are Useful For
9:53
Example 1
11:29
Example 2
15:30
Example 3
26:22
Example 4
32:05
Example 5
39:19
VI. Trigonometric Functions
Angles

39m 5s

Intro
0:00
Degrees
0:22
Circle is 360 Degrees
0:48
Splitting a Circle
1:13
Radians
2:08
Circle is 2 Pi Radians
2:31
One Radian
2:52
Half-Circle and Right Angle
4:00
Converting Between Degrees and Radians
6:24
Formulas for Degrees and Radians
6:52
Coterminal, Complementary, Supplementary Angles
7:23
Coterminal Angles
7:30
Complementary Angles
9:40
Supplementary Angles
10:08
Example 1: Dividing a Circle
10:38
Example 2: Converting Between Degrees and Radians
11:56
Example 3: Quadrants and Coterminal Angles
14:18
Extra Example 1: Common Angle Conversions
-1
Extra Example 2: Quadrants and Coterminal Angles
-2
Sine and Cosine Functions

43m 16s

Intro
0:00
Sine and Cosine
0:15
Unit Circle
0:22
Coordinates on Unit Circle
1:03
Right Triangles
1:52
Adjacent, Opposite, Hypotenuse
2:25
Master Right Triangle Formula: SOHCAHTOA
2:48
Odd Functions, Even Functions
4:40
Example: Odd Function
4:56
Example: Even Function
7:30
Example 1: Sine and Cosine
10:27
Example 2: Graphing Sine and Cosine Functions
14:39
Example 3: Right Triangle
21:40
Example 4: Odd, Even, or Neither
26:01
Extra Example 1: Right Triangle
-1
Extra Example 2: Graphing Sine and Cosine Functions
-2
Sine and Cosine Values of Special Angles

33m 5s

Intro
0:00
45-45-90 Triangle and 30-60-90 Triangle
0:08
45-45-90 Triangle
0:21
30-60-90 Triangle
2:06
Mnemonic: All Students Take Calculus (ASTC)
5:21
Using the Unit Circle
5:59
New Angles
6:21
Other Quadrants
9:43
Mnemonic: All Students Take Calculus
10:13
Example 1: Convert, Quadrant, Sine/Cosine
13:11
Example 2: Convert, Quadrant, Sine/Cosine
16:48
Example 3: All Angles and Quadrants
20:21
Extra Example 1: Convert, Quadrant, Sine/Cosine
-1
Extra Example 2: All Angles and Quadrants
-2
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D

52m 3s

Intro
0:00
Amplitude and Period of a Sine Wave
0:38
Sine Wave Graph
0:58
Amplitude: Distance from Middle to Peak
1:18
Peak: Distance from Peak to Peak
2:41
Phase Shift and Vertical Shift
4:13
Phase Shift: Distance Shifted Horizontally
4:16
Vertical Shift: Distance Shifted Vertically
6:48
Example 1: Amplitude/Period/Phase and Vertical Shift
8:04
Example 2: Amplitude/Period/Phase and Vertical Shift
17:39
Example 3: Find Sine Wave Given Attributes
25:23
Extra Example 1: Amplitude/Period/Phase and Vertical Shift
-1
Extra Example 2: Find Cosine Wave Given Attributes
-2
Tangent and Cotangent Functions

36m 4s

Intro
0:00
Tangent and Cotangent Definitions
0:21
Tangent Definition
0:25
Cotangent Definition
0:47
Master Formula: SOHCAHTOA
1:01
Mnemonic
1:16
Tangent and Cotangent Values
2:29
Remember Common Values of Sine and Cosine
2:46
90 Degrees Undefined
4:36
Slope and Menmonic: ASTC
5:47
Uses of Tangent
5:54
Example: Tangent of Angle is Slope
6:09
Sign of Tangent in Quadrants
7:49
Example 1: Graph Tangent and Cotangent Functions
10:42
Example 2: Tangent and Cotangent of Angles
16:09
Example 3: Odd, Even, or Neither
18:56
Extra Example 1: Tangent and Cotangent of Angles
-1
Extra Example 2: Tangent and Cotangent of Angles
-2
Secant and Cosecant Functions

27m 18s

Intro
0:00
Secant and Cosecant Definitions
0:17
Secant Definition
0:18
Cosecant Definition
0:33
Example 1: Graph Secant Function
0:48
Example 2: Values of Secant and Cosecant
6:49
Example 3: Odd, Even, or Neither
12:49
Extra Example 1: Graph of Cosecant Function
-1
Extra Example 2: Values of Secant and Cosecant
-2
Inverse Trigonometric Functions

32m 58s

Intro
0:00
Arcsine Function
0:24
Restrictions between -1 and 1
0:43
Arcsine Notation
1:26
Arccosine Function
3:07
Restrictions between -1 and 1
3:36
Cosine Notation
3:53
Arctangent Function
4:30
Between -Pi/2 and Pi/2
4:44
Tangent Notation
5:02
Example 1: Domain/Range/Graph of Arcsine
5:45
Example 2: Arcsin/Arccos/Arctan Values
10:46
Example 3: Domain/Range/Graph of Arctangent
17:14
Extra Example 1: Domain/Range/Graph of Arccosine
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Computations of Inverse Trigonometric Functions

31m 8s

Intro
0:00
Inverse Trigonometric Function Domains and Ranges
0:31
Arcsine
0:41
Arccosine
1:14
Arctangent
1:41
Example 1: Arcsines of Common Values
2:44
Example 2: Odd, Even, or Neither
5:57
Example 3: Arccosines of Common Values
12:24
Extra Example 1: Arctangents of Common Values
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
VII. Trigonometric Identities
Pythagorean Identity

19m 11s

Intro
0:00
Pythagorean Identity
0:17
Pythagorean Triangle
0:27
Pythagorean Identity
0:45
Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity
1:14
Example 2: Find Angle Given Cosine and Quadrant
4:18
Example 3: Verify Trigonometric Identity
8:00
Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem
-1
Extra Example 2: Find Angle Given Cosine and Quadrant
-2
Identity Tan(squared)x+1=Sec(squared)x

23m 16s

Intro
0:00
Main Formulas
0:19
Companion to Pythagorean Identity
0:27
For Cotangents and Cosecants
0:52
How to Remember
0:58
Example 1: Prove the Identity
1:40
Example 2: Given Tan Find Sec
3:42
Example 3: Prove the Identity
7:45
Extra Example 1: Prove the Identity
-1
Extra Example 2: Given Sec Find Tan
-2
Addition and Subtraction Formulas

52m 52s

Intro
0:00
Addition and Subtraction Formulas
0:09
How to Remember
0:48
Cofunction Identities
1:31
How to Remember Graphically
1:44
Where to Use Cofunction Identities
2:52
Example 1: Derive the Formula for cos(A-B)
3:08
Example 2: Use Addition and Subtraction Formulas
16:03
Example 3: Use Addition and Subtraction Formulas to Prove Identity
25:11
Extra Example 1: Use cos(A-B) and Cofunction Identities
-1
Extra Example 2: Convert to Radians and use Formulas
-2
Double Angle Formulas

29m 5s

Intro
0:00
Main Formula
0:07
How to Remember from Addition Formula
0:18
Two Other Forms
1:35
Example 1: Find Sine and Cosine of Angle using Double Angle
3:16
Example 2: Prove Trigonometric Identity using Double Angle
9:37
Example 3: Use Addition and Subtraction Formulas
12:38
Extra Example 1: Find Sine and Cosine of Angle using Double Angle
-1
Extra Example 2: Prove Trigonometric Identity using Double Angle
-2
Half-Angle Formulas

43m 55s

Intro
0:00
Main Formulas
0:09
Confusing Part
0:34
Example 1: Find Sine and Cosine of Angle using Half-Angle
0:54
Example 2: Prove Trigonometric Identity using Half-Angle
11:51
Example 3: Prove the Half-Angle Formula for Tangents
18:39
Extra Example 1: Find Sine and Cosine of Angle using Half-Angle
-1
Extra Example 2: Prove Trigonometric Identity using Half-Angle
-2
VIII. Applications of Trigonometry
Trigonometry in Right Angles

25m 43s

Intro
0:00
Master Formula for Right Angles
0:11
SOHCAHTOA
0:15
Only for Right Triangles
1:26
Example 1: Find All Angles in a Triangle
2:19
Example 2: Find Lengths of All Sides of Triangle
7:39
Example 3: Find All Angles in a Triangle
11:00
Extra Example 1: Find All Angles in a Triangle
-1
Extra Example 2: Find Lengths of All Sides of Triangle
-2
Law of Sines

56m 40s

Intro
0:00
Law of Sines Formula
0:18
SOHCAHTOA
0:27
Any Triangle
0:59
Graphical Representation
1:25
Solving Triangle Completely
2:37
When to Use Law of Sines
2:55
ASA, SAA, SSA, AAA
2:59
SAS, SSS for Law of Cosines
7:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
8:44
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:30
Example 3: How Many Triangles Satisfy Conditions, Solve Completely
28:32
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely
-2
Law of Cosines

49m 5s

Intro
0:00
Law of Cosines Formula
0:23
Graphical Representation
0:34
Relates Sides to Angles
1:00
Any Triangle
1:20
Generalization of Pythagorean Theorem
1:32
When to Use Law of Cosines
2:26
SAS, SSS
2:30
Heron's Formula
4:49
Semiperimeter S
5:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
5:53
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:19
Example 3: Find Area of a Triangle Given All Side Lengths
26:33
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: Length of Third Side and Area of Triangle
-2
Finding the Area of a Triangle

27m 37s

Intro
0:00
Master Right Triangle Formula and Law of Cosines
0:19
SOHCAHTOA
0:27
Law of Cosines
1:23
Heron's Formula
2:22
Semiperimeter S
2:37
Example 1: Area of Triangle with Two Sides and One Angle
3:12
Example 2: Area of Triangle with Three Sides
6:11
Example 3: Area of Triangle with Three Sides, No Heron's Formula
8:50
Extra Example 1: Area of Triangle with Two Sides and One Angle
-1
Extra Example 2: Area of Triangle with Two Sides and One Angle
-2
Word Problems and Applications of Trigonometry

34m 25s

Intro
0:00
Formulas to Remember
0:11
SOHCAHTOA
0:15
Law of Sines
0:55
Law of Cosines
1:48
Heron's Formula
2:46
Example 1: Telephone Pole Height
4:01
Example 2: Bridge Length
7:48
Example 3: Area of Triangular Field
14:20
Extra Example 1: Kite Height
-1
Extra Example 2: Roads to a Town
-2
IX. Systems of Equations and Inequalities
Systems of Linear Equations

55m 40s

Intro
0:00
Introduction
0:04
Graphs as Location of 'True'
1:49
All Locations that Make the Function True
2:25
Understand the Relationship Between Solutions and the Graph
3:43
Systems as Graphs
4:07
Equations as Lines
4:20
Intersection Point
5:19
Three Possibilities for Solutions
6:17
Independent
6:24
Inconsistent
6:36
Dependent
7:06
Solving by Substitution
8:37
Solve for One Variable
9:07
Substitute into the Second Equation
9:34
Solve for Both Variables
10:12
What If a System is Inconsistent or Dependent?
11:08
No Solutions
11:25
Infinite Solutions
12:30
Solving by Elimination
13:56
Example
14:22
Determining the Number of Solutions
16:30
Why Elimination Makes Sense
17:25
Solving by Graphing Calculator
19:59
Systems with More than Two Variables
23:22
Example 1
25:49
Example 2
30:22
Example 3
34:11
Example 4
38:55
Example 5
46:01
(Non-) Example 6
53:37
Systems of Linear Inequalities

1h 13s

Intro
0:00
Introduction
0:04
Inequality Refresher-Solutions
0:46
Equation Solutions vs. Inequality Solutions
1:02
Essentially a Wide Variety of Answers
1:35
Refresher--Negative Multiplication Flips
1:43
Refresher--Negative Flips: Why?
3:19
Multiplication by a Negative
3:43
The Relationship Flips
3:55
Refresher--Stick to Basic Operations
4:34
Linear Equations in Two Variables
6:50
Graphing Linear Inequalities
8:28
Why It Includes a Whole Section
8:43
How to Show The Difference Between Strict and Not Strict Inequalities
10:08
Dashed Line--Not Solutions
11:10
Solid Line--Are Solutions
11:24
Test Points for Shading
11:42
Example of Using a Point
12:41
Drawing Shading from the Point
13:14
Graphing a System
14:53
Set of Solutions is the Overlap
15:17
Example
15:22
Solutions are Best Found Through Graphing
18:05
Linear Programming-Idea
19:52
Use a Linear Objective Function
20:15
Variables in Objective Function have Constraints
21:24
Linear Programming-Method
22:09
Rearrange Equations
22:21
Graph
22:49
Critical Solution is at the Vertex of the Overlap
23:40
Try Each Vertice
24:35
Example 1
24:58
Example 2
28:57
Example 3
33:48
Example 4
43:10
Nonlinear Systems

41m 1s

Intro
0:00
Introduction
0:06
Substitution
1:12
Example
1:22
Elimination
3:46
Example
3:56
Elimination is Less Useful for Nonlinear Systems
4:56
Graphing
5:56
Using a Graphing Calculator
6:44
Number of Solutions
8:44
Systems of Nonlinear Inequalities
10:02
Graph Each Inequality
10:06
Dashed and/or Solid
10:18
Shade Appropriately
11:14
Example 1
13:24
Example 2
15:50
Example 3
22:02
Example 4
29:06
Example 4, cont.
33:40
X. Vectors and Matrices
Vectors

1h 9m 31s

Intro
0:00
Introduction
0:10
Magnitude of the Force
0:22
Direction of the Force
0:48
Vector
0:52
Idea of a Vector
1:30
How Vectors are Denoted
2:00
Component Form
3:20
Angle Brackets and Parentheses
3:50
Magnitude/Length
4:26
Denoting the Magnitude of a Vector
5:16
Direction/Angle
7:52
Always Draw a Picture
8:50
Component Form from Magnitude & Angle
10:10
Scaling by Scalars
14:06
Unit Vectors
16:26
Combining Vectors - Algebraically
18:10
Combining Vectors - Geometrically
19:54
Resultant Vector
20:46
Alternate Component Form: i, j
21:16
The Zero Vector
23:18
Properties of Vectors
24:20
No Multiplication (Between Vectors)
28:30
Dot Product
29:40
Motion in a Medium
30:10
Fish in an Aquarium Example
31:38
More Than Two Dimensions
33:12
More Than Two Dimensions - Magnitude
34:18
Example 1
35:26
Example 2
38:10
Example 3
45:48
Example 4
50:40
Example 4, cont.
56:07
Example 5
1:01:32
Dot Product & Cross Product

35m 20s

Intro
0:00
Introduction
0:08
Dot Product - Definition
0:42
Dot Product Results in a Scalar, Not a Vector
2:10
Example in Two Dimensions
2:34
Angle and the Dot Product
2:58
The Dot Product of Two Vectors is Deeply Related to the Angle Between the Two Vectors
2:59
Proof of Dot Product Formula
4:14
Won't Directly Help Us Better Understand Vectors
4:18
Dot Product - Geometric Interpretation
4:58
We Can Interpret the Dot Product as a Measure of How Long and How Parallel Two Vectors Are
7:26
Dot Product - Perpendicular Vectors
8:24
If the Dot Product of Two Vectors is 0, We Know They are Perpendicular to Each Other
8:54
Cross Product - Definition
11:08
Cross Product Only Works in Three Dimensions
11:09
Cross Product - A Mnemonic
12:16
The Determinant of a 3 x 3 Matrix and Standard Unit Vectors
12:17
Cross Product - Geometric Interpretations
14:30
The Right-Hand Rule
15:17
Cross Product - Geometric Interpretations Cont.
17:00
Example 1
18:40
Example 2
22:50
Example 3
24:04
Example 4
26:20
Bonus Round
29:18
Proof: Dot Product Formula
29:24
Proof: Dot Product Formula, cont.
30:38
Matrices

54m 7s

Intro
0:00
Introduction
0:08
Definition of a Matrix
3:02
Size or Dimension
3:58
Square Matrix
4:42
Denoted by Capital Letters
4:56
When are Two Matrices Equal?
5:04
Examples of Matrices
6:44
Rows x Columns
6:46
Talking About Specific Entries
7:48
We Use Capitals to Denote a Matrix and Lower Case to Denotes Its Entries
8:32
Using Entries to Talk About Matrices
10:08
Scalar Multiplication
11:26
Scalar = Real Number
11:34
Example
12:36
Matrix Addition
13:08
Example
14:22
Matrix Multiplication
15:00
Example
18:52
Matrix Multiplication, cont.
19:58
Matrix Multiplication and Order (Size)
25:26
Make Sure Their Orders are Compatible
25:27
Matrix Multiplication is NOT Commutative
28:20
Example
30:08
Special Matrices - Zero Matrix (0)
32:48
Zero Matrix Has 0 for All of its Entries
32:49
Special Matrices - Identity Matrix (I)
34:14
Identity Matrix is a Square Matrix That Has 1 for All Its Entries on the Main Diagonal and 0 for All Other Entries
34:15
Example 1
36:16
Example 2
40:00
Example 3
44:54
Example 4
50:08
Determinants & Inverses of Matrices

47m 12s

Intro
0:00
Introduction
0:06
Not All Matrices Are Invertible
1:30
What Must a Matrix Have to Be Invertible?
2:08
Determinant
2:32
The Determinant is a Real Number Associated With a Square Matrix
2:38
If the Determinant of a Matrix is Nonzero, the Matrix is Invertible
3:40
Determinant of a 2 x 2 Matrix
4:34
Think in Terms of Diagonals
5:12
Minors and Cofactors - Minors
6:24
Example
6:46
Minors and Cofactors - Cofactors
8:00
Cofactor is Closely Based on the Minor
8:01
Alternating Sign Pattern
9:04
Determinant of Larger Matrices
10:56
Example
13:00
Alternative Method for 3x3 Matrices
16:46
Not Recommended
16:48
Inverse of a 2 x 2 Matrix
19:02
Inverse of Larger Matrices
20:00
Using Inverse Matrices
21:06
When Multiplied Together, They Create the Identity Matrix
21:24
Example 1
23:45
Example 2
27:21
Example 3
32:49
Example 4
36:27
Finding the Inverse of Larger Matrices
41:59
General Inverse Method - Step 1
43:25
General Inverse Method - Step 2
43:27
General Inverse Method - Step 2, cont.
43:27
General Inverse Method - Step 3
45:15
Using Matrices to Solve Systems of Linear Equations

58m 34s

Intro
0:00
Introduction
0:12
Augmented Matrix
1:44
We Can Represent the Entire Linear System With an Augmented Matrix
1:50
Row Operations
3:22
Interchange the Locations of Two Rows
3:50
Multiply (or Divide) a Row by a Nonzero Number
3:58
Add (or Subtract) a Multiple of One Row to Another
4:12
Row Operations - Keep Notes!
5:50
Suggested Symbols
7:08
Gauss-Jordan Elimination - Idea
8:04
Gauss-Jordan Elimination - Idea, cont.
9:16
Reduced Row-Echelon Form
9:18
Gauss-Jordan Elimination - Method
11:36
Begin by Writing the System As An Augmented Matrix
11:38
Gauss-Jordan Elimination - Method, cont.
13:48
Cramer's Rule - 2 x 2 Matrices
17:08
Cramer's Rule - n x n Matrices
19:24
Solving with Inverse Matrices
21:10
Solving Inverse Matrices, cont.
25:28
The Mighty (Graphing) Calculator
26:38
Example 1
29:56
Example 2
33:56
Example 3
37:00
Example 3, cont.
45:04
Example 4
51:28
XI. Alternate Ways to Graph
Parametric Equations

53m 33s

Intro
0:00
Introduction
0:06
Definition
1:10
Plane Curve
1:24
The Key Idea
2:00
Graphing with Parametric Equations
2:52
Same Graph, Different Equations
5:04
How Is That Possible?
5:36
Same Graph, Different Equations, cont.
5:42
Here's Another to Consider
7:56
Same Plane Curve, But Still Different
8:10
A Metaphor for Parametric Equations
9:36
Think of Parametric Equations As a Way to Describe the Motion of An Object
9:38
Graph Shows Where It Went, But Not Speed
10:32
Eliminating Parameters
12:14
Rectangular Equation
12:16
Caution
13:52
Creating Parametric Equations
14:30
Interesting Graphs
16:38
Graphing Calculators, Yay!
19:18
Example 1
22:36
Example 2
28:26
Example 3
37:36
Example 4
41:00
Projectile Motion
44:26
Example 5
47:00
Polar Coordinates

48m 7s

Intro
0:00
Introduction
0:04
Polar Coordinates Give Us a Way To Describe the Location of a Point
0:26
Polar Equations and Functions
0:50
Plotting Points with Polar Coordinates
1:06
The Distance of the Point from the Origin
1:09
The Angle of the Point
1:33
Give Points as the Ordered Pair (r,θ)
2:03
Visualizing Plotting in Polar Coordinates
2:32
First Way We Can Plot
2:39
Second Way We Can Plot
2:50
First, We'll Look at Visualizing r, Then θ
3:09
Rotate the Length Counter-Clockwise by θ
3:38
Alternatively, We Can Visualize θ, Then r
4:06
'Polar Graph Paper'
6:17
Horizontal and Vertical Tick Marks Are Not Useful for Polar
6:42
Use Concentric Circles to Helps Up See Distance From the Pole
7:08
Can Use Arc Sectors to See Angles
7:57
Multiple Ways to Name a Point
9:17
Examples
9:30
For Any Angle θ, We Can Make an Equivalent Angle
10:44
Negative Values for r
11:58
If r Is Negative, We Go In The Direction Opposite the One That The Angle θ Points Out
12:22
Another Way to Name the Same Point: Add π to θ and Make r Negative
13:44
Converting Between Rectangular and Polar
14:37
Rectangular Way to Name
14:43
Polar Way to Name
14:52
The Rectangular System Must Have a Right Angle Because It's Based on a Rectangle
15:08
Connect Both Systems Through Basic Trigonometry
15:38
Equation to Convert From Polar to Rectangular Coordinate Systems
16:55
Equation to Convert From Rectangular to Polar Coordinate Systems
17:13
Converting to Rectangular is Easy
17:20
Converting to Polar is a Bit Trickier
17:21
Draw Pictures
18:55
Example 1
19:50
Example 2
25:17
Example 3
31:05
Example 4
35:56
Example 5
41:49
Polar Equations & Functions

38m 16s

Intro
0:00
Introduction
0:04
Equations and Functions
1:16
Independent Variable
1:21
Dependent Variable
1:30
Examples
1:46
Always Assume That θ Is In Radians
2:44
Graphing in Polar Coordinates
3:29
Graph is the Same Way We Graph 'Normal' Stuff
3:32
Example
3:52
Graphing in Polar - Example, Cont.
6:45
Tips for Graphing
9:23
Notice Patterns
10:19
Repetition
13:39
Graphing Equations of One Variable
14:39
Converting Coordinate Types
16:16
Use the Same Conversion Formulas From the Previous Lesson
16:23
Interesting Graphs
17:48
Example 1
18:03
Example 2
18:34
Graphing Calculators, Yay!
19:07
Plot Random Things, Alter Equations You Understand, Get a Sense for How Polar Stuff Works
19:11
Check Out the Appendix
19:26
Example 1
21:36
Example 2
28:13
Example 3
34:24
Example 4
35:52
XII. Complex Numbers and Polar Coordinates
Polar Form of Complex Numbers

40m 43s

Intro
0:00
Polar Coordinates
0:49
Rectangular Form
0:52
Polar Form
1:25
R and Theta
1:51
Polar Form Conversion
2:27
R and Theta
2:35
Optimal Values
4:05
Euler's Formula
4:25
Multiplying Two Complex Numbers in Polar Form
6:10
Multiply r's Together and Add Exponents
6:32
Example 1: Convert Rectangular to Polar Form
7:17
Example 2: Convert Polar to Rectangular Form
13:49
Example 3: Multiply Two Complex Numbers
17:28
Extra Example 1: Convert Between Rectangular and Polar Forms
-1
Extra Example 2: Simplify Expression to Polar Form
-2
DeMoivre's Theorem

57m 37s

Intro
0:00
Introduction to DeMoivre's Theorem
0:10
n nth Roots
3:06
DeMoivre's Theorem: Finding nth Roots
3:52
Relation to Unit Circle
6:29
One nth Root for Each Value of k
7:11
Example 1: Convert to Polar Form and Use DeMoivre's Theorem
8:24
Example 2: Find Complex Eighth Roots
15:27
Example 3: Find Complex Roots
27:49
Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem
-1
Extra Example 2: Find Complex Fourth Roots
-2
XIII. Counting & Probability
Counting

31m 36s

Intro
0:00
Introduction
0:08
Combinatorics
0:56
Definition: Event
1:24
Example
1:50
Visualizing an Event
3:02
Branching line diagram
3:06
Addition Principle
3:40
Example
4:18
Multiplication Principle
5:42
Example
6:24
Pigeonhole Principle
8:06
Example
10:26
Draw Pictures
11:06
Example 1
12:02
Example 2
14:16
Example 3
17:34
Example 4
21:26
Example 5
25:14
Permutations & Combinations

44m 3s

Intro
0:00
Introduction
0:08
Permutation
0:42
Combination
1:10
Towards a Permutation Formula
2:38
How Many Ways Can We Arrange the Letters A, B, C, D, and E?
3:02
Towards a Permutation Formula, cont.
3:34
Factorial Notation
6:56
Symbol Is '!'
6:58
Examples
7:32
Permutation of n Objects
8:44
Permutation of r Objects out of n
9:04
What If We Have More Objects Than We Have Slots to Fit Them Into?
9:46
Permutation of r Objects Out of n, cont.
10:28
Distinguishable Permutations
14:46
What If Not All Of the Objects We're Permuting Are Distinguishable From Each Other?
14:48
Distinguishable Permutations, cont.
17:04
Combinations
19:04
Combinations, cont.
20:56
Example 1
23:10
Example 2
26:16
Example 3
28:28
Example 4
31:52
Example 5
33:58
Example 6
36:34
Probability

36m 58s

Intro
0:00
Introduction
0:06
Definition: Sample Space
1:18
Event = Something Happening
1:20
Sample Space
1:36
Probability of an Event
2:12
Let E Be An Event and S Be The Corresponding Sample Space
2:14
'Equally Likely' Is Important
3:52
Fair and Random
5:26
Interpreting Probability
6:34
How Can We Interpret This Value?
7:24
We Can Represent Probability As a Fraction, a Decimal, Or a Percentage
8:04
One of Multiple Events Occurring
9:52
Mutually Exclusive Events
10:38
What If The Events Are Not Mutually Exclusive?
12:20
Taking the Possibility of Overlap Into Account
13:24
An Event Not Occurring
17:14
Complement of E
17:22
Independent Events
19:36
Independent
19:48
Conditional Events
21:28
What Is The Events Are Not Independent Though?
21:30
Conditional Probability
22:16
Conditional Events, cont.
23:51
Example 1
25:27
Example 2
27:09
Example 3
28:57
Example 4
30:51
Example 5
34:15
XIV. Conic Sections
Parabolas

41m 27s

Intro
0:00
What is a Parabola?
0:20
Definition of a Parabola
0:29
Focus
0:59
Directrix
1:15
Axis of Symmetry
3:08
Vertex
3:33
Minimum or Maximum
3:44
Standard Form
4:59
Horizontal Parabolas
5:08
Vertex Form
5:19
Upward or Downward
5:41
Example: Standard Form
6:06
Graphing Parabolas
8:31
Shifting
8:51
Example: Completing the Square
9:22
Symmetry and Translation
12:18
Example: Graph Parabola
12:40
Latus Rectum
17:13
Length
18:15
Example: Latus Rectum
18:35
Horizontal Parabolas
18:57
Not Functions
20:08
Example: Horizontal Parabola
21:21
Focus and Directrix
24:11
Horizontal
24:48
Example 1: Parabola Standard Form
25:12
Example 2: Graph Parabola
30:00
Example 3: Graph Parabola
33:13
Example 4: Parabola Equation
37:28
Circles

21m 3s

Intro
0:00
What are Circles?
0:08
Example: Equidistant
0:17
Radius
0:32
Equation of a Circle
0:44
Example: Standard Form
1:11
Graphing Circles
1:47
Example: Circle
1:56
Center Not at Origin
3:07
Example: Completing the Square
3:51
Example 1: Equation of Circle
6:44
Example 2: Center and Radius
11:51
Example 3: Radius
15:08
Example 4: Equation of Circle
16:57
Ellipses

46m 51s

Intro
0:00
What Are Ellipses?
0:11
Foci
0:23
Properties of Ellipses
1:43
Major Axis, Minor Axis
1:47
Center
1:54
Length of Major Axis and Minor Axis
3:21
Standard Form
5:33
Example: Standard Form of Ellipse
6:09
Vertical Major Axis
9:14
Example: Vertical Major Axis
9:46
Graphing Ellipses
12:51
Complete the Square and Symmetry
13:00
Example: Graphing Ellipse
13:16
Equation with Center at (h, k)
19:57
Horizontal and Vertical
20:14
Difference
20:27
Example: Center at (h, k)
20:55
Example 1: Equation of Ellipse
24:05
Example 2: Equation of Ellipse
27:57
Example 3: Equation of Ellipse
32:32
Example 4: Graph Ellipse
38:27
Hyperbolas

38m 15s

Intro
0:00
What are Hyperbolas?
0:12
Two Branches
0:18
Foci
0:38
Properties
2:00
Transverse Axis and Conjugate Axis
2:06
Vertices
2:46
Length of Transverse Axis
3:14
Distance Between Foci
3:31
Length of Conjugate Axis
3:38
Standard Form
5:45
Vertex Location
6:36
Known Points
6:52
Vertical Transverse Axis
7:26
Vertex Location
7:50
Asymptotes
8:36
Vertex Location
8:56
Rectangle
9:28
Diagonals
10:29
Graphing Hyperbolas
12:58
Example: Hyperbola
13:16
Equation with Center at (h, k)
16:32
Example: Center at (h, k)
17:21
Example 1: Equation of Hyperbola
19:20
Example 2: Equation of Hyperbola
22:48
Example 3: Graph Hyperbola
26:05
Example 4: Equation of Hyperbola
36:29
Conic Sections

18m 43s

Intro
0:00
Conic Sections
0:16
Double Cone Sections
0:24
Standard Form
1:27
General Form
1:37
Identify Conic Sections
2:16
B = 0
2:50
X and Y
3:22
Identify Conic Sections, Cont.
4:46
Parabola
5:17
Circle
5:51
Ellipse
6:31
Hyperbola
7:10
Example 1: Identify Conic Section
8:01
Example 2: Identify Conic Section
11:03
Example 3: Identify Conic Section
11:38
Example 4: Identify Conic Section
14:50
XV. Sequences, Series, & Induction
Introduction to Sequences

57m 45s

Intro
0:00
Introduction
0:06
Definition: Sequence
0:28
Infinite Sequence
2:08
Finite Sequence
2:22
Length
2:58
Formula for the nth Term
3:22
Defining a Sequence Recursively
5:54
Initial Term
7:58
Sequences and Patterns
10:40
First, Identify a Pattern
12:52
How to Get From One Term to the Next
17:38
Tips for Finding Patterns
19:52
More Tips for Finding Patterns
24:14
Even More Tips
26:50
Example 1
30:32
Example 2
34:54
Fibonacci Sequence
34:55
Example 3
38:40
Example 4
45:02
Example 5
49:26
Example 6
51:54
Introduction to Series

40m 27s

Intro
0:00
Introduction
0:06
Definition: Series
1:20
Why We Need Notation
2:48
Simga Notation (AKA Summation Notation)
4:44
Thing Being Summed
5:42
Index of Summation
6:21
Lower Limit of Summation
7:09
Upper Limit of Summation
7:23
Sigma Notation, Example
7:36
Sigma Notation for Infinite Series
9:08
How to Reindex
10:58
How to Reindex, Expanding
12:56
How to Reindex, Substitution
16:46
Properties of Sums
19:42
Example 1
23:46
Example 2
25:34
Example 3
27:12
Example 4
29:54
Example 5
32:06
Example 6
37:16
Arithmetic Sequences & Series

31m 36s

Intro
0:00
Introduction
0:05
Definition: Arithmetic Sequence
0:47
Common Difference
1:13
Two Examples
1:19
Form for the nth Term
2:14
Recursive Relation
2:33
Towards an Arithmetic Series Formula
5:12
Creating a General Formula
10:09
General Formula for Arithmetic Series
14:23
Example 1
15:46
Example 2
17:37
Example 3
22:21
Example 4
24:09
Example 5
27:14
Geometric Sequences & Series

39m 27s

Intro
0:00
Introduction
0:06
Definition
0:48
Form for the nth Term
2:42
Formula for Geometric Series
5:16
Infinite Geometric Series
11:48
Diverges
13:04
Converges
14:48
Formula for Infinite Geometric Series
16:32
Example 1
20:32
Example 2
22:02
Example 3
26:00
Example 4
30:48
Example 5
34:28
Mathematical Induction

49m 53s

Intro
0:00
Introduction
0:06
Belief Vs. Proof
1:22
A Metaphor for Induction
6:14
The Principle of Mathematical Induction
11:38
Base Case
13:24
Inductive Step
13:30
Inductive Hypothesis
13:52
A Remark on Statements
14:18
Using Mathematical Induction
16:58
Working Example
19:58
Finding Patterns
28:46
Example 1
30:17
Example 2
37:50
Example 3
42:38
The Binomial Theorem

1h 13m 13s

Intro
0:00
Introduction
0:06
We've Learned That a Binomial Is An Expression That Has Two Terms
0:07
Understanding Binomial Coefficients
1:20
Things We Notice
2:24
What Goes In the Blanks?
5:52
Each Blank is Called a Binomial Coefficient
6:18
The Binomial Theorem
6:38
Example
8:10
The Binomial Theorem, cont.
10:46
We Can Also Write This Expression Compactly Using Sigma Notation
12:06
Proof of the Binomial Theorem
13:22
Proving the Binomial Theorem Is Within Our Reach
13:24
Pascal's Triangle
15:12
Pascal's Triangle, cont.
16:12
Diagonal Addition of Terms
16:24
Zeroth Row
18:04
First Row
18:12
Why Do We Care About Pascal's Triangle?
18:50
Pascal's Triangle, Example
19:26
Example 1
21:26
Example 2
24:34
Example 3
28:34
Example 4
32:28
Example 5
37:12
Time for the Fireworks!
43:38
Proof of the Binomial Theorem
43:44
We'll Prove This By Induction
44:04
Proof (By Induction)
46:36
Proof, Base Case
47:00
Proof, Inductive Step - Notation Discussion
49:22
Induction Step
49:24
Proof, Inductive Step - Setting Up
52:26
Induction Hypothesis
52:34
What We What To Show
52:44
Proof, Inductive Step - Start
54:18
Proof, Inductive Step - Middle
55:38
Expand Sigma Notations
55:48
Proof, Inductive Step - Middle, cont.
58:40
Proof, Inductive Step - Checking In
1:01:08
Let's Check In With Our Original Goal
1:01:12
Want to Show
1:01:18
Lemma - A Mini Theorem
1:02:18
Proof, Inductive Step - Lemma
1:02:52
Proof of Lemma: Let's Investigate the Left Side
1:03:08
Proof, Inductive Step - Nearly There
1:07:54
Proof, Inductive Step - End!
1:09:18
Proof, Inductive Step - End!, cont.
1:11:01
XVI. Preview of Calculus
Idea of a Limit

40m 22s

Intro
0:00
Introduction
0:05
Motivating Example
1:26
Fuzzy Notion of a Limit
3:38
Limit is the Vertical Location a Function is Headed Towards
3:44
Limit is What the Function Output is Going to Be
4:15
Limit Notation
4:33
Exploring Limits - 'Ordinary' Function
5:26
Test Out
5:27
Graphing, We See The Answer Is What We Would Expect
5:44
Exploring Limits - Piecewise Function
6:45
If We Modify the Function a Bit
6:49
Exploring Limits - A Visual Conception
10:08
Definition of a Limit
12:07
If f(x) Becomes Arbitrarily Close to Some Number L as x Approaches Some Number c, Then the Limit of f(x) As a Approaches c is L.
12:09
We Are Not Concerned with f(x) at x=c
12:49
We Are Considering x Approaching From All Directions, Not Just One Side
13:10
Limits Do Not Always Exist
15:47
Finding Limits
19:49
Graphs
19:52
Tables
21:48
Precise Methods
24:53
Example 1
26:06
Example 2
27:39
Example 3
30:51
Example 4
33:11
Example 5
37:07
Formal Definition of a Limit

57m 11s

Intro
0:00
Introduction
0:06
New Greek Letters
2:42
Delta
3:14
Epsilon
3:46
Sometimes Called the Epsilon-Delta Definition of a Limit
3:56
Formal Definition of a Limit
4:22
What does it MEAN!?!?
5:00
The Groundwork
5:38
Set Up the Limit
5:39
The Function is Defined Over Some Portion of the Reals
5:58
The Horizontal Location is the Value the Limit Will Approach
6:28
The Vertical Location L is Where the Limit Goes To
7:00
The Epsilon-Delta Part
7:26
The Hard Part is the Second Part of the Definition
7:30
Second Half of Definition
10:04
Restrictions on the Allowed x Values
10:28
The Epsilon-Delta Part, cont.
13:34
Sherlock Holmes and Dr. Watson
15:08
The Adventure of the Delta-Epsilon Limit
15:16
Setting
15:18
We Begin By Setting Up the Game As Follows
15:52
The Adventure of the Delta-Epsilon, cont.
17:24
This Game is About Limits
17:46
What If I Try Larger?
19:39
Technically, You Haven't Proven the Limit
20:53
Here is the Method
21:18
What We Should Concern Ourselves With
22:20
Investigate the Left Sides of the Expressions
25:24
We Can Create the Following Inequalities
28:08
Finally…
28:50
Nothing Like a Good Proof to Develop the Appetite
30:42
Example 1
31:02
Example 1, cont.
36:26
Example 2
41:46
Example 2, cont.
47:50
Finding Limits

32m 40s

Intro
0:00
Introduction
0:08
Method - 'Normal' Functions
2:04
The Easiest Limits to Find
2:06
It Does Not 'Break'
2:18
It Is Not Piecewise
2:26
Method - 'Normal' Functions, Example
3:38
Method - 'Normal' Functions, cont.
4:54
The Functions We're Used to Working With Go Where We Expect Them To Go
5:22
A Limit is About Figuring Out Where a Function is 'Headed'
5:42
Method - Canceling Factors
7:18
One Weird Thing That Often Happens is Dividing By 0
7:26
Method - Canceling Factors, cont.
8:16
Notice That The Two Functions Are Identical With the Exception of x=0
8:20
Method - Canceling Factors, cont.
10:00
Example
10:52
Method - Rationalization
12:04
Rationalizing a Portion of Some Fraction
12:05
Conjugate
12:26
Method - Rationalization, cont.
13:14
Example
13:50
Method - Piecewise
16:28
The Limits of Piecewise Functions
16:30
Example 1
17:42
Example 2
18:44
Example 3
20:20
Example 4
22:24
Example 5
24:24
Example 6
27:12
Continuity & One-Sided Limits

32m 43s

Intro
0:00
Introduction
0:06
Motivating Example
0:56
Continuity - Idea
2:14
Continuous Function
2:18
All Parts of Function Are Connected
2:28
Function's Graph Can Be Drawn Without Lifting Pencil
2:36
There Are No Breaks or Holes in Graph
2:56
Continuity - Idea, cont.
3:38
We Can Interpret the Break in the Continuity of f(x) as an Issue With the Function 'Jumping'
3:52
Continuity - Definition
5:16
A Break in Continuity is Caused By the Limit Not Matching Up With What the Function Does
5:18
Discontinuous
6:02
Discontinuity
6:10
Continuity and 'Normal' Functions
6:48
Return of the Motivating Example
8:14
One-Sided Limit
8:48
One-Sided Limit - Definition
9:16
Only Considers One Side
9:20
Be Careful to Keep Track of Which Symbol Goes With Which Side
10:06
One-Sided Limit - Example
10:50
There Does Not Necessarily Need to Be a Connection Between Left or Right Side Limits
11:16
Normal Limits and One-Sided Limits
12:08
Limits of Piecewise Functions
14:12
'Breakover' Points
14:22
We Find the Limit of a Piecewise Function By Checking If the Left and Right Side Limits Agree With Each Other
15:34
Example 1
16:40
Example 2
18:54
Example 3
22:00
Example 4
26:36
Limits at Infinity & Limits of Sequences

32m 49s

Intro
0:00
Introduction
0:06
Definition: Limit of a Function at Infinity
1:44
A Limit at Infinity Works Very Similarly to How a Normal Limit Works
2:38
Evaluating Limits at Infinity
4:08
Rational Functions
4:17
Examples
4:30
For a Rational Function, the Question Boils Down to Comparing the Long Term Growth Rates of the Numerator and Denominator
5:22
There are Three Possibilities
6:36
Evaluating Limits at Infinity, cont.
8:08
Does the Function Grow Without Bound? Will It 'Settle Down' Over Time?
10:06
Two Good Ways to Think About This
10:26
Limit of a Sequence
12:20
What Value Does the Sequence Tend to Do in the Long-Run?
12:41
The Limit of a Sequence is Very Similar to the Limit of a Function at Infinity
12:52
Numerical Evaluation
14:16
Numerically: Plug in Numbers and See What Comes Out
14:24
Example 1
16:42
Example 2
21:00
Example 3
22:08
Example 4
26:14
Example 5
28:10
Example 6
31:06
Instantaneous Slope & Tangents (Derivatives)

51m 13s

Intro
0:00
Introduction
0:08
The Derivative of a Function Gives Us a Way to Talk About 'How Fast' the Function If Changing
0:16
Instantaneous Slop
0:22
Instantaneous Rate of Change
0:28
Slope
1:24
The Vertical Change Divided by the Horizontal
1:40
Idea of Instantaneous Slope
2:10
What If We Wanted to Apply the Idea of Slope to a Non-Line?
2:14
Tangent to a Circle
3:52
What is the Tangent Line for a Circle?
4:42
Tangent to a Curve
5:20
Towards a Derivative - Average Slope
6:36
Towards a Derivative - Average Slope, cont.
8:20
An Approximation
11:24
Towards a Derivative - General Form
13:18
Towards a Derivative - General Form, cont.
16:46
An h Grows Smaller, Our Slope Approximation Becomes Better
18:44
Towards a Derivative - Limits!
20:04
Towards a Derivative - Limits!, cont.
22:08
We Want to Show the Slope at x=1
22:34
Towards a Derivative - Checking Our Slope
23:12
Definition of the Derivative
23:54
Derivative: A Way to Find the Instantaneous Slope of a Function at Any Point
23:58
Differentiation
24:54
Notation for the Derivative
25:58
The Derivative is a Very Important Idea In Calculus
26:04
The Important Idea
27:34
Why Did We Learn the Formal Definition to Find a Derivative?
28:18
Example 1
30:50
Example 2
36:06
Example 3
40:24
The Power Rule
44:16
Makes It Easier to Find the Derivative of a Function
44:24
Examples
45:04
n Is Any Constant Number
45:46
Example 4
46:26
Area Under a Curve (Integrals)

45m 26s

Intro
0:00
Introduction
0:06
Integral
0:12
Idea of Area Under a Curve
1:18
Approximation by Rectangles
2:12
The Easiest Way to Find Area is With a Rectangle
2:18
Various Methods for Choosing Rectangles
4:30
Rectangle Method - Left-Most Point
5:12
The Left-Most Point
5:16
Rectangle Method - Right-Most Point
5:58
The Right-Most Point
6:00
Rectangle Method - Mid-Point
6:42
Horizontal Mid-Point
6:48
Rectangle Method - Maximum (Upper Sum)
7:34
Maximum Height
7:40
Rectangle Method - Minimum
8:54
Minimum Height
9:02
Evaluating the Area Approximation
10:08
Split the Interval Into n Sub-Intervals
10:30
More Rectangles, Better Approximation
12:14
The More We Us , the Better Our Approximation Becomes
12:16
Our Approximation Becomes More Accurate as the Number of Rectangles n Goes Off to Infinity
12:44
Finding Area with a Limit
13:08
If This Limit Exists, It Is Called the Integral From a to b
14:08
The Process of Finding Integrals is Called Integration
14:22
The Big Reveal
14:40
The Integral is Based on the Antiderivative
14:46
The Big Reveal - Wait, Why?
16:28
The Rate of Change for the Area is Based on the Height of the Function
16:50
Height is the Derivative of Area, So Area is Based on the Antiderivative of Height
17:50
Example 1
19:06
Example 2
22:48
Example 3
29:06
Example 3, cont.
35:14
Example 4
40:14
XVII. Appendix: Graphing Calculators
Buying a Graphing Calculator

10m 41s

Intro
0:00
Should You Buy?
0:06
Should I Get a Graphing Utility?
0:20
Free Graphing Utilities - Web Based
0:38
Personal Favorite: Desmos
0:58
Free Graphing Utilities - Offline Programs
1:18
GeoGebra
1:31
Microsoft Mathematics
1:50
Grapher
2:18
Other Graphing Utilities - Tablet/Phone
2:48
Should You Buy a Graphing Calculator?
3:22
The Only Real Downside
4:10
Deciding on Buying
4:20
If You Plan on Continuing in Math and/or Science
4:26
If Money is Not Particularly Tight for You
4:32
If You Don't Plan to Continue in Math and Science
5:02
If You Do Plan to Continue and Money Is Tight
5:28
Which to Buy
5:44
Which Graphing Calculator is Best?
5:46
Too Many Factors
5:54
Do Your Research
6:12
The Old Standby
7:10
TI-83 (Plus)
7:16
TI-84 (Plus)
7:18
Tips for Purchasing
9:17
Buy Online
9:19
Buy Used
9:35
Ask Around
10:09
Graphing Calculator Basics

10m 51s

Intro
0:00
Read the Manual
0:06
Skim It
0:20
Play Around and Experiment
0:34
Syntax
0:40
Definition of Syntax in English and Math
0:46
Pay Careful Attention to Your Syntax When Working With a Calculator
2:08
Make Sure You Use Parentheses to Indicate the Proper Order of Operations
2:16
Think About the Results
3:54
Settings
4:58
You'll Almost Never Need to Change the Settings on Your Calculator
5:00
Tell Calculator In Settings Whether the Angles Are In Radians or Degrees
5:26
Graphing Mode
6:32
Error Messages
7:10
Don't Panic
7:11
Internet Search
7:32
So Many Things
8:14
More Powerful Than You Realize
8:18
Other Things Your Graphing Calculator Can Do
8:24
Playing Around
9:16
Graphing Functions, Window Settings, & Table of Values

10m 38s

Intro
0:00
Graphing Functions
0:18
Graphing Calculator Expects the Variable to Be x
0:28
Syntax
0:58
The Syntax We Choose Will Affect How the Function Graphs
1:00
Use Parentheses
1:26
The Viewing Window
2:00
One of the Most Important Ideas When Graphing Is To Think About The Viewing Window
2:01
For Example
2:30
The Viewing Window, cont.
2:36
Window Settings
3:24
Manually Choose Window Settings
4:20
x Min
4:40
x Max
4:42
y Min
4:44
y Max
4:46
Changing the x Scale or y Scale
5:08
Window Settings, cont.
5:44
Table of Values
7:38
Allows You to Quickly Churn Out Values for Various Inputs
7:42
For example
7:44
Changing the Independent Variable From 'Automatic' to 'Ask'
8:50
Finding Points of Interest

9m 45s

Intro
0:00
Points of Interest
0:06
Interesting Points on the Graph
0:11
Roots/Zeros (Zero)
0:18
Relative Minimums (Min)
0:26
Relative Maximums (Max)
0:32
Intersections (Intersection)
0:38
Finding Points of Interest - Process
1:48
Graph the Function
1:49
Adjust Viewing Window
2:12
Choose Point of Interest Type
2:54
Identify Where Search Should Occur
3:04
Give a Guess
3:36
Get Result
4:06
Advanced Technique: Arbitrary Solving
5:10
Find Out What Input Value Causes a Certain Output
5:12
For Example
5:24
Advanced Technique: Calculus
7:18
Derivative
7:22
Integral
7:30
But How Do You Show Work?
8:20
Parametric & Polar Graphs

7m 8s

Intro
0:00
Change Graph Type
0:08
Located in General 'Settings'
0:16
Graphing in Parametric
1:06
Set Up Both Horizontal Function and Vertical Function
1:08
For Example
2:04
Graphing in Polar
4:00
For Example
4:28
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Lecture Comments (4)

1 answer

Last reply by: Professor Selhorst-Jones
Sat Oct 24, 2015 1:06 AM

Post by Fadumo Kediye on October 18, 2015





P(x) = 2x^3 + ax^2 +bx + 6 is divided by x + 2, the remainder is -12. If x - 1 is a factor of the polynomial, find the values of a and b.

1 answer

Last reply by: Professor Selhorst-Jones
Mon Aug 26, 2013 11:12 PM

Post by Humberto Coello on August 23, 2013

How are there no roots for X2 + 4 = 0
isn't it equal to (x+2)(x-2)=0
and don't we get the roots x=2, x=-2 from it?

Intermediate Value Theorem and Polynomial Division

  • There is no general root formula for all polynomials. There are formulas for degree 3 and 4 polynomials, but they're so long and complicated, we wonÕt even look at them. Thus, if we want to find the precise roots of a high degree polynomial, we're stuck factoring.
  • If you manage to find a root of a polynomial (say, by pure luck or an educated guess), then you now know one of the factors of the polynomial. Each root of a polynomial automatically implies a factor:
    f(k) = 0     ⇔     (x−k)  is a factor of f(x).
    Once you know a factor, it becomes that much easier to find the other factors.
  • The intermediate value theorem says that If:
    • f(x) is a polynomial (or any continuous function),
    • a, b are real numbers such that a < b,
    • u is a real number such that f(a) < u < f(b);
    Then: there exists some c ∈ (a,b) such that f(c) = u. [The pictures in the video will greatly help in explaining this theorem.]
  • This means we can use the intermediate value theorem to help us find roots. If we know that f(a) and f(b) are opposite signs (one +, other −), then we know there must be a root in (a,b). [Note that there they may be more than one. It guarantees the existence of at least one root, but does not say the precise number.] Knowing there is a root in some interval makes it that much easier to guess it, check to make sure, then use it to find a factor.
  • The division algorithm states that if f(x) and d(x) are polynomials where the degree of f is greater than or equal to d and d(x) ≠ 0, then there exist polynomials q(x) and r(x) such that
    f(x) = d(x)  · q(x)  +  r(x) .
    Alternatively, we can write this as
    f(x)

    d(x)


     
    = q(x) +r(x)

    d(x)


     
    .
  • In the above, d(x) is what we divide by, q(x) is the quotient (what comes out of the division), and r(x) is the remainder. Thus if after we use the division algorithm we get that r(x) = 0, we know that d(x) divides evenly into f(x), or, in other words, d(x) is a factor of f(x).
  • We can use the division algorithm through polynomial long division. [This is a rather difficult technique to describe with words, but much easier to see in action. Check out the video if you haven't already to understand how to do this.] The process for long division goes like this:

      1. Divide the first term of the dividend (thing being divided) by the first term of the divisor (thing doing the division). Write the result above.
      2. Multiply the entire divisor by the result, then subtract that from the dividend (so that the first term of one lines up with first term of the other).
      3. Bring down the next term from the dividend.
      4. Repeat the process (divide first terms, multiply, subtract, bring down) until finished with the entire dividend.
  • There is also a shortcut method that goes a bit faster if the divisor is in the form (x−k). This is called synthetic division. [Again, this is a rather difficult technique to describe with words, but much easier to see in action. Check out the video if you haven't already to understand how to do this.] The process for synthetic division goes like this:

      1. Write out the coefficients of the polynomial in order.
      2. Write k in the top-left corner [from the factor you're dividing by, (x−k)].
      3. Start on the left of the coefficients and work towards the right. Every step will have a "vertical" part and a "diagonal" part.
      4. On the "vertical" part, add the two numbers above and below each other to produce another number. On the "diagonal" part, multiply the result from the "vertical" by k.
      5. The final number produced by the process is the remainder. The other numbers are the resulting coefficients of the quotient.
  • If you need to do division on a polynomial that is "missing" a variable raised to a certain exponent, make sure to fill that space in with a 0 multiplying the appropriate variable with exponent. For example, if you wanted to divide x4 + 5x + 7, notice that it's "missing" x3 and x2. Before you can divide the polynomial, you have to put something in for those "missing" parts. We do that by filling it in with a 0 multiplying them:
    x4 + 5x + 7     ⇒     x4 + 0x3 + 0 x2 + 5x + 7.

Intermediate Value Theorem and Polynomial Division


f(x) = 3x3 + 9x2 − 30x − 72
Use the fact that f(0) = −72 and f(5)=378 to help you guess a root.
  • By the intermediate value theorem, we know there must be a root (or more than one root) between x=0 and x=5. This is because f(0) is negative, while f(5) is positive. Since a polynomial can't "jump", the function must cross f(x) = 0 for some x between 0 and 5.
  • Try guessing an x and plugging it in. Let's try x=2:
    f(2) = 3(2)3+9(2)2−30(2)−72 = −72
    Now we know that x=2 is not a root. However, we do know that there must be a root between x=2 and x=5 because f(2) is negative and f(5) is positive.
  • Try guessing another x, using the information from our previous guess to help us decide where to guess. Let's try x=3:
    f(3) = 3(3)3+9(3)2−30(3)−72 = 0
    Now we know that x=3 is a root! If we wanted to, we could convert that root into a factor of (x−3) and use it to help us factor the rest of the polynomial and find the rest of the roots.
There is a root at x=3. [ NOTICE: While this method of guessing works on some problems (like this problem), for other problems the root might be something you would not normally guess, like 1.87 or −√5. In such a case, the intermediate value theorem and guessing can help us get a sense for where the roots are, but we probably won't be able to find the precise value for the roots.]

f(x) = x5+7x4−15x3−105x2+50x+350
Use the fact that f(0) = 350 and f(−10)=−25 650 to help you guess a root.
  • By the intermediate value theorem, we know there must be a root (or more than one root) between x=−10 and x=0. This is because f(−10) is negative, while f(0) is positive. Since a polynomial can't "jump", the function must cross f(x) = 0 for some x between −10 and 0.
  • Try guessing an x and plugging it in. Let's try x=−5:
    f(−5) = (−5)5+7(−5)4−15(−5)3−105(−5)2+50(−5)+350 = 600
    Now we know that x=−5 is not a root. However, we do know that there must be a root between x=−10 and x=−5 because f(−10) is negative and f(−5) is positive.
  • Try guessing another x, using the information from our previous guess to help us decide where to guess. Let's try x=−8:
    f(−8) = (−8)5+7(−8)4−15(−8)3−105(−8)2+50(−8)+350 = −3 186
    Now we know that x=−8 is not a root. However, we do know that there must be a root between x=−8 and x=−5 because f(−8) is negative and f(−5) is positive.
  • Try guessing another x, using the information from our previous guess to help us decide where to guess. Let's try x=−7:
    f(−7) = (−7)5+7(−7)4−15(−7)3−105(−7)2+50(−7)+350 = 0
    Now we know that x=3 is a root! If we wanted to, we could convert that root into a factor of (x+7) and use it to help us factor the rest of the polynomial and find the rest of the roots.
There is a root at x=−7. [ NOTICE: If you attempted this problem without looking at the steps (good for you!), you might have noticed that there are roots somewhere around −2.2 and −3.2, but you were not able to guess them. This is because the polynomial has additional roots of ±√5 ≈ ±2.2361 and ±√{10} ≈ ±3.1623. This shows the issue with guessing: while guessing works great when the root is a nice round number, it can be very difficult (or impossible) to guess roots that are not.]
Use polynomial long division to divide:
(x2+5x−24) ÷(x−3)
  • We set up polynomial long division very similarly to how we set up long division many years ago in grade school:
    x
    −3

    x2
    +5x
    −24
  • Begin by figuring out how many times the first term of the divisor (the thing doing the dividing: x−3) go into the first term of the dividend (the thing being divided: x2+5x−24). Thus, our question is what do we get when we divide x2 by x? The result is [(x2)/x] = x, and we write that above the second term of the dividend because we have two terms in the divisor. [We can imagine this as if we lined up the dividend and the divisor, one over the other, then put our first result at the back of where they line up. This is very similar to how we did long division way back when we learned it in primary school.]
    x
    x
    −3

    x2
    +5x
    −24
  • Next, we multiply our divisor by the result that we just obtained. In this case, that's (x−3) ·x = x2−3x. Write this below the dividend, lining the first terms up with each other.
    x
    x
    −3

    x2
    +5x
    −24
    x2
    −3x
  • Finally, subtract what we just wrote from the dividend above and write the result below. [It's critical to remember that you must subtract every term from the one above! It can be difficult to remember this, so you might want to write a big subtraction sign, then distribute it to each term to help you see what you need to do. The important thing is to remember to subtract what you got from the previous step.]
    x
    x
    −3

    x2
    +5x
    −24
    x2
    −3x
    8x
    That completes the first round of long division, but we need to repeat the process until we reach the end of the dividend. Immediately after doing the subtraction, bring down the next term from the dividend so you can continue with the division:
    x
    x
    −3

    x2
    +5x
    −24
    x2
    −3x
    8x
    −24
  • We're now on to the next round of long division. This round is very similar to the previous round, except this time we're dividing what we just found previously. Once again, how many times does the first term of the divisor divide in? [8x/x] = 8, which we now write over the next open spot on the very top.
    x
    +8
    x
    −3

    x2
    +5x
    −24
    x2
    −3x
    8x
    −24
  • Next, multiply the divisor by the result we just obtained: (x−3)·8 = 8x−24. Write this at the bottom, lining up the first terms.
    x
    +8
    x
    −3

    x2
    +5x
    −24
    x2
    −3x
    8x
    −24
    8x
    −24
  • Finally, subtract what we just found from the line above it.
    x
    +8
    x
    −3

    x2
    +5x
    −24
    x2
    −3x
    8x
    −24
    8x
    −24
    0
  • The result of this last subtraction is the remainder. In this case, it is 0, so we know that (x−3) divides evenly into (x2+5x−24). We can multiply the quotient (the result: x+8) by the divisor and, since the remainder was 0, we'll get back the original dividend. It's always a good idea to check long division problems because it's easy to make a mistake somewhere. Check simply by multiplying your answer (the quotient) with the divisor (what you've been dividing by):
    Check:        (x−3)(x+8) = x2 + 5x − 24   
    Because we get back our dividend (the thing being divided), we know that we did the division correctly.
x+8
Use synthetic division to divide:
(x2+5x−24) ÷(x−3)
  • To use synthetic division, our divisor (thing doing the dividing) must be in the form (x−k). In this case, our divisor is (x−3), so we can use synthetic division. The first step is to figure out what the value of k is. Here we have k=3.
  • Next, figure out the value of the coefficients on each term in the polynomial being divided (the dividend). If the polynomial is in the form ax2 + bx + c, where a, b, and c are all coefficients, for our polynomial, we would have a = 1, b=5, and c=−24.
  • Once you have found k and the coefficients to the polynomial, set them up in the following structure:
    k

    a
    b
    c

    So, plugging in the values we found for this problem, we have
    3

    1
    5
    −24

  • Once you've set up the structure, begin by bringing down the first number from the right hand side.
    3

    1
    5
    −24

    1
  • Next, multiply that number by whatever value k is (the number on the left hand side, in this case, 3). Write that number under the next term on the right side.
    3

    1
    5
    −24

    3
    1
  • Add the two numbers that are above one another and write that number directly below them.
    3

    1
    5
    −24

    3
    1
    8
  • Continue this process of multiplying on the diagonal, then adding on the vertical. Next, we have 8·3.
    3

    1
    5
    −24

    3
    24
    1
    8
  • And then we add the two numbers together.
    3

    1
    5
    −24

    3
    24
    1
    8
    0
  • The process is now completed. The final number that we get out (the number in the bottom right, bolded below) is the remainder.
    3

    1
    5
    −24

    3
    24
    1
    8
    0
    Because the remainder is 0, we know the divisor (x−3) divides evenly into the dividend (x2 + 5x − 24). We can find the quotient (the result of the division) by looking at the non-remainder numbers. In this case, we found  1    8  and these make up the coefficients to the quotient:
    x+8
    [If you're a little unsure how to turn the numbers into a polynomial, work from the right and move left. The right-most number is the constant, then one to the left is the coefficient for the variable to the power of 1, then another to the left is the coefficient for the variable to the power of 2, and just keep going as such until you run out of numbers. In this case we only had two numbers, so we only made it up to x1.]
  • It's always a good idea to check your answer, and we can do so by multiplying the divisor by the quotient (our answer):
    Check:        (x−3)(x+8) = x2 + 5x − 24   
    Because we get back our dividend (the thing being divided), we know that we did the division correctly.
x+8
Use polynomial long division to divide:
(x3−18x+8)÷(x−4)
  • (Notice: A detailed explanation of how to do long division will not be given in these steps. The steps will be shown, but no careful explanation of why each step occurs. If you already understand the mechanics of polynomial long division, you'll be able to understand what follows. If polynomial long division does not currently make sense, first go try to understand how it works. Watch the lesson carefully, see how it is used, and check out the steps to the first problem about long division in this question set. That problem has very detailed steps that explain how it works. Come back to this problem once you understand the mechanics of polynomial long division.)
  • Set up the long division. However, there is an issue: (x3−18x+8) does not have an x2 term. This will cause problems later on, so give it an x2 term. Since it has 0 x2, we have
    (x3−18x+8) = (x3+0x2−18x+8)
    Now we're ready to write out the long division.
    x
    −4

    x3
    +0x2
    −18x
    +8
  • Divide the first term by the first term of the divisor.
    x2
    x
    −4

    x3
    +0x2
    −18x
    +8
  • Multiply the result by the divisor, then subtract and bring down the next term from the dividend.
    x2
    x
    −4

    x3
    +0x2
    −18x
    +8
    x3
    −4x2
    4x2
    −18x
  • Divide the first term by the first term of the divisor.
    x2
    +4x
    x
    −4

    x3
    +0x2
    −18x
    +8
    x3
    −4x2
    4x2
    −18x
  • Multiply the result by the divisor, then subtract and bring down the next term from the dividend.
    x2
    +4x
    x
    −4

    x3
    +0x2
    −18x
    +8
    x3
    −4x2
    4x2
    −18x
    4x2
    −16x
    −2x
    +8
  • Divide the first term by the first term of the divisor.
    x2
    +4x
    −2
    x
    −4

    x3
    +0x2
    −18x
    +8
    x3
    −4x2
    4x2
    −18x
    4x2
    −16x
    −2x
    +8
  • Multiply the result by the divisor, then subtract and bring down the next term from the dividend.
    x2
    +4x
    −2
    x
    −4

    x3
    +0x2
    −18x
    +8
    x3
    −4x2
    4x2
    −18x
    4x2
    −16x
    −2x
    +8
    −2x
    +8
    0
  • The result of the last subtraction is the remainder. We got 0, so the divisor divides evenly into the dividend. Our answer is x2+4x−2.
  • It's really easy to make a mistake on long division problems, so always check by expanding it.
    Check:        (x−4)(x2+4x−2) = x3 −18x +8   
x2+4x−2
Use synthetic division to divide:
(x3−18x+8)÷(x−4)
  • (Notice: A detailed explanation of how to do synthetic division will not be given in these steps. The steps will be shown, but no careful explanation of why each step occurs. If you already understand the mechanics of synthetic division, you'll be able to understand what follows. If synthetic division does not currently make sense, first go try to understand how it works. Watch the lesson carefully, see how it is used, and check out the steps to the first problem about synthetic division in this question set. That problem has very detailed steps that explain how it works. Come back to this problem once you understand the mechanics of synthetic division.)
  • To use synthetic division, our divisor must be in the form (x−k). In this case, our divisor is (x−4), so we can use synthetic division.
  • Set up the synthetic division structure. However, there is an issue: (x3−18x+8) does not have an x2 term. This will cause problems later on, so give it an x2 term. Since it has 0 x2, we have
    (x3−18x+8) = (x3+0x2−18x+8)
    Now we're ready to write out the structure.
    4

    1
    0
    −18
    8

  • Start from the left, add on verticals, multiply (by k=4) on diagonals.
    4

    1
    0
    −18
    8

    4
    1

  • 4

    1
    0
    −18
    8

    4
    16
    1
    4

  • 4

    1
    0
    −18
    8

    4
    16
    −8
    1
    4
    −2

  • 4

    1
    0
    −18
    8

    4
    16
    −8
    1
    4
    −2
    0
    This is the last step, and the final number in the bottom right is the remainder. It's 0 in this case, so we know that the divisor divides in evenly. We can find the quotient from the non-remainder numbers on the bottom:
    x2 + 4x −2
  • It's always a good idea to check your work, which we can do by expanding the divisor and quotient together
    Check:        (x−4)(x2+4x−2) = x3 −18x +8   
x2+4x−2
Use polynomial long division to divide:
(2x2+3x−8)÷(x+3)
[Hint: This will not divide evenly. You will get a remainder. Use that in your answer.]
  • (Notice: A detailed explanation of how to do long division will not be given in these steps. The steps will be shown, but no careful explanation of why each step occurs. If you already understand the mechanics of polynomial long division, you'll be able to understand what follows. If polynomial long division does not currently make sense, first go try to understand how it works. Watch the lesson carefully, see how it is used, and check out the steps to the first problem about long division in this question set. That problem has very detailed steps that explain how it works. Come back to this problem once you understand the mechanics of polynomial long division.)
  • Set up the long division.
    x
    +3

    2x2
    +3x
    −8
  • Divide first terms.
    2x
    x
    +3

    2x2
    +3x
    −8
  • Multiply and subtract/bring down next term.
    2x
    x
    +3

    2x2
    +3x
    −8
    2x2
    +6x
    −3x
    −8
  • Divide first terms.
    2x
    −3
    x
    +3

    2x2
    +3x
    −8
    2x2
    +6x
    −3x
    −8
  • Multiply and subtract. We're at the end of the dividend, so whatever results from the subtraction is our remainder.
    2x
    −3
    x
    +3

    2x2
    +3x
    −8
    2x2
    +6x
    −3x
    −8
    −3x
    −9
    1
  • Our quotient is 2x−3 but we also have a remainder of 1. Thus, (x+3) "cleanly" divides out a (2x−3) factor if we also have it divide the remainder. Thus our final answer is
    2x −3 + 1

    x+3
    .
  • As always, it's a good idea to check and make sure that our answer is correct. Furthermore, it will help us see how that strange part with the fraction needs to be a part of the answer:
    Check:       
    (x+3)
    2x−3+ 1

    x+3

    =
    (x+3)(2x−3) + (x+3)
    1

    x+3

    =
    (2x2 + 3x − 9) + 1
    =
    2x2 + 3x − 8    
2x −3 + [1/(x+3)]
Use synthetic division to divide:
(−4x3−16x2−35)÷(x+5)
[Hint: This will not divide evenly. You will get a remainder. Use that in your answer.]
  • (Notice: A detailed explanation of how to do synthetic division will not be given in these steps. The steps will be shown, but no careful explanation of why each step occurs. If you already understand the mechanics of synthetic division, you'll be able to understand what follows. If synthetic division does not currently make sense, first go try to understand how it works. Watch the lesson carefully, see how it is used, and check out the steps to the first problem about synthetic division in this question set. That problem has very detailed steps that explain how it works. Come back to this problem once you understand the mechanics of synthetic division.)
  • To use synthetic division, our divisor must be in the form (x−k). In this case, our divisor is (x+5), so we can use synthetic division. However, notice that k=−5 because the format is (x−k) and so we must have a negative value for k to get the +5 to show up.
  • Set up the synthetic division structure. However, there is an issue: (−4x3−16x2−35) does not have an x term. This will cause problems later on, so give it an x term. Since it has 0 x, we have
    (−4x3−16x2−35) = (−4x3−16x2+0x−35)
    Now we're ready to write out the structure.
    −5

    −4
    −16
    0
    −35

  • Start from the left, add on verticals, multiply (by k=−5) on diagonals.
    −5

    −4
    −16
    0
    −35

    20
    −4

  • −5

    −4
    −16
    0
    −35

    20
    −20
    −4
    4

  • −5

    −4
    −16
    0
    −35

    20
    −20
    100
    −4
    4
    −20

  • −5

    −4
    −16
    0
    −35

    20
    −20
    100
    −4
    4
    −20
    65
    This is the last step, and the final number in the bottom right is the remainder. In this case, our remainder is 65. The numbers before it   [−4    4    −20] are the coefficients of the quotient, but we also need the remainder to appear and be divided to find the answer to our division:
    −4x2 + 4x −20 + 65

    x+5
  • As always, it's a good idea to check and make sure that our answer is correct. Furthermore, it will help us see how that strange part with the fraction needs to be a part of the answer:
    Check:       
    (x+5)
    −4x2 + 4x −20 + 65

    x+5

    =
    (x+5)(−4x2 + 4x −20) + (x+5)
    65

    x+5

    =
    (−4x3 − 16x2 − 100) + 65
    =
    −4x3 − 16x2 − 35    
−4x2 + 4x −20 + [65/(x+5)]
Divide:    (x4+8x3−3x2+32x−28)÷(x2+4)
  • Begin by noticing that synthetic division can not be used for this problem. This is because synthetic division can only be used if the divisor is in the form (x−k), but (x2+4) is not in that format. Therefore we must use long division, so set up the long division structure. However, there is a slight issue: the divisor (x2+4) does not have an x term. While we can do the problem without one, it will make it a bit easier to see how things line up if it has one. Write it instead as (x2+0x+4).
    x2
    +0x
    +4

    x4
    +8x3
    −3x2
    +32x
    −28
  • Divide the first term by the first term of the divisor.
    x2
    x2
    +0x
    +4

    x4
    +8x3
    −3x2
    +32x
    −28
  • Multiply and subtract/bring down next term.
    x2
    x2
    +0x
    +4

    x4
    +8x3
    −3x2
    +32x
    −28
    x4
    +0x3
    + 4x2
    8x3
    −7x2
    + 32x
  • Repeat process until finished.
    x2
    +8x
    x2
    +0x
    +4

    x4
    +8x3
    −3x2
    +32x
    −28
    x4
    +0x3
    + 4x2
    8x3
    −7x2
    + 32x

  • x2
    +8x
    x2
    +0x
    +4

    x4
    +8x3
    −3x2
    +32x
    −28
    x4
    +0x3
    + 4x2
    8x3
    −7x2
    + 32x
    8x3
    +0x2
    + 32x
    −7x2
    +0x
    −28

  • x2
    +8x
    −7
    x2
    +0x
    +4

    x4
    +8x3
    −3x2
    +32x
    −28
    x4
    +0x3
    + 4x2
    8x3
    −7x2
    + 32x
    8x3
    +0x2
    + 32x
    −7x2
    +0x
    −28

  • x2
    +8x
    −7
    x2
    +0x
    +4

    x4
    +8x3
    −3x2
    +32x
    −28
    x4
    +0x3
    + 4x2
    8x3
    −7x2
    + 32x
    8x3
    +0x2
    + 32x
    −7x2
    +0x
    −28
    −7x2
    +0x
    −28
    0
    The process is now complete with a remainder of 0, so we see that (x2+4) divides in evenly. The quotient of x2+8x−7 is our answer.
  • It's always a good idea to check your work, so multiply the divisor and quotient to make sure we have the correct answer.
    Check:     (x2+4)(x2+8x−7) = x4 + 8x3 − 3x2 + 32x − 28    
x2+8x−7
Find all the roots to x4−6x3−5x2+54x−36, using the fact that (x2−9) is one of its factors.
  • No matter what we do, we need to factor the polynomial to find its roots. There are multiple ways to approach factoring it. First, since (x2−9) is a factor, we could put it in the form
    (x2−9)( x2 +  x +  ) = x4−6x3−5x2+54x−36 .
    Using logic, we could eventually figure out what has to go in those blanks, but depending on the problem it might be somewhat difficult. Next, we could attempt to use synthetic division, but we can't because (x2−9) is not in the form (x−k). We could break it into two different factors of (x−3) and (x+3), then divide one after the other, but that takes multiple steps. Instead, for this problem, we'll do it with polynomial long division. [However, the other two methods above would work just fine if you wanted to do them instead.]
  • Set up the long division:
    x2
    +0x
    −9

    x4
    −6x3
    −5x2
    +54x
    −36

  • x2
    x2
    +0x
    −9

    x4
    −6x3
    −5x2
    +54x
    −36
    x4
    +0x3
    −9x2
    −6x3
    +4x2
    + 54x

  • x2
    −6x
    x2
    +0x
    −9

    x4
    −6x3
    −5x2
    +54x
    −36
    x4
    +0x3
    −9x2
    −6x3
    +4x2
    + 54x
    −6x3
    +0x2
    + 54x
    4x2
    +0x
    −36

  • x2
    −6x
    +4
    x2
    +0x
    −9

    x4
    −6x3
    −5x2
    +54x
    −36
    x4
    +0x3
    −9x2
    −6x3
    +4x2
    + 54x
    −6x3
    +0x2
    + 54x
    4x2
    +0x
    −36
    4x2
    +0x
    −36
    0
    Our quotient is (x2 − 6x + 4) when we divide out (x2−9). [Notice that the remainder came out to be 0. This is an automatic check on our work: since the problem said (x2−9) was a factor of x4−6x3−5x2+54x−36, we know it must divide in evenly. If it didn't come out with a remainder of 0, we would have made a mistake somewhere.]
  • Using long division, we have factored the polynomial as
    x4−6x3−5x2+54x−36 = (x2−9)(x2 − 6x + 4).
    We can easily factor (x2−9) further as (x+3)(x−3), so
    (x+3)(x−3)(x2 − 6x + 4),
    which means we have roots at x+3 = 0 and x−3=0, so x=−3,  3. However, we still need to figure out if (x2−6x+4) has any roots. Looking at it, we realize that it would be very difficult to factor because it will involve non-integer numbers in the factors.
  • Luckily, we have an easy way to find the roots of a quadratic polynomial: the quadratic formula! [See the lesson on the quadratic formula if you don't know how to use it yet. Set it up using (x2−6x+4) to find its roots:
    x =
    −(−6) ±


    (−6)2 − 4 ·1 ·4

    2·1
  • Working through it, we get
    x  = 
    6 ±


    20

    2
     =  6 ±2√5

    2
     = 3±√5.
    Therefore, the roots of (x2 − 6x+4) are x = 3 − √5,   3 + √5.
  • Assemble these roots with the roots we found earlier to have all the roots of the polynomial.
Roots: x = −3,     3−√5,     3,     3 + √5

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Intermediate Value Theorem and Polynomial Division

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:05
  • Reminder: Roots Imply Factors 1:32
  • The Intermediate Value Theorem 3:41
    • The Basis: U between a and b
    • U is on the Function
  • Intermediate Value Theorem, Proof Sketch 5:51
    • If Not True, the Graph Would Have to Jump
    • But Graph is Defined as Continuous
  • Finding Roots with the Intermediate Value Theorem 7:01
    • Picking a and b to be of Different Signs
    • Must Be at Least One Root
  • Dividing a Polynomial 8:16
    • Using Roots and Division to Factor
  • Long Division Refresher 9:08
  • The Division Algorithm 12:18
    • How It Works to Divide Polynomials
    • The Parts of the Equation
    • Rewriting the Equation
  • Polynomial Long Division 16:20
    • Polynomial Long Division In Action
    • One Step at a Time
  • Synthetic Division 22:46
    • Setup
  • Synthetic Division, Example 24:44
  • Which Method Should We Use 26:39
    • Advantages of Synthetic Method
    • Advantages of Long Division
  • Example 1 29:24
  • Example 2 31:27
  • Example 3 36:22
  • Example 4 40:55

Transcription: Intermediate Value Theorem and Polynomial Division

Hi--welcome back to Educator.com.0000

Today, we are going to talk about the intermediate value theorem and polynomial division.0002

Previously, we talked about how we need to factor a polynomial to find its roots.0006

But we recently saw the quadratic formula, which gave us the root of any quadratic without even having to factor at all.0009

So, maybe we don't need factoring; maybe there are formulas that allow us to find the roots for any polynomial; wouldn't that be great?0015

We would just be able to plug things in, put out some arithmetic, and we would have answers, no matter what polynomial we were dealing with.0021

Not really--while there are root formulas for polynomials of degree 3 and 4, they are so long and complicated, we are not even going to look at them.0028

The formula for finding the roots of a cubic, of a degree 3 polynomial, is really long and really complex.0036

And it is just something that we don't really want to look at right now.0044

And a degree 4 would be even worse; so we are just not going to worry about them.0049

And then, not only that--they just simply don't exist for degree 5 or higher.0054

So, if you are looking at a degree 5 or higher, there is no such formula for any degree 5 or higher thing.0060

It was proven in 1824 that no such formula can exist, that would be able to do that.0067

Thus, it looks like we are stuck factoring, if you want to find the precise roots of a higher-degree polynomial.0073

If we are working with a higher-degree polynomial, and we need to know its roots for some reason, we have to figure out a way to factor it.0077

In this lesson, we are going to learn some methods to help factor these complicated polynomials.0082

We will first learn a theorem to help us guess where roots are located, and then a technique for helping us break apart big polynomials.0086

All right, let's go: In the lesson Roots of Polynomials, we mentioned the theorem that every root implies a factor.0092

That is, if f(x) is a polynomial, then if we have f(k) = 0 (k is a root), then that means we know x - k is a factor of f(x),0102

because if k = 0, then that means that x = k causes a root; so x - k = 0; and thus, we have a factor from our normal factor breakdown.0112

For example, if we have g(x) = x3 - 3x2 - 4x + 12,0123

and we happen to realize that when we plug in a 2, that all turns into a 00128

(and it does), then we would know that g(x) is equal to (x - 2), this thing becoming (x - 2), times (_x2 + _x + _).0131

We know that there is some way to factor that polynomial where something is going to go in those blanks.0147

2 is a root; this theorem tells us that (x - 2) must be a factor.0152

It doesn't tell us what will be left; but it does make the polynomial one step easier.0156

We know we can pull up (x - 2), so then we can use some logic, play some games, and figure out what has to go in those blanks.0162

But notice: it doesn't directly tell us what is going to be there.0168

If we are lucky, we can sometimes find a root or two purely by guessing.0173

We might think, "Well, I don't know where the roots are; but let's try -5; let's try √2; let's try π."0176

We might just try something, and surprisingly, it ends up working; that is great.0185

But is this always going to end up being the case?0189

If we manage to pick something where we figure out the root, and then we figure out something that gets us a root;0193

we plug in a number, and we get 0, and we know we have a root.0201

And if we have a root, that means we have a factor.0204

Knowing a factor makes it that much easier to factor the whole polynomial.0207

But it is hard to guess correctly every time; guessing is guessing--you can't guess every single time.0211

Luckily, there is a theorem that will give us a better idea of where the roots are located.0218

The intermediate value theorem will help us find roots; it goes like this:0223

If, first, f(x) is a polynomial (for example, we have this nice red curve here; that is our f(x));0227

then a and b are real numbers, such that a < b.0234

What that means is just that a and b are going from left to right; a is on the left side, and then we make it up to b.0238

That is what this a < b is--just that we know an order that we are going in.0247

And then, u is a real number, such that f(a) < u < f(b).0251

So, we look and figure out that part; we see that, at a, we are at some height f(a); and at b, we are at some height f(b).0256

That is how we get that graph in the first place.0267

Then, u is just something between those two heights; so u is just some height level,0269

where we put an imaginary horizontal line that ends up saying,0276

"Here is an intermediate value between f(a) and f(b), some intermediate height between those two."0280

The intermediate value theorem tells us that there exists some c contained in ab, such that f(c) = u.0289

So, we are guaranteed the existence of some c that is going to end up giving us this height, u.0298

Basically, if we have some height u, and that height u crosses between two different heights0305

that go...we have two points going from left to right, so we are going from left to right, and we cross over some height during that thing;0316

we start lower, and then we end above; we are guaranteed that we had to actually cross it.0332

We had to go across it; and since we had to go across it, there must be some c where we do that crossing--where we end up landing on that height.0338

There is something that will give us that intermediate value.0348

Why does this have to be true? Because polynomials are continuous; there are no breaks in their graphs.0352

The only way f could possibly manage to not end up being on this height u--the only way f could dodge the height u-- is by jumping this intermediate height.0358

The only way that we have this...the graph is going; the graph is going; the graph is going; the graph is going; the graph is going.0369

And then, all of a sudden, it would have to jump over that height to be able to manage to not end up touching it.0375

If our graph touches the height at any point, then we have whatever point is directly below where it touched that height;0381

that is the location that intersects that intermediate value.0388

That is the thing that is going to fulfill our intermediate value theorem.0391

So, on any polynomial or any continuous function (in fact, the intermediate value theorem is true for any continuous function,0395

but we are just focusing on polynomials), they can't jump; polynomials--continuous functions--they are not allowed to jump.0400

There are no breaks in their graphs; so since there are no breaks, they have to end up crossing over this height.0406

Since they cross over this height, there is some place on the graph...we just look directly below that,0412

and that guarantees us our c, where f(c) is going to be equal to u; there we go.0416

This means we can use the intermediate value theorem to help us find roots.0423

If we know that f(a) and f(b) are opposite signs--so, for example, if we know that at a, f(a) is positive--0427

we have a positive for f(a)--and then, at b, we know that we are negative--we have a negative for f(b)--0434

then we know that there has to be a root.0440

Why? Well, we have to have some f that is going to make it from here somehow to here.0442

It has to manage to get both of those things.0448

So, the only way it can do it is by crossing over at some location.0451

It might cross over multiple times, but it has to cross over somewhere.0455

Otherwise, it is not going to be able to make it to that point that we know is below y = 0.0462

Since f must cross over y = 0, we are guaranteed the existence of this c at some point where it ends up crossing over.0467

Now, this does not mean that there is only one root; like we just saw in that second thing I drew, it could cross over multiple times.0478

This theorem guarantees the existence of at least one root, but there could be multiple roots.0485

if it bounces back and forth over that y = 0 on the way to making it to the second point.0491

OK, so say we find a root of a polynomial by a combination of luck and the intermediate value theorem.0498

We somehow manage to figure them out, or the problem just tells us a root directly from the beginning.0503

In either case, with the root, we now know a factor; a root tells us a factor.0508

But how can we actually break up the polynomial if we know a factor?0513

How do we divide a polynomial by a factor?0516

For example, say we know x = 3 is a root of this polynomial.0519

Then we know that there is some way to divide out x - 3 so that we have x - 3 pulled out,0523

and some _x3 + _x2 + _x + _--0527

there is some other polynomial that is going to go with it, that the two will multiply.0531

Otherwise, it would not have divided out cleanly.0534

It couldn't be a factor unless there was going to be this other polynomial where it does divide out cleanly.0536

So, how do we actually find out that thing that happens after we divide out this polynomial?0541

How do we do polynomial division?0545

To explore this idea, let's refresh ourselves on long division from when we were young.0548

So, long ago, in grade school and primary school, we were used to doing problems like 1456 divided by 3.0553

Let's break out long division: we have 1456, so the first thing we do is see how many times 3 goes into 1.0559

3 goes into 1 0 times; so it is 0 times 3; that gets us 0 down here; we subtract by 0; nothing interesting happens yet.0570

1; and then we bring down the 4; so we have 14 now.0577

How many times does 3 go into 14? It goes in 4 times.0583

3 times 4 gets us 12; we subtract by 12: minus 12; so minus 12...we get 2.0586

Then, we bring down the next one in the running; let's keep the colors consistent.0596

We have 5 coming down, so we now have 25; how many times does 3 go into 25?0602

It goes in 8 times...16, 24...so now we subtract by 24; 25 - 24 gets us 1.0607

We bring down the 6; we have 16; how many times does 3 go into 16? It goes in 5; we get 15.0617

So, minus 15...we get 1; now, we don't have any more numbers to go here.0625

There is nothing else, so that means we are left with a remainder of 1.0633

We have whatever our very last thing was, once we ran out of stuff; that becomes our remainder: 485 with a remainder of 1.0640

If we wanted to express this, we could say 1456 divided by 3; another way of thinking of that is 1456 is equal to 3(485) + 1.0647

Or, alternately, if we wanted to, we could say 1456/3 is equal to 485 plus the remainder, also divided,0660

because we know that we get the 485 cleanly, but the 1 is a remainder.0670

So, it doesn't come out cleanly; so it comes out as 1/3.0675

You could also see the connection between these two things, because we simply divide both sides by 3;0677

and that is how we are getting from one place to the other place.0683

That is what we are getting by going through long division.0686

Really quickly, let's also look at this 1456 = 3(485) + 1; we call this right here the dividend; the thing being divided is the dividend.0691

This one here is the thing doing the dividing; we call the thing doing the dividing the divisor.0705

Then, what we get out of it is our quotient; what comes out of division is the quotient.0715

And finally, what we have left at the very end is the remainder.0724

All right, these are some special terms; you might not remember those from grade school.0732

But these are the terms that we use to talk about it.0736

Why does that matter? because we are now going to want to be able to express it in a more abstract, interesting way,0738

where we are talking, not just about real numbers, but being able to talk about polynomials.0743

We found that 1456 divided by 3 became 3 times 485 plus 1.0747

Clearly, we can do this method for any two numbers; and it turns out that we can do a very similar idea for polynomials.0752

We call this the division algorithm--this idea of being able to do this.0757

And it says that if f(x) and d(x) are both polynomials, and the degree of f is greater than or equal to d--0761

that is to say, f is a bigger polynomial than d--and d(x) is not equal to 0 (why does d(x) not equal 0?0770

because we are not allowed to divide by 0, so if d(x) is just simply 0 all the time, forever,0776

then we can't divide by it, because we are not allowed to divide by 0)--given these things (f(x) and d(x), both polynomials;0781

f(x) is a bigger polynomial--that is to say, higher degree than d--and d(x) is not simply 0 everywhere),0786

then there exist polynomials q(x) and r(x) such that f(x) = d(x) times q(x) plus r(x).0792

So, how is this parallel? f(x) is the thing being divided.0800

The thing being divided is our dividend, once again.0804

The thing doing the dividing is the divisor.0808

What results after we have done that division is the quotient.0818

And finally, what is left at the end is our remainder.0826

So, the remainder is right here, and our quotient is right here.0836

So, we have parallels in this idea of 1456/3; we have this same thing coming up here.0844

1456/3 is not equal to 3(485); it becomes this idea; so it is not actually equal: 1456/3 is 485 plus 1/3; but it becomes this idea.0850

So, the dividend here, the thing that we are breaking up, in this idea, is 1456; let's just knock this out, so we don't get confused by it.0864

Our divisor, the thing doing the dividing, is 3; what we get in the end, our quotient, is 485; and the remainder is that 1; 1 is left out of it.0874

Now, we could also have an alternative form where we write this as f(x)/d(x) = [q(x) + r(x)]/d(x),0884

where we just turn this into dividing; we get between these two by dividing by...not 3...but dividing by d(x), dividing by our polynomial.0892

So, basically the same thing is happening over here, so this would not be 3 times 45; it should be 1/3.0906

1456/3 is equal to 485 + 1/3, because that is 1456/3.0913

We have a real connection between these two things.0920

The division algorithm is giving us this idea that if we have some polynomial f(x),0924

we can break it into the divisor, times the quotient, plus some remainder, which, alternatively, we can express as0928

the polynomial that we are dividing, divided by its divisor, is equal to the quotient, plus the remainder, also divided by the divisor.0934

This is effectively a way of looking at f(x) dividing d(x)--seeing what is happening here.0941

Now, notice: r(x) is the remainder, so in the case when r(x) = 0, then that means we have no remainder,0945

which we describe as d(x) dividing evenly; so when it divides evenly into f(x), then that means d(x) is simply a factor.0954

5 divides evenly into 15, so that means 5 is a factor of 15.0964

How do we actually use the division algorithm to break apart a polynomial?0970

Let's look at two methods: we will first look at long division, and then we will look at synthetic division.0974

First, long division: we will just take a quick run at how we actually use polynomial long division.0978

And it works a lot like long division that we are already used to.0983

So, let's see it in action first; and then we will talk about how it just worked.0986

We have x4 - 5x3 - 7x2 + 29x + 30, divided by x - 3.0989

So, we are dividing x - 3; it is dividing that polynomial: x4 - 5x3 - 7x2 + 29x + 30.0996

OK, the first thing we do is ask, "All right, how many times does x - 3 go into x4 - 5x3?"1013

Well, really, we are just concerned with the front part; so just look at the first term, x.1021

How many times does x go into x4?1025

Well, x4 divided by x would be x3; so the x3 goes here.1028

Now, that part might be a little confusing: why didn't we end up having it go at the front?1033

Well, think of it like this: if we have 12 divide into 24, does 2 show up at the front?1037

No, 2 doesn't show up at the front; 2 shows up on the side, because 12 is 2 digits long; so we end up being at the second-place digit, as well.1043

It is 2 digits long, so we go with the second-place digit.1052

So, the same thing is going on over here: x - 3 is two terms long, so we end up being at the second term, as well.1054

All right, so all of the ideas...we are going to knock them out really quickly, now that we have them explained.1061

So, that is why we are not starting at the very first place--because we have to start out at where they line up appropriately.1066

We check first term to first term, but then we go as far wide as that thing dividing is.1074

So, x3 is what we get out of x4 divided by x.1078

So, now we take x3, and we multiply (x - 3), just as we did in long division.1083

x3 times (x - 3) becomes x4 - 3x3.1087

Now, we also, in long division, subtracted now; so subtract.1092

Let's put that subtraction over it; minus, minus, 2 minus's become a plus...so we have -x4 attacking x4, so we have 0 here.1097

And 3x3 + -5x3 becomes -2x3.1105

And then, the next thing we do is bring down the -7x2.1115

So, -7x2: now we ask ourselves, "How many times does x go into -2x3?"1119

Well, that is going to go in -2x2, so we get -2x2.1127

-2x2 times x - 3 becomes -2x3; -2x2 times -3 becomes + 6x2.1133

Now, we subtract by all this stuff; we distribute that; that becomes positive; this becomes negative.1142

We now add these things together, so -2x3 + 2x3 becomes 0 once again.1149

-7x2 - 6x2 becomes -13x2.1154

The next thing we do is bring down the 29x, so + 29x.1159

How many times does x go into -13x2? That goes in -13x.1164

The -13x times x - 3 gets us -13x2 + 39x; that is this whole quantity; subtracting that whole thing,1171

we distribute that, and it becomes addition there, and subtraction there.1182

So, we have -13x2 + 13x2; that becomes 0 once again; 29x - 39x becomes -10x.1185

Once again, we bring down the 30; so we have + 30 here.1193

And now, how many times does x go into -10x? x goes in -10 times.1199

So, -10 times x - 3 is -10 + 30; we subtract this whole thing; we distribute that; and we get 0 and 0.1205

So, we end up having a remainder of 0, which is to say it goes in evenly.1215

So, if that is the case, we now have x3 - 2x2 - 13x - 10 as what is left over after we divide out x - 3.1219

So, we know our original x4 - 5x3 - 7x2 + 29x + 30 factors as...1230

let's write this in blue, just so we don't get it confused...(x - 3)(x3 - 2x2 - 13x - 10).1236

That is what we have gotten out of it; cool.1248

To help us understand how that worked, let's look at the steps one at a time.1252

You begin by dividing the first term in the dividend by the first term in the divisor.1255

So, our dividend is this thing right here; its first term is x4; the first term of our divisor,1260

the thing doing the dividing, is x; so how many times does x go into x4?1266

Well, x4 divided by x...if we are confused by the exponents, we have x times x times x times x, over x;1269

so, one pair of them knock each other out; so we have x3 now, x times x times x; great.1279

That is why the x3 goes here; it is the very first thing that happens.1287

The next thing: we take x3, and we multiply it onto (x - 3).1291

So, x3(x - 3) becomes x4 - 3x3.1298

You multiply the entire divisor (this right here) by the result, our x3.1304

And then, we subtract what we just had from the dividend.1310

x4 - 3x3: we subtract that from x4 - 5x3.1315

This gets distributed, so we get a negative here, a plus here; and so that becomes -2x3.1319

The next thing we do is bring down the next term; so our next term to deal with is this 7x2.1325

It gets brought down, and we have -2x3 - 7x2.1331

And then, once again, we do the same thing: how many times does x go into -2x3?1337

It goes in -2x2; so then, it is -2x2(x - 3); and we get -2x3 + 6x2.1343

We subtract that, and we keep doing this process until we are finally at the end.1354

We might have a remainder if it doesn't come out to be 0 after the very last step.1358

Or if it comes out to be 0, we are good; we don't have a remainder.1362

All right, there is also a shortcut method that goes a bit faster if the divisor is in the form x - k.1365

And notice: it has to be in the form x - k; if it is in a different form, like x2 + something, we can't do it.1372

Now, notice that you could deal with x + 3; it would just mean that k is equal to -3, so that is OK.1378

It just needs to be x, and then a constant; so that is the important thing if we are going to use synthetic division.1386

So, it goes like this: we let a, b, c, d, e be the coefficients of the polynomial being divided.1391

For example, if we have ax4 + bx3 + cx2 + dx + e,1396

then we set it up as follows; this k right here is on the outside of our little bracket thing.1401

And then, we set them up: a, b, c, d, e.1407

Now, the very first step: every vertical arrow--you bring whatever is above down below the line.1410

So, a, since there is nothing underneath it...we add terms on vertical arrows, so they come down adding together.1417

So, a comes down; there is nothing below it, so it becomes just a.1423

Then, you multiply by k on the diagonal arrow; so we have a; it comes up; we multiply by k, and so we get k times a.1426

Then, once again, we are doing another vertical arrow where we are adding.1435

We go down: k times a...b + ka becomes ka + b.1439

The next thing that will happen (you will probably want to simplify it, just to make it easier, but) we multiply that whole thing by k once again.1446

And we keep up the process until we get to the very last thing.1452

And the very last thing is our remainder.1456

All the terms preceding that, all of the terms in these green circles, are the coefficients of the quotient.1459

So, if this is one, then we will have a constant here, starting from the right; and this will be x's coefficient;1468

this will be x2's coefficient; this will be x3's coefficient.1473

And that makes sense: since we started with that to the fourth,1476

and we were dividing by something in degree 1, we should be left with something of degree 3.1479

All right, let's see it in action now.1483

Once again, dividing the same thing, we have x4 - 5x3 - 7x2 + 29x + 30.1485

So, we have x - 3; remember, it is x - k, so that means our k is equal to 3, because it is already doing the subtraction.1491

We have 3; and we set this up; our first coefficient here is just a 1; 1 goes here.1500

What is our next coefficient? -5; -5 goes here.1511

What is our next coefficient? -7; -7 goes here.1516

Our next coefficient is 29; what is our next coefficient? 30, and that is our last one, because we just hit the constant.1519

All right, so on the vertical parts, we add; so 1 + _ (underneath it) becomes 1.1526

Then, 3 times 1 becomes 3; -5 plus 3 becomes -2; -2 times 3 is -6; -6 plus -7 becomes -13.1533

3 times -13 becomes -39; 29 plus -39 becomes -10; 3 times -10 becomes -30; 30 + -30 becomes 0.1546

Now, remember: this very last one is our remainder; so our remainder is 0, so it went in evenly, which is great,1558

because since we just did this with polynomial long division, and we saw it went in evenly, it had better go in evenly here, as well.1564

So, this is our constant right here (working from the right); this is our x; this our x2; this is our x3.1571

So, we get x3 - 2x2 - 13x - 10; that is what is remaining.1578

So, we could multiply that by x - 3; and then this whole expression here would be exactly what we started with in here, before we did the division.1587

Great; all right, so which of these two methods should we use?1598

At this point, we have seen both polynomial long division and synthetic division.1602

And so, which is the better method--which one should we use when we have to divide polynomials?1606

Now, synthetic division, as you just saw, has the advantage of being fast--it goes pretty quickly.1610

But it can only be used when you are dividing by (x - k); remember, it has to be in this form1614

of linear things dividing only: x and plus a constant or minus a constant.1618

Ultimately, it is just a trick for one very specific kind of problem, where you have some long polynomial,1623

and you are dividing by a linear factor--by something x ± constant.1628

Long division, on the other hand, while slower, is useful in dividing any polynomial.1633

We can use it for dividing any polynomial at all.1638

I think it is easier to remember, because it goes just like the long division that we are used to from long, long ago.1641

There is a slight change in the way we are doing it, but it is pretty much the exact same format.1646

How many times does it fit it? Multiply how many times it fits in by what you started with, and then subtract that.1651

And just repeat endlessly until you get to the end.1655

And then, lastly, it is connected to some deep ideas in mathematics.1660

Now, you probably won't end up seeing those deep ideas in mathematics until you get to some pretty heavy college courses.1663

But I think it is really cool how something you are learning at this stage can be connected to some really, really amazing ideas in later parts of mathematics.1668

So, personally, I would recommend using long division.1676

I think long division is the clear winner for the better one of these to use,1679

unless you are doing a lot of the (x - k) type divisions, or the problem specifically says to do it in synthetic.1682

If your teacher or the book says you have to do this problem in synthetic, then you have to do it in synthetic, because you are being told to do that.1689

But I think long division is easier to remember; it is more useful in more situations;1694

and it is connected to some really deep ideas that help you actually understand1699

what is going on in mathematics, as opposed to just being a trick.1701

Honestly, the only reason we are learning synthetic division in this lesson--in this course--1704

is because so many other teachers and books teach it.1709

I personally don't think it is that great.1712

It is a useful trick; it is really useful in the specific case of linear division.1715

If you had to do a lot of division by linear factors, it would be really great.1718

But we are just sort of seeing that we can break up polynomials, so I think the better thing is long division.1721

It is easier to remember; you can actually pull it out on an exam after you haven't done it for two months,1727

and you will remember, "Oh, yes, it is just like long division," which by now is burned into your memory from learning it so long ago.1731

And so, synthetic division is really just watered-down long division; I would recommend keeping long division in your memory.1737

It is interesting; it is not that hard to remember; it is useful in any situation; and it is connected to some deep stuff.1743

And synthetic division is really only useful for this one specific situation.1748

So, it is really just a trick; I am not a big fan of tricks, because it is easy to forget them and easy to make mistakes with them.1751

But long division is connected to deep ideas, and it is already in your memory; you just have to figure out,1756

"How do I apply that same idea to a new format?"1761

All right, let's see some examples: Let f(x) = 2x3 + 4x2 - 50x - 100.1764

Use the fact that f(-3) = 32, and f(-1) = -48, to help you guess a root.1771

This sounds a lot like the intermediate value theorem: notice, 32 starts positive; this one is negative.1777

So, that means that between these two things, at -3, we are somewhere really positive.1782

At -1, we are somewhere really negative; so we know that somewhere on the way, it manages to cross; so we know we have a root there.1788

So, how are we going to guess it? Well, we might as well try the first thing that is in the middle of them.1796

So, let's give a try to f(-2); now notice, there is no guarantee that f(-2) is going to be the answer.1800

It could be f(-2.7); it could be f(-1.005); it could be something that is actually going to require square roots to truly express.1807

But we can get a better sense of where it is; and we are students--they are probably going to make it not too hard on us.1817

So, let's try -2; let's guess it; let's see what happens.1823

We plug in -2; we have 2 being plugged in, so (-2)3 + 4(-2)2 - 50(-2) - 100.1826

OK, f(-2) is going to be equal to 2 times...what is -2 cubed?1837

-2 times -2 is 4, times -2 is -8; keep that negative sign; plus 4 times -2 squared (is positive 4);1842

minus 50 times -2; these will cancel out to plus signs; we will get 50 times 2, which is 100; minus 100.1853

So, those cancel out (-100 + 100).1861

2 times -8 is -16, plus 4 times 4 is 16; they end up being not too difficult on us.1865

And sure enough, we get = 0; so we just found a root: f(-2) = 0, so we have a root, or a zero, however you want to say it, at x = -2.1873

Great; there is our answer.1886

All right, Example 2: f(x) is an even-degree polynomial, and there exists some a and b,1887

such that f(a) and f(b) have opposite signs (one positive, the other negative).1893

Why is it impossible for f, our polynomial, to have just one root?1897

OK, so to do this, we need to figure out how we are going to do it.1902

Well, we first think, "Oh, one positive; the other negative; that sounds a lot like the intermediate value theorem that they just introduced to us."1905

So, it is likely that we are going to end up using that.1912

Let's think in terms of that: f(a) and f(b)...we have two possibilities: f(a) could be positive, while f(b) is negative;1914

or it could end up being the case that f(a) is the negative one, while f(b) is the positive one.1924

They didn't tell us which one; so we have to think about all of the cases.1932

Now, how can we see what this is?1935

We have an even-degree polynomial...let's start doing this by drawing.1937

We could have a world where we have a positive a (there is some a here, and then some b here,1941

where it is negative); and then we could also have another world where we have f(a) start as being negative somewhere,1951

and then f(b) is positive somewhere; they don't necessarily have to be on opposite sides of the y-axis.1961

But we are just trying to get an idea of what it is going to look like.1966

We know that we are somewhere on the left, and then we go up to the right.1968

So, how is this going to work?1972

And then, we could draw in a polynomial now; we could try to draw in a picture.1973

And we might say, "OK, we have a polynomial coming like this."1978

And we remember, "Oh, the polynomial could also come from the bottom."1981

So, now we have another two possibilities.1984

The polynomial is coming from up, or the polynomial is coming from down.1987

There is polyup, if it is coming from above and then going down, or the polynomial that starts below, so polydown.1992

It is coming from the bottom part, and it is going up.2001

And maybe that was a little bit confusing as a way to phrase it; but we have one of two possibilities.2004

The polynomial is coming in from either the top, or it is coming in from the bottom; polynomial at the top/polynomial at the bottom.2007

Those are our two possibilities; OK.2016

So, we could be coming from the top; and let's put in our points, as well, again--the same points, positive to negative.2020

Or we could be coming from the bottom, like this.2026

Then, on our negative to positive, we could have negative down here and positive here.2030

And once again, we could be coming from the top or...oops, I put that on the higher one...we could be coming from the bottom.2037

So now, let's see how it goes.2047

Well, we know for sure that the polynomial has to end up cutting through here, because we are told that it has that value.2048

And then, it has to also cut through here.2054

In this one, it cuts through here; and then, it has to get somehow to here, so it cuts through here.2057

And that is basically the idea of the theorem: we are coming from the bottom, and we go up and come down.2063

So, we have already hit more than one root; we have two roots minimum here already.2068

What about this one? Well, we go down, and now we have to go up to this one, as well.2075

So, we go up, and we are done already; we have two roots for this one.2081

In this one, we come up; we go through this one, and then we come up; and we go through this one.2084

So, these are the two where we still are unsure what has to happen next.2089

Well, remember: what do we know about even-degree polynomials?2093

They mentioned specifically that it is an even degree; even degree always means that the two ends,2096

if we have it going down this way...then it means that on the right side, it goes down here, as well.2104

On the other hand, if it goes up on the left side, then it has to go up on the right side.2109

Hard-to-see yellow...we will cover that with a little bit of black, so we can make sure we can see it.2112

So, if we are an even degree, they have to be the same direction on both the left and the right side.2118

They could both go down, or they could both go up; but it has to, in the end, eventually go off in that way.2123

So, who knows what happens for a while here?2128

It could do various stuff; but eventually, at some point later on, it has to come back down,2131

which means that it has to end up crossing the x-axis a second time.2136

So, this one has to be true, as well; the same basic idea is going on over here.2140

Who knows what it is going to do for a while.2147

But because it is an even-degree polynomial, we know it eventually has to do the same thing.2148

So, it is going to have to come back up; so it is going to cross here and here; so it checks out.2153

All of our four possible cases, +/- or -/+, combined with coming from the top or coming from the bottom--2158

the four possible cases--no matter what, by drawing out these pictures, we see that it is impossible,2164

because it is either going to have to hit the two just to make it there;2168

or because it has the even degree, it is going to be forced to come back up and reverse what it has done previously.2172

And we are getting that from the intermediate value theorem.2178

Great; Example 3: Let f(x) = x5 - 3x4 + x3 - 20x + 60, and d(x) = x - 3.2182

And we want to use synthetic division to find f(x)/d(x).2191

All right, the first thing to notice: x5, x4, x3...there is no x2!2195

So, we need to figure out what x2 is, so we can effectively put it in as 0x2.2202

Remember: the coefficient that must be on the x2 to keep it from appearing, to make it disappear, is a 0.2207

So, what we really have is the secret 0x2 + 60, because we have to have coefficients2213

for every single thing, from the highest degree on down, to use synthetic division.2218

So, what is our k? Well, it is x - k for synthetic division; so our k equals 3.2223

We have 3 here; and now we just need to place in all of our various coefficients.2229

Our various coefficients: we have a 1 at the front; we have a -3 in front of x4;2235

we have a hidden 1 in front of the x3; we have a 0 on our completely-hidden x2.2241

We have a -20 on our x, and we have a 60 on the very end for our constant.2247

That is all of them; we have made it all the way out to the constant.2254

So remember, on vertical arrows, when we go down, we add.2257

So, it is adding on vertical arrows: 1 + _ becomes just 1; and then, on these, it is multiplying by whatever our k is.2260

So, 1 times 3...we get 3; we add -3 and 3; we get 0; 0 times 3...we get 0 still;2271

0 + 1 is 1; 1 times 3 is 3; 0 + 3 is 3; 3 times 3 is 9; -20 + 9 is -11; -11 times 3 is -33; positive 27.2279

Now, remember: the very last spot is always the remainder; so this right here is our remainder of 27.2291

From there on, -11 is our constant; we work from the right to the left.2299

Then come our x coefficient, our x2 coefficient, our x3 coefficient, and our x4 coefficient.2304

And it makes sense that it is going to be one degree lower on the thing that eventually comes out of it, the quotient.2311

So, we write this thing out now; we have x4 + 0x3, so we will just omit that;2316

plus 1x2 + 3x - 11; but we can't forget that remainder of 27.2323

So, we have a remainder of + 27; but the remainder has to be divided, because that is the one part where it didn't divide out evenly.2331

So, 27/(x - 3)--that is what we originally divided by; this is f(x)/d(x).2338

Great; and that is our answer, that thing in parentheses right there.2348

And now, if we wanted to do a check, we could come by, and we could multiply by x - 3.2351

If you divide out the number, and then you multiply it back in, you should be exactly where you started.2358

So, we multiply by x - 3; for x4, we get x5; minus 3x4...x2...2362

+ x3 - 3x2 + 3x(x)...+ 3x2...+ 3x(-3), so - 9x; -11(x), so -11x...2371

minus 3 times...-11 times -3 becomes positive 33; and then finally, all of 27/(x - 3) times (x - 3)...2384

As opposed to distributing it to the two pieces, we say, "x - 3 and x - 3; they cancel out," and we are just left with 27.2394

Now, we work through it, and we check out that this all works.2400

So, we have x5; there are no other x5's, so we have just x5 that comes down.2404

3x4...do we have any other x4's?...no, no other x4's, so it is - 3x4.2408

x3: do we have any other x3's?...no, no other x3's, so it is + x3.2414

- 3x2: are they any others?...yes, they cancel each other out, so it is 0x2.2420

- 9x - 11x; that becomes - 20x; let's just knock them out, so we can see what we are doing; + 33 + 27 is + 60.2425

Great; we end up getting what we originally started with; it checks out, so our answer in red is definitely correct.2433

So, remember: that remainder is the one thing where it didn't come out evenly, so it has to be this "divide by,"2442

whatever your remainder is here, divided by the thing you are dividing by,2449

because it is the one thing that didn't come out evenly.2453

All right, the final example: Find all roots of x4 - 2x3 - 11x2 - 8x - 602456

by using the fact that x2 + 4 is one of its factors.2462

The first thing that is going to make this easier for us: if we want to keep breaking this down into factors,2466

if we want to find the answers, if we want to find what it is--we have to factor it, so that we can get to the roots.2469

So, we want to factor this larger thing: we know that we can pull out x2 + 4.2476

If we are going to pull it out, can we use synthetic division?...no, because it is not in the form x - k.2481

We have this x2, so we have to use polynomial long division.2486

So, x2 + 4: we plug in x4 - 2x3 - 11x2 - 8x - 60.2490

Great; x2 + 4 goes into x4...oh, but notice: do we have an x in here?2502

We don't, so it is once again + 0x; so let's rewrite this; it is not just x2 + 4.2509

We can see this as a three-termed thing, where one has actually disappeared; + 0x + 4.2514

Notice that they are the same thing; but it will help us see what we are doing.2520

How many times did x2 go into x4? It goes in x2.2523

But we don't put it here; we put it here, where it would line up for three different terms: 1 term, 2 terms, 3 terms.2525

It lines up on the third term over here, -11x, so it will be x2 here.2534

x2 times (x2 + 0x + 4): we get x4; this is just blank still; + 4x2.2540

Now, we subtract by that; we put our subtraction onto both of our pieces, so now we are adding.2549

x4 - x4 becomes 0; -11x2 - 4x2 becomes -15x2.2554

We bring down our -2x3; we bring down our -8x; so we have everything.2560

-2x3 - 15x2 - 8x: once again, we just ask,2566

"How many times does the first term go into the first term here?"2572

-2x3 divided by x2 gets us just -2x2.2575

Oops, I'm sorry: we are dividing by x2, because -2 is just x1.2582

So, -2 times x2 gets us -2x3; and then, -2x times 4 becomes -8x.2586

We subtract by this; we distribute to start our subtraction, so that becomes positive; that becomes positive.2594

Now we are adding: -2x3 + 2x3 becomes 0.2599

-8x + 8x becomes nothing; and now we bring down the thing that didn't get touched, the -15x2 and the -60.2603

And we have -15x2 - 60; and hopefully, it will line up perfectly.2612

In fact, we know it has to line up perfectly, because we were told explicitly that it is one of the factors.2617

So, there should be no remainder; otherwise, something went wrong.2621

x2 + 0x + 4; how many times does that fit into -15x2 - 60?2624

Once again, we just look at the first part: -15x2 divided by x2 becomes just -15.2628

-15x2...multiplying it out...4 times -15, minus 60; we now subtract by all that.2634

Subtraction distributes and cancels those into plus signs.2640

We add 0 and 0; we have a remainder of 0, which is good; that should just be the case, because we were told it was a factor.2643

So, we get x2 - 2x - 15.2650

What is our polynomial--what is another way of stating this polynomial?2654

We could also say this as (x2 + 4)(x2 - 2x - 15).2658

Let's keep breaking this up: x2 - 2x - 15...how can we factor that?2666

x2 + 4...can we factor that any more?...no, we can't; it is irreducible.2670

If we were to try to set that to 0, we would have to have x2, when squared, become a negative number.2674

There are no real numbers that do that, so that one is irreducible; we are not going to get any roots out of that; no real roots were there.2679

x2 - 2x - 09 how can we factor this?2685

Just 1 is in front of the x2, so that part is easy; it is going to be x and x, and then... what about the next part, -15?2688

We could factor that into...one of them is going to have to be negative;2695

we factor it into 5 and 3; 5 and 3 have a difference of 2, so let's make it -5 and +3.2697

We check that: x2 + 3x - 5x...-2x; -5 times 3 is -15; great.2704

So, at this point, we set everything to 0; x2 + 4 will provide no answers.2710

x2 + 4 = 0...nothing there; there are no answers there.2715

x - 5 = 0; turn this one in--that gets us x = 5; x + 3 = 0 (this gives us all of the roots): x = - 3.2721

Our answers, all of the roots for this, are x = -3 and 5.2732

And we were able to figure this by being able to break down a much more complicated polynomial2737

into something that was manageable, something we can totally factor; and we had to do that through polynomial long division.2741

All right, cool--I hope you got a good idea of how this all works.2746

Just remember: polynomial long division is probably your best choice.2748

Just think of it the same way that you approach just doing normal long division with plain numbers that you did many, many years ago.2751

It is basically the same thing, just with a slightly different format.2758

How many times does it fit in? Multiply; subtract; repeat; repeat; repeat; get to a remainder.2761

All right, we will see you at Educator.com later--goodbye!2766