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INSTRUCTORS Carleen Eaton Grant Fraser
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Dr. Carleen Eaton

Dr. Carleen Eaton

Binomial Theorem

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Table of Contents

I. Equations and Inequalities
Expressions and Formulas

22m 23s

Intro
0:00
Order of Operations
0:19
Variable
0:27
Algebraic Expression
0:46
Term
0:57
Example: Algebraic Expression
1:25
Evaluate Inside Grouping Symbols
1:55
Evaluate Powers
2:30
Multiply/Divide Left to Right
2:55
Add/Subtract Left to Right
3:35
Monomials
4:40
Examples of Monomials
4:52
Constant
5:27
Coefficient
5:46
Degree
6:25
Power
7:15
Polynomials
8:02
Examples of Polynomials
8:24
Binomials, Trinomials, Monomials
8:53
Term
9:21
Like Terms
10:02
Formulas
11:00
Example: Pythagorean Theorem
11:15
Example 1: Evaluate the Algebraic Expression
11:50
Example 2: Evaluate the Algebraic Expression
14:38
Example 3: Area of a Triangle
19:11
Example 4: Fahrenheit to Celsius
20:41
Properties of Real Numbers

20m 15s

Intro
0:00
Real Numbers
0:07
Number Line
0:15
Rational Numbers
0:46
Irrational Numbers
2:24
Venn Diagram of Real Numbers
4:03
Irrational Numbers
5:00
Rational Numbers
5:19
Real Number System
5:27
Natural Numbers
5:32
Whole Numbers
5:53
Integers
6:19
Fractions
6:46
Properties of Real Numbers
7:15
Commutative Property
7:34
Associative Property
8:07
Identity Property
9:04
Inverse Property
9:53
Distributive Property
11:03
Example 1: What Set of Numbers?
12:21
Example 2: What Properties Are Used?
13:56
Example 3: Multiplicative Inverse
16:00
Example 4: Simplify Using Properties
17:18
Solving Equations

19m 10s

Intro
0:00
Translations
0:06
Verbal Expressions and Algebraic Expressions
0:13
Example: Sum of Two Numbers
0:19
Example: Square of a Number
1:33
Properties of Equality
3:20
Reflexive Property
3:30
Symmetric Property
3:42
Transitive Property
4:01
Addition Property
5:01
Subtraction Property
5:37
Multiplication Property
6:02
Division Property
6:30
Solving Equations
6:58
Example: Using Properties
7:18
Solving for a Variable
8:25
Example: Solve for Z
8:34
Example 1: Write Algebraic Expression
10:15
Example 2: Write Verbal Expression
11:31
Example 3: Solve the Equation
14:05
Example 4: Simplify Using Properties
17:26
Solving Absolute Value Equations

17m 31s

Intro
0:00
Absolute Value Expressions
0:09
Distance from Zero
0:18
Example: Absolute Value Expression
0:24
Absolute Value Equations
1:50
Example: Absolute Value Equation
2:00
Example: Isolate Expression
3:13
No Solution
3:46
Empty Set
3:58
Example: No Solution
4:12
Number of Solutions
4:46
Check Each Solution
4:57
Example: Two Solutions
5:05
Example: No Solution
6:18
Example: One Solution
6:28
Example 1: Evaluate for X
7:16
Example 2: Write Verbal Expression
9:08
Example 3: Solve the Equation
12:18
Example 4: Simplify Using Properties
13:36
Solving Inequalities

17m 14s

Intro
0:00
Properties of Inequalities
0:08
Addition Property
0:17
Example: Using Numbers
0:30
Subtraction Property
1:03
Example: Using Numbers
1:19
Multiplication Properties
1:44
C>0 (Positive Number)
1:50
Example: Using Numbers
2:05
C<0 (Negative Number)
2:40
Example: Using Numbers
3:10
Division Properties
4:11
C>0 (Positive Number)
4:15
Example: Using Numbers
4:27
C<0 (Negative Number)
5:21
Example: Using Numbers
5:32
Describing the Solution Set
6:10
Example: Set Builder Notation
6:26
Example: Graph (Closed Circle)
7:08
Example: Graph (Open Circle)
7:30
Example 1: Solve the Inequality
7:58
Example 2: Solve the Inequality
9:06
Example 3: Solve the Inequality
10:10
Example 4: Solve the Inequality
13:12
Solving Compound and Absolute Value Inequalities

25m

Intro
0:00
Compound Inequalities
0:08
And and Or
0:13
Example: And
0:22
Example: Or
1:12
And Inequality
1:41
Intersection
1:49
Example: Numbers
2:08
Example: Inequality
2:43
Or Inequality
4:35
Example: Union
4:45
Example: Inequality
5:53
Absolute Value Inequalities
7:19
Definition of Absolute Value
7:33
Examples: Compound Inequalities
8:30
Example: Complex Inequality
12:21
Example 1: Solve the Inequality
12:54
Example 2: Solve the Inequality
17:21
Example 3: Solve the Inequality
18:54
Example 4: Solve the Inequality
22:15
II. Linear Relations and Functions
Relations and Functions

32m 5s

Intro
0:00
Coordinate Plane
0:20
X-Coordinate and Y-Coordinate
0:30
Example: Coordinate Pairs
0:37
Quadrants
1:20
Relations
2:14
Domain and Range
2:19
Set of Ordered Pairs
2:29
As a Table
2:51
Functions
4:21
One Element in Range
4:32
Example: Mapping
4:43
Example: Table and Map
6:26
One-to-One Functions
8:01
Example: One-to-One
8:22
Example: Not One-to-One
9:18
Graphs of Relations
11:01
Discrete and Continuous
11:12
Example: Discrete
11:22
Example: Continous
12:30
Vertical Line Test
14:09
Example: S Curve
14:29
Example: Function
16:15
Equations, Relations, and Functions
17:03
Independent Variable and Dependent Variable
17:16
Function Notation
19:11
Example: Function Notation
19:23
Example 1: Domain and Range
20:51
Example 2: Discrete or Continous
23:03
Example 3: Discrete or Continous
25:53
Example 4: Function Notation
30:05
Linear Equations

14m 46s

Intro
0:00
Linear Equations and Functions
0:07
Linear Equation
0:19
Example: Linear Equation
0:29
Example: Linear Function
1:07
Standard Form
2:02
Integer Constants with No Common Factor
2:08
Example: Standard Form
2:27
Graphing with Intercepts
4:05
X-Intercept and Y-Intercept
4:12
Example: Intercepts
4:26
Example: Graphing
5:14
Example 1: Linear Function
7:53
Example 2: Linear Function
9:10
Example 3: Standard Form
10:04
Example 4: Graph with Intercepts
12:25
Slope

23m 7s

Intro
0:00
Definition of Slope
0:07
Change in Y / Change in X
0:26
Example: Slope of Graph
0:37
Interpretation of Slope
3:07
Horizontal Line (0 Slope)
3:13
Vertical Line (Undefined Slope)
4:52
Rises to Right (Positive Slope)
6:36
Falls to Right (Negative Slope)
6:53
Parallel Lines
7:18
Example: Not Vertical
7:30
Example: Vertical
7:58
Perpendicular Lines
8:31
Example: Perpendicular
8:42
Example 1: Slope of Line
10:32
Example 2: Graph Line
11:45
Example 3: Parallel to Graph
13:37
Example 4: Perpendicular to Graph
17:57
Writing Linear Functions

23m 5s

Intro
0:00
Slope Intercept Form
0:11
m and b
0:28
Example: Graph Using Slope Intercept
0:43
Point Slope Form
2:41
Relation to Slope Formula
3:03
Example: Point Slope Form
4:36
Parallel and Perpendicular Lines
6:28
Review of Parallel and Perpendicular Lines
6:31
Example: Parallel
7:50
Example: Perpendicular
9:58
Example 1: Slope Intercept Form
11:07
Example 2: Slope Intercept Form
13:07
Example 3: Parallel
15:49
Example 4: Perpendicular
18:42
Special Functions

31m 5s

Intro
0:00
Step Functions
0:07
Example: Apple Prices
0:30
Absolute Value Function
4:55
Example: Absolute Value
5:05
Piecewise Functions
9:08
Example: Piecewise
9:27
Example 1: Absolute Value Function
14:00
Example 2: Absolute Value Function
20:39
Example 3: Piecewise Function
22:26
Example 4: Step Function
25:25
Graphing Inequalities

21m 42s

Intro
0:00
Graphing Linear Inequalities
0:07
Shaded Region
0:19
Using Test Points
0:32
Graph Corresponding Linear Function
0:46
Dashed or Solid Lines
0:59
Use Test Point
1:21
Example: Linear Inequality
1:58
Graphing Absolute Value Inequalities
4:50
Graph Corresponding Equations
4:59
Use Test Point
5:20
Example: Absolute Value Inequality
5:38
Example 1: Linear Inequality
9:17
Example 2: Linear Inequality
11:56
Example 3: Linear Inequality
14:29
Example 4: Absolute Value Inequality
17:06
III. Systems of Equations and Inequalities
Solving Systems of Equations by Graphing

17m 13s

Intro
0:00
Systems of Equations
0:09
Example: Two Equations
0:24
Solving by Graphing
0:53
Point of Intersection
1:09
Types of Systems
2:29
Independent (Single Solution)
2:34
Dependent (Infinite Solutions)
3:05
Inconsistent (No Solution)
4:23
Example 1: Solve by Graphing
5:20
Example 2: Solve by Graphing
9:10
Example 3: Solve by Graphing
12:27
Example 4: Solve by Graphing
14:54
Solving Systems of Equations Algebraically

23m 53s

Intro
0:00
Solving by Substitution
0:08
Example: System of Equations
0:36
Solving by Multiplication
7:22
Extra Step of Multiplying
7:38
Example: System of Equations
8:00
Inconsistent and Dependent Systems
11:14
Variables Drop Out
11:48
Inconsistent System (Never True)
12:01
Constant Equals Constant
12:53
Dependent System (Always True)
13:11
Example 1: Solve Algebraically
13:58
Example 2: Solve Algebraically
15:52
Example 3: Solve Algebraically
17:54
Example 4: Solve Algebraically
21:40
Solving Systems of Inequalities By Graphing

27m 12s

Intro
0:00
Solving by Graphing
0:08
Graph Each Inequality
0:25
Overlap
0:35
Corresponding Linear Equations
1:03
Test Point
1:23
Example: System of Inequalities
1:51
No Solution
7:06
Empty Set
7:26
Example: No Solution
7:34
Example 1: Solve by Graphing
10:27
Example 2: Solve by Graphing
13:30
Example 3: Solve by Graphing
17:19
Example 4: Solve by Graphing
23:23
Solving Systems of Equations in Three Variables

28m 53s

Intro
0:00
Solving Systems in Three Variables
0:17
Triple of Values
0:31
Example: Three Variables
0:56
Number of Solutions
5:55
One Solution
6:08
No Solution
6:24
Infinite Solutions
7:06
Example 1: Solve 3 Variables
7:59
Example 2: Solve 3 Variables
13:50
Example 3: Solve 3 Variables
19:54
Example 4: Solve 3 Variables
25:50
IV. Matrices
Basic Matrix Concepts

11m 34s

Intro
0:00
What is a Matrix
0:26
Brackets
0:46
Designation
1:21
Element
1:47
Matrix Equations
1:59
Dimensions
2:27
Rows (m) and Columns (n)
2:37
Examples: Dimensions
2:43
Special Matrices
4:22
Row Matrix
4:32
Column Matrix
5:00
Zero Matrix
6:00
Equal Matrices
6:30
Example: Corresponding Elements
6:36
Example 1: Matrix Dimension
8:12
Example 2: Matrix Dimension
9:03
Example 3: Zero Matrix
9:38
Example 4: Row and Column Matrix
10:26
Matrix Operations

21m 36s

Intro
0:00
Matrix Addition
0:18
Same Dimensions
0:25
Example: Adding Matrices
1:04
Matrix Subtraction
3:42
Same Dimensions
3:48
Example: Subtracting Matrices
4:04
Scalar Multiplication
6:08
Scalar Constant
6:24
Example: Multiplying Matrices
6:32
Properties of Matrix Operations
8:23
Commutative Property
8:41
Associative Property
9:08
Distributive Property
9:44
Example 1: Matrix Addition
10:24
Example 2: Matrix Subtraction
11:58
Example 3: Scalar Multiplication
14:23
Example 4: Matrix Properties
16:09
Matrix Multiplication

29m 36s

Intro
0:00
Dimension Requirement
0:17
n = p
0:24
Resulting Product Matrix (m x q)
1:21
Example: Multiplication
1:54
Matrix Multiplication
3:38
Example: Matrix Multiplication
4:07
Properties of Matrix Multiplication
10:46
Associative Property
11:00
Associative Property (Scalar)
11:28
Distributive Property
12:06
Distributive Property (Scalar)
12:30
Example 1: Possible Matrices
13:31
Example 2: Multiplying Matrices
17:08
Example 3: Multiplying Matrices
20:41
Example 4: Matrix Properties
24:41
Determinants

33m 13s

Intro
0:00
What is a Determinant
0:13
Square Matrices
0:23
Vertical Bars
0:41
Determinant of a 2x2 Matrix
1:21
Second Order Determinant
1:37
Formula
1:45
Example: 2x2 Determinant
1:58
Determinant of a 3x3 Matrix
2:50
Expansion by Minors
3:08
Third Order Determinant
3:19
Expanding Row One
4:06
Example: 3x3 Determinant
6:40
Diagonal Method for 3x3 Matrices
13:24
Example: Diagonal Method
13:36
Example 1: Determinant of 2x2
18:59
Example 2: Determinant of 3x3
20:03
Example 3: Determinant of 3x3
25:35
Example 4: Determinant of 3x3
29:22
Cramer's Rule

28m 25s

Intro
0:00
System of Two Equations in Two Variables
0:16
One Variable
0:50
Determinant of Denominator
1:14
Determinants of Numerators
2:23
Example: System of Equations
3:34
System of Three Equations in Three Variables
7:06
Determinant of Denominator
7:17
Determinants of Numerators
7:52
Example 1: Two Equations
8:57
Example 2: Two Equations
13:21
Example 3: Three Equations
17:11
Example 4: Three Equations
23:43
Identity and Inverse Matrices

22m 25s

Intro
0:00
Identity Matrix
0:13
Example: 2x2 Identity Matrix
0:30
Example: 4x4 Identity Matrix
0:50
Properties of Identity Matrices
1:24
Example: Multiplying Identity Matrix
2:52
Matrix Inverses
5:30
Writing Matrix Inverse
6:07
Inverse of a 2x2 Matrix
6:39
Example: 2x2 Matrix
7:31
Example 1: Inverse Matrix
10:18
Example 2: Find the Inverse Matrix
13:04
Example 3: Find the Inverse Matrix
17:53
Example 4: Find the Inverse Matrix
20:44
Solving Systems of Equations Using Matrices

22m 32s

Intro
0:00
Matrix Equations
0:11
Example: System of Equations
0:21
Solving Systems of Equations
4:01
Isolate x
4:16
Example: Using Numbers
5:10
Multiplicative Inverse
5:54
Example 1: Write as Matrix Equation
7:18
Example 2: Use Matrix Equations
9:12
Example 3: Use Matrix Equations
15:06
Example 4: Use Matrix Equations
19:35
V. Quadratic Functions and Inequalities
Graphing Quadratic Functions

31m 48s

Intro
0:00
Quadratic Functions
0:12
A is Zero
0:27
Example: Parabola
0:45
Properties of Parabolas
2:08
Axis of Symmetry
2:11
Vertex
2:32
Example: Parabola
2:48
Minimum and Maximum Values
9:02
Positive or Negative
9:28
Upward or Downward
9:58
Example: Minimum
10:31
Example: Maximum
11:16
Example 1: Axis of Symmetry, Vertex, Graph
12:41
Example 2: Axis of Symmetry, Vertex, Graph
17:25
Example 3: Minimum or Maximum
21:47
Example 4: Minimum or Maximum
27:09
Solving Quadratic Equations by Graphing

27m 3s

Intro
0:00
Quadratic Equations
0:16
Standard Form
0:18
Example: Quadratic Equation
0:47
Solving by Graphing
1:41
Roots (x-Intercepts)
1:48
Example: Number of Solutions
2:12
Estimating Solutions
9:23
Example: Integer Solutions
9:30
Example: Estimating
9:53
Example 1: Solve by Graphing
10:52
Example 2: Solve by Graphing
15:10
Example 1: Solve by Graphing
17:50
Example 1: Solve by Graphing
20:54
Solving Quadratic Equations by Factoring

19m 53s

Intro
0:00
Factoring Techniques
0:15
Greatest Common Factor (GCF)
0:37
Difference of Two Squares
1:48
Perfect Square Trinomials
2:30
General Trinomials
3:09
Zero Product Rule
5:22
Example: Zero Product
5:53
Example 1: Solve by Factoring
7:46
Example 1: Solve by Factoring
9:48
Example 1: Solve by Factoring
12:34
Example 1: Solve by Factoring
15:28
Imaginary and Complex Numbers

35m 45s

Intro
0:00
Properties of Square Roots
0:10
Product Property
0:26
Example: Product Property
0:56
Quotient Property
2:17
Example: Quotient Property
2:35
Imaginary Numbers
3:12
Imaginary i
3:51
Examples: Imaginary Number
4:22
Complex Numbers
7:23
Real Part and Imaginary Part
7:33
Examples: Complex Numbers
7:57
Equality
9:37
Example: Equal Complex Numbers
9:52
Addition and Subtraction
10:12
Examples: Adding Complex Numbers
10:25
Complex Plane
13:32
Horizontal Axis (Real)
13:49
Vertical Axis (Imaginary)
13:59
Example: Labeling
14:11
Multiplication
15:57
Example: FOIL Method
16:03
Division
18:37
Complex Conjugates
18:45
Conjugate Pairs
19:10
Example: Dividing Complex Numbers
20:00
Example 1: Simplify Complex Number
24:50
Example 2: Simplify Complex Number
27:56
Example 3: Multiply Complex Numbers
29:27
Example 3: Dividing Complex Numbers
31:48
Completing the Square

27m 11s

Intro
0:00
Square Root Property
0:12
Example: Perfect Square
0:38
Example: Perfect Square Trinomial
3:00
Completing the Square
4:39
Constant Term
4:50
Example: Complete the Square
5:04
Solve Equations
6:42
Add to Both Sides
6:59
Example: Complete the Square
7:07
Equations Where a Not Equal to 1
10:58
Divide by Coefficient
11:08
Example: Complete the Square
11:24
Complex Solutions
14:05
Real and Imaginary
14:14
Example: Complex Solution
14:35
Example 1: Square Root Property
18:31
Example 2: Complete the Square
19:15
Example 3: Complete the Square
20:40
Example 4: Complete the Square
23:56
Quadratic Formula and the Discriminant

22m 48s

Intro
0:00
Quadratic Formula
0:21
Standard Form
0:29
Example: Quadratic Formula
0:57
One Rational Root
3:00
Example: One Root
3:31
Complex Solutions
6:16
Complex Conjugate
6:28
Example: Complex Solution
7:15
Discriminant
9:42
Positive Discriminant
10:03
Perfect Square (Rational)
10:51
Not Perfect Square (2 Irrational)
11:27
Negative Discriminant
12:28
Zero Discriminant
12:57
Example 1: Quadratic Formula
13:50
Example 2: Quadratic Formula
16:03
Example 3: Quadratic Formula
19:00
Example 4: Discriminant
21:33
Analyzing the Graphs of Quadratic Functions

30m 7s

Intro
0:00
Vertex Form
0:12
H and K
0:32
Axis of Symmetry
0:36
Vertex
0:42
Example: Origin
1:00
Example: k = 2
2:12
Example: h = 1
4:27
Significance of Coefficient a
7:13
Example: |a| > 1
7:25
Example: |a| < 1
8:18
Example: |a| > 0
8:51
Example: |a| < 0
9:05
Writing Quadratic Equations in Vertex Form
10:22
Standard Form to Vertex Form
10:35
Example: Standard Form
11:02
Example: a Term Not 1
14:42
Example 1: Vertex Form
19:47
Example 2: Vertex Form
22:09
Example 3: Vertex Form
24:32
Example 4: Vertex Form
28:23
Graphing and Solving Quadratic Inequalities

27m 5s

Intro
0:00
Graphing Quadratic Inequalities
0:11
Test Point
0:18
Example: Quadratic Inequality
0:29
Solving Quadratic Inequalities
3:57
Example: Parameter
4:24
Example 1: Graph Inequality
11:16
Example 2: Solve Inequality
14:27
Example 3: Graph Inequality
19:14
Example 4: Solve Inequality
23:48
VI. Polynomial Functions
Properties of Exponents

19m 29s

Intro
0:00
Simplifying Exponential Expressions
0:09
Monomial Simplest Form
0:19
Negative Exponents
1:07
Examples: Simple
1:34
Properties of Exponents
3:06
Negative Exponents
3:13
Mutliplying Same Base
3:24
Dividing Same Base
3:45
Raising Power to a Power
4:33
Parentheses (Multiplying)
5:11
Parentheses (Dividing)
5:47
Raising to 0th Power
6:15
Example 1: Simplify Exponents
7:59
Example 2: Simplify Exponents
10:41
Example 3: Simplify Exponents
14:11
Example 4: Simplify Exponents
18:04
Operations on Polynomials

13m 27s

Intro
0:00
Adding and Subtracting Polynomials
0:13
Like Terms and Like Monomials
0:23
Examples: Adding Monomials
1:14
Multiplying Polynomials
3:40
Distributive Property
3:44
Example: Monomial by Polynomial
4:06
Example 1: Simplify Polynomials
5:47
Example 2: Simplify Polynomials
6:28
Example 3: Simplify Polynomials
8:38
Example 4: Simplify Polynomials
10:47
Dividing Polynomials

31m 11s

Intro
0:00
Dividing by a Monomial
0:13
Example: Numbers
0:26
Example: Polynomial by a Monomial
1:18
Long Division
2:28
Remainder Term
2:41
Example: Dividing with Numbers
3:04
Example: With Polynomials
5:01
Example: Missing Terms
7:58
Synthetic Division
11:44
Restriction
12:04
Example: Divisor in Form
12:20
Divisor in Synthetic Division
15:54
Example: Coefficient to 1
16:07
Example 1: Divide Polynomials
17:10
Example 2: Divide Polynomials
19:08
Example 3: Synthetic Division
21:42
Example 4: Synthetic Division
25:09
Polynomial Functions

22m 30s

Intro
0:00
Polynomial in One Variable
0:13
Leading Coefficient
0:27
Example: Polynomial
1:18
Degree
1:31
Polynomial Functions
2:57
Example: Function
3:13
Function Values
3:33
Example: Numerical Values
3:53
Example: Algebraic Expressions
5:11
Zeros of Polynomial Functions
5:50
Odd Degree
6:04
Even Degree
7:29
End Behavior
8:28
Even Degrees
9:09
Example: Leading Coefficient +/-
9:23
Odd Degrees
12:51
Example: Leading Coefficient +/-
13:00
Example 1: Degree and Leading Coefficient
15:03
Example 2: Polynomial Function
15:56
Example 3: Polynomial Function
17:34
Example 4: End Behavior
19:53
Analyzing Graphs of Polynomial Functions

33m 29s

Intro
0:00
Graphing Polynomial Functions
0:11
Example: Table and End Behavior
0:39
Location Principle
4:43
Zero Between Two Points
5:03
Example: Location Principle
5:21
Maximum and Minimum Points
8:40
Relative Maximum and Relative Minimum
9:16
Example: Number of Relative Max/Min
11:11
Example 1: Graph Polynomial Function
11:57
Example 2: Graph Polynomial Function
16:19
Example 3: Graph Polynomial Function
23:27
Example 4: Graph Polynomial Function
28:35
Solving Polynomial Functions

21m 10s

Intro
0:00
Factoring Polynomials
0:06
Greatest Common Factor (GCF)
0:25
Difference of Two Squares
1:14
Perfect Square Trinomials
2:07
General Trinomials
2:57
Grouping
4:32
Sum and Difference of Two Cubes
6:03
Examples: Two Cubes
6:14
Quadratic Form
8:22
Example: Quadratic Form
8:44
Example 1: Factor Polynomial
12:03
Example 2: Factor Polynomial
13:54
Example 3: Quadratic Form
15:33
Example 4: Solve Polynomial Function
17:24
Remainder and Factor Theorems

31m 21s

Intro
0:00
Remainder Theorem
0:07
Checking Work
0:22
Dividend and Divisor in Theorem
1:12
Example: f(a)
2:05
Synthetic Substitution
5:43
Example: Polynomial Function
6:15
Factor Theorem
9:54
Example: Numbers
10:16
Example: Confirm Factor
11:27
Factoring Polynomials
14:48
Example: 3rd Degree Polynomial
15:07
Example 1: Remainder Theorem
19:17
Example 2: Other Factors
21:57
Example 3: Remainder Theorem
25:52
Example 4: Other Factors
28:21
Roots and Zeros

31m 27s

Intro
0:00
Number of Roots
0:08
Not Nature of Roots
0:18
Example: Real and Complex Roots
0:25
Descartes' Rule of Signs
2:05
Positive Real Roots
2:21
Example: Positve
2:39
Negative Real Roots
5:44
Example: Negative
6:06
Finding the Roots
9:59
Example: Combination of Real and Complex
10:07
Conjugate Roots
13:18
Example: Conjugate Roots
13:50
Example 1: Solve Polynomial
16:03
Example 2: Solve Polynomial
18:36
Example 3: Possible Combinations
23:13
Example 4: Possible Combinations
27:11
Rational Zero Theorem

31m 16s

Intro
0:00
Equation
0:08
List of Possibilities
0:16
Equation with Constant and Leading Coefficient
1:04
Example: Rational Zero
2:46
Leading Coefficient Equal to One
7:19
Equation with Leading Coefficient of One
7:34
Example: Coefficient Equal to 1
8:45
Finding Rational Zeros
12:58
Division with Remainder Zero
13:32
Example 1: Possible Rational Zeros
14:20
Example 2: Possible Rational Zeros
16:02
Example 3: Possible Rational Zeros
19:58
Example 4: Find All Zeros
22:06
VII. Radical Expressions and Inequalities
Operations on Functions

34m 30s

Intro
0:00
Arithmetic Operations
0:07
Domain
0:16
Intersection
0:24
Denominator is Zero
0:49
Example: Operations
1:02
Composition of Functions
7:18
Notation
7:48
Right to Left
8:18
Example: Composition
8:48
Composition is Not Commutative
17:23
Example: Not Commutative
17:51
Example 1: Function Operations
20:55
Example 2: Function Operations
24:34
Example 3: Compositions
27:51
Example 4: Function Operations
31:09
Inverse Functions and Relations

22m 42s

Intro
0:00
Inverse of a Relation
0:14
Example: Ordered Pairs
0:56
Inverse of a Function
3:24
Domain and Range Switched
3:52
Example: Inverse
4:28
Procedure to Construct an Inverse Function
6:42
f(x) to y
6:42
Interchange x and y
6:59
Solve for y
7:06
Write Inverse f(x) for y
7:14
Example: Inverse Function
7:25
Example: Inverse Function 2
8:48
Inverses and Compositions
10:44
Example: Inverse Composition
11:46
Example 1: Inverse Relation
14:49
Example 2: Inverse of Function
15:40
Example 3: Inverse of Function
17:06
Example 4: Inverse Functions
18:55
Square Root Functions and Inequalities

30m 4s

Intro
0:00
Square Root Functions
0:07
Examples: Square Root Function
0:16
Example: Not Square Root Function
0:46
Radicand
1:12
Example: Restriction
1:31
Graphing Square Root Functions
3:42
Example: Graphing
3:49
Square Root Inequalities
8:47
Same Technique
9:00
Example: Square Root Inequality
9:20
Example 1: Graph Square Root Function
15:19
Example 2: Graph Square Root Function
18:03
Example 3: Graph Square Root Function
22:41
Example 4: Square Root Inequalities
25:37
nth Roots

20m 46s

Intro
0:00
Definition of the nth Root
0:07
Example: 5th Root
0:20
Example: 6th Root
0:51
Principal nth Root
1:39
Example: Principal Roots
2:06
Using Absolute Values
5:58
Example: Square Root
6:18
Example: 6th Root
8:40
Example: Negative
10:15
Example 1: Simplify Radicals
12:23
Example 2: Simplify Radicals
13:29
Example 3: Simplify Radicals
16:07
Example 4: Simplify Radicals
18:18
Operations with Radical Expressions

41m 11s

Intro
0:00
Properties of Radicals
0:16
Quotient Property
0:29
Example: Quotient
1:00
Example: Product Property
1:47
Simplifying Radical Expressions
3:24
Radicand No nth Powers
3:47
Radicand No Fractions
6:33
No Radicals in Denominator
7:16
Rationalizing Denominators
8:27
Example: Radicand nth Power
9:05
Conjugate Radical Expressions
11:47
Conjugates
12:07
Example: Conjugate Radical Expression
13:11
Adding and Subtracting Radicals
16:12
Same Index, Same Radicand
16:20
Example: Like Radicals
16:28
Multiplying Radicals
19:04
Distributive Property
19:10
Example: Multiplying Radicals
19:20
Example 1: Simplify Radical
24:11
Example 2: Simplify Radicals
28:43
Example 3: Simplify Radicals
32:00
Example 4: Simplify Radical
36:34
Rational Exponents

30m 45s

Intro
0:00
Definition 1
0:20
Example: Using Numbers
0:39
Example: Non-Negative
2:46
Example: Odd
3:34
Definition 2
4:32
Restriction
4:52
Example: Relate to Definition 1
5:04
Example: m Not 1
5:31
Simplifying Expressions
7:53
Multiplication
8:31
Division
9:29
Multiply Exponents
10:08
Raised Power
11:05
Zero Power
11:29
Negative Power
11:49
Simplified Form
13:52
Complex Fraction
14:16
Negative Exponents
14:40
Example: More Complicated
15:14
Example 1: Write as Radical
19:03
Example 2: Write with Rational Exponents
20:40
Example 3: Complex Fraction
22:09
Example 4: Complex Fraction
26:22
Solving Radical Equations and Inequalities

31m 27s

Intro
0:00
Radical Equations
0:11
Variables in Radicands
0:22
Example: Radical Equation
1:06
Example: Complex Equation
2:42
Extraneous Roots
7:21
Squaring Technique
7:35
Double Check
7:44
Example: Extraneous
8:21
Eliminating nth Roots
10:04
Isolate and Raise Power
10:14
Example: nth Root
10:27
Radical Inequalities
11:27
Restriction: Index is Even
11:53
Example: Radical Inequality
12:29
Example 1: Solve Radical Equation
15:41
Example 2: Solve Radical Equation
17:44
Example 3: Solve Radical Inequality
20:24
Example 4: Solve Radical Equation
24:34
VIII. Rational Equations and Inequalities
Multiplying and Dividing Rational Expressions

40m 54s

Intro
0:00
Simplifying Rational Expressions
0:22
Algebraic Fraction
0:29
Examples: Rational Expressions
0:49
Example: GCF
1:33
Example: Simplify Rational Expression
2:26
Factoring -1
4:04
Example: Simplify with -1
4:19
Multiplying and Dividing Rational Expressions
6:59
Multiplying and Dividing
7:28
Example: Multiplying Rational Expressions
8:36
Example: Dividing Rational Expressions
11:20
Factoring
14:01
Factoring Polynomials
14:19
Example: Factoring
14:35
Complex Fractions
18:22
Example: Numbers
18:37
Example: Algebraic Complex Fractions
19:25
Example 1: Simplify Rational Expression
25:56
Example 2: Simplify Rational Expression
29:34
Example 3: Simplify Rational Expression
31:39
Example 4: Simplify Rational Expression
37:50
Adding and Subtracting Rational Expressions

55m 4s

Intro
0:00
Least Common Multiple (LCM)
0:27
Examples: LCM of Numbers
0:43
Example: LCM of Polynomials
4:02
Adding and Subtracting
7:55
Least Common Denominator (LCD)
8:07
Example: Numbers
8:17
Example: Rational Expressions
11:03
Equivalent Fractions
15:22
Simplifying Complex Fractions
21:19
Example: Previous Lessons
21:36
Example: More Complex
22:53
Example 1: Find LCM
28:30
Example 2: Add Rational Expressions
31:44
Example 3: Subtract Rational Expressions
39:18
Example 4: Simplify Rational Expression
38:26
Graphing Rational Functions

57m 13s

Intro
0:00
Rational Functions
0:18
Restriction
0:34
Example: Rational Function
0:51
Breaks in Continuity
2:52
Example: Continuous Function
3:10
Discontinuities
3:30
Example: Excluded Values
4:37
Graphs and Discontinuities
5:02
Common Binomial Factor (Hole)
5:08
Example: Common Factor
5:31
Asymptote
10:06
Example: Vertical Asymptote
11:08
Horizontal Asymptotes
20:00
Example: Horizontal Asymptote
20:25
Example 1: Holes and Vertical Asymptotes
26:12
Example 2: Graph Rational Faction
28:35
Example 3: Graph Rational Faction
39:23
Example 4: Graph Rational Faction
47:28
Direct, Joint, and Inverse Variation

20m 21s

Intro
0:00
Direct Variation
0:07
Constant of Variation
0:25
Graph of Constant Variation
1:26
Slope is Constant k
1:35
Example: Straight Lines
1:41
Joint Variation
2:48
Three Variables
2:52
Inverse Variation
3:38
Rewritten Form
3:52
Examples in Biology
4:22
Graph of Inverse Variation
4:51
Asymptotes are Axes
5:12
Example: Inverse Variation
5:40
Proportions
10:11
Direct Variation
10:25
Inverse Variation
11:32
Example 1: Type of Variation
12:42
Example 2: Direct Variation
14:13
Example 3: Joint Variation
16:24
Example 4: Graph Rational Faction
18:50
Solving Rational Equations and Inequalities

55m 14s

Intro
0:00
Rational Equations
0:15
Example: Algebraic Fraction
0:26
Least Common Denominator
0:49
Example: Simple Rational Equation
1:22
Example: Solve Rational Equation
5:40
Extraneous Solutions
9:31
Doublecheck
10:00
No Solution
10:38
Example: Extraneous
10:44
Rational Inequalities
14:01
Excluded Values
14:31
Solve Related Equation
14:49
Find Intervals
14:58
Use Test Values
15:25
Example: Rational Inequality
15:51
Example: Rational Inequality 2
17:07
Example 1: Rational Equation
28:50
Example 2: Rational Equation
33:51
Example 3: Rational Equation
38:19
Example 4: Rational Inequality
46:49
IX. Exponential and Logarithmic Relations
Exponential Functions

35m 58s

Intro
0:00
What is an Exponential Function?
0:12
Restriction on b
0:31
Base
0:46
Example: Exponents as Bases
0:56
Variables as Exponents
1:12
Example: Exponential Function
1:50
Graphing Exponential Functions
2:33
Example: Using Table
2:49
Properties
11:52
Continuous and One to One
12:00
Domain is All Real Numbers
13:14
X-Axis Asymptote
13:55
Y-Intercept
14:02
Reflection Across Y-Axis
14:31
Growth and Decay
15:06
Exponential Growth
15:10
Real Life Examples
15:41
Example: Growth
15:52
Example: Decay
16:12
Real Life Examples
16:30
Equations
17:32
Bases are Same
18:05
Examples: Variables as Exponents
18:20
Inequalities
21:29
Property
21:51
Example: Inequality
22:37
Example 1: Graph Exponential Function
24:05
Example 2: Growth or Decay
27:50
Example 3: Exponential Equation
29:31
Example 4: Exponential Inequality
32:54
Logarithms and Logarithmic Functions

45m 54s

Intro
0:00
What are Logarithms?
0:08
Restrictions
0:15
Written Form
0:26
Logarithms are Exponents
0:52
Example: Logarithms
1:49
Logarithmic Functions
5:14
Same Restrictions
5:30
Inverses
5:53
Example: Logarithmic Function
6:24
Graph of the Logarithmic Function
9:20
Example: Using Table
9:35
Properties
15:09
Continuous and One to One
15:14
Domain
15:36
Range
15:56
Y-Axis is Asymptote
16:02
X Intercept
16:12
Inverse Property
16:57
Compositions of Functions
17:10
Equations
18:30
Example: Logarithmic Equation
19:13
Inequalities
20:36
Properties
20:47
Example: Logarithmic Inequality
21:40
Equations with Logarithms on Both Sides
24:43
Property
24:51
Example: Both Sides
25:23
Inequalities with Logarithms on Both Sides
26:52
Property
27:02
Example: Both Sides
28:05
Example 1: Solve Log Equation
31:52
Example 2: Solve Log Equation
33:53
Example 3: Solve Log Equation
36:15
Example 4: Solve Log Inequality
39:19
Properties of Logarithms

28m 43s

Intro
0:00
Product Property
0:08
Example: Product
0:46
Quotient Property
2:40
Example: Quotient
2:59
Power Property
3:51
Moved Exponent
4:07
Example: Power
4:37
Equations
5:15
Example: Use Properties
5:58
Example 1: Simplify Log
11:17
Example 2: Single Log
15:54
Example 3: Solve Log Equation
18:48
Example 4: Solve Log Equation
22:13
Common Logarithms

25m 23s

Intro
0:00
What are Common Logarithms?
0:10
Real World Applications
0:16
Base Not Written
0:27
Example: Base 10
0:39
Equations
1:47
Example: Same Base
1:56
Example: Different Base
2:37
Inequalities
6:07
Multiplying/Dividing Inequality
6:21
Example: Log Inequality
6:54
Change of Base
12:45
Base 10
13:24
Example: Change of Base
14:05
Example 1: Log Equation
15:21
Example 2: Common Logs
17:13
Example 3: Log Equation
18:22
Example 4: Log Inequality
21:52
Base e and Natural Logarithms

21m 14s

Intro
0:00
Number e
0:09
Natural Base
0:21
Growth/Decay
0:33
Example: Exponential Function
0:53
Natural Logarithms
1:11
ln x
1:19
Inverse and Identity Function
1:39
Example: Inverse Composition
1:55
Equations and Inequalities
4:39
Extraneous Solutions
5:30
Examples: Natural Log Equations
5:48
Example 1: Natural Log Equation
9:08
Example 2: Natural Log Equation
10:37
Example 3: Natural Log Inequality
16:54
Example 4: Natural Log Inequality
18:16
Exponential Growth and Decay

24m 30s

Intro
0:00
Decay
0:17
Decreases by Fixed Percentage
0:23
Rate of Decay
0:56
Example: Finance
1:34
Scientific Model of Decay
3:37
Exponential Decay
3:45
Radioactive Decay
4:13
Example: Half Life
5:33
Growth
9:06
Increases by Fixed Percentage
9:18
Example: Finance
10:09
Scientific Model of Growth
11:35
Population Growth
12:04
Example: Growth
12:20
Example 1: Computer Price
14:00
Example 2: Stock Price
15:46
Example 3: Medicine Disintegration
19:10
Example 4: Population Growth
22:33
X. Conic Sections
Midpoint and Distance Formulas

32m 42s

Intro
0:00
Midpoint Formula
0:15
Example: Midpoint
0:30
Distance Formula
2:30
Example: Distance
2:52
Example 1: Midpoint and Distance
4:58
Example 2: Midpoint and Distance
8:07
Example 3: Median Length
18:51
Example 4: Perimeter and Area
23:36
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
Solving Quadratic Systems

47m 4s

Intro
0:00
Linear Quadratic Systems
0:22
Example: Linear Quadratic System
0:45
Solutions
2:49
Graphs of Possible Solutions
3:10
Quadratic Quadratic System
4:10
Example: Elimination
4:21
Solutions
11:39
Example: 0, 1, 2, 3, 4 Solutions
11:50
Systems of Quadratic Inequalities
12:48
Example: Quadratic Inequality
13:09
Example 1: Solve Quadratic System
21:42
Example 2: Solve Quadratic System
29:13
Example 3: Solve Quadratic System
35:02
Example 4: Solve Quadratic Inequality
40:29
XI. Sequences and Series
Arithmetic Sequences

21m 16s

Intro
0:00
Sequences
0:10
General Form of Sequence
0:16
Example: Finite/Infinite Sequences
0:33
Arithmetic Sequences
0:28
Common Difference
2:41
Example: Arithmetic Sequence
2:50
Formula for the nth Term
3:51
Example: nth Term
4:32
Equation for the nth Term
6:37
Example: Using Formula
6:56
Arithmetic Means
9:47
Example: Arithmetic Means
10:16
Example 1: nth Term
12:38
Example 2: Arithmetic Means
13:49
Example 3: Arithmetic Means
16:12
Example 4: nth Term
18:26
Arithmetic Series

21m 36s

Intro
0:00
What are Arithmetic Series?
0:11
Common Difference
0:28
Example: Arithmetic Sequence
0:43
Example: Arithmetic Series
1:09
Finite/Infinite Series
1:36
Sum of Arithmetic Series
2:27
Example: Sum
3:21
Sigma Notation
5:53
Index
6:14
Example: Sigma Notation
7:14
Example 1: First Term
9:00
Example 2: Three Terms
10:52
Example 3: Sum of Series
14:14
Example 4: Sum of Series
18:13
Geometric Sequences

23m 3s

Intro
0:00
Geometric Sequences
0:11
Common Difference
0:38
Common Ratio
1:08
Example: Geometric Sequence
2:38
nth Term of a Geometric Sequence
4:41
Example: nth Term
4:56
Geometric Means
6:51
Example: Geometric Mean
7:09
Example 1: 9th Term
12:04
Example 2: Geometric Means
15:18
Example 3: nth Term
18:32
Example 4: Three Terms
20:59
Geometric Series

22m 43s

Intro
0:00
What are Geometric Series?
0:11
List of Numbers
0:24
Example: Geometric Series
1:12
Sum of Geometric Series
2:16
Example: Sum of Geometric Series
2:41
Sigma Notation
4:21
Lower Index, Upper Index
4:38
Example: Sigma Notation
4:57
Another Sum Formula
6:08
Example: n Unknown
6:28
Specific Terms
7:41
Sum Formula
7:56
Example: Specific Term
8:11
Example 1: Sum of Geometric Series
10:02
Example 2: Sum of 8 Terms
14:15
Example 3: Sum of Geometric Series
18:23
Example 4: First Term
20:16
Infinite Geometric Series

18m 32s

Intro
0:00
What are Infinite Geometric Series
0:10
Example: Finite
0:29
Example: Infinite
0:51
Partial Sums
1:09
Formula
1:37
Sum of an Infinite Geometric Series
2:39
Convergent Series
2:58
Example: Sum of Convergent Series
3:28
Sigma Notation
7:31
Example: Sigma
8:17
Repeating Decimals
8:42
Example: Repeating Decimal
8:53
Example 1: Sum of Infinite Geometric Series
12:15
Example 2: Repeating Decimal
13:24
Example 3: Sum of Infinite Geometric Series
15:14
Example 4: Repeating Decimal
16:48
Recursion and Special Sequences

14m 34s

Intro
0:00
Fibonacci Sequence
0:05
Background of Fibonacci
0:23
Recursive Formula
0:37
Fibonacci Sequence
0:52
Example: Recursive Formula
2:18
Iteration
3:49
Example: Iteration
4:30
Example 1: Five Terms
7:08
Example 2: Three Terms
9:00
Example 3: Five Terms
10:38
Example 4: Three Iterates
12:41
Binomial Theorem

48m 30s

Intro
0:00
Pascal's Triangle
0:06
Expand Binomial
0:13
Pascal's Triangle
4:26
Properties
6:52
Example: Properties of Binomials
6:58
Factorials
9:11
Product
9:28
Example: Factorial
9:45
Binomial Theorem
11:08
Example: Binomial Theorem
13:48
Finding a Specific Term
18:36
Example: Specific Term
19:26
Example 1: Expand
24:39
Example 2: Fourth Term
30:26
Example 3: Five Terms
36:13
Example 4: Three Iterates
45:07
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Lecture Comments (6)

0 answers

Post by Suhani Pant on July 27, 2013

Thank you so much for this amazing series on Algebra II. I greatly appreciate your hard-work and effort.

1 answer

Last reply by: Dr Carleen Eaton
Mon Apr 8, 2013 4:35 PM

Post by Kenneth Montfort on March 7, 2013

You are super great teacher - thank you so much for making algebra 2 super understandable :) it was an awesome course.

1 answer

Last reply by: Dr Carleen Eaton
Sat Jan 5, 2013 9:55 AM

Post by Angel La Fayette on January 2, 2013

The "2" was left out @ 38:13.

0 answers

Post by Timothy miranda on July 3, 2010

Thanks for all the help! This was a great course!

Binomial Theorem

  • The expansion of (a + b)n has n + 1 terms.
  • In the kth term of (a + b)n , the exponent of b is k – 1 and the exponent of is n – (k – 1).
  • In the kth term, the coefficient has k – 1 factors in the numerator and the denominator.
  • You can always find the coefficients of the expansion or of a particular term by writing out Pascal’s triangle and using row n.
  • The coefficients of the expansion are symmetrical around the middle term. So once you have found the first half of the coefficients, you already have the coefficients for the rest of the terms. They are the same ones you have found but listed in reverse order.

Binomial Theorem

Expand (x + y)5
  • You can expand this problem using several properties of the binomial theorem.
  • Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
  • (x + y)5 = ___(xy) + ___(xy) + ___(xy) + ___(xy) + ____(xy) + ____(xy)
  • Step 2 - Distribute the power n, notice how you start at 5 with x, and you start at 0 with y
  • The sum of the powers of x and y always equal n, in this case 5
  • (x + y)5 = ___(xy) + ___(xy) + ___(xy) + ___(xy) + ____(xy) + ____(xy)
  • (x + y)5 = ___(x5y0) + ___(x4y1) + ___(x3y2) + ___(x2y3) + ____(x1y4) + ____(x0y5)
  • Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
  • (x + y)5 = 1(x5y0) + 5(x4y1) + 10(x3y2) + 10(x2y3) + 5(x1y4) + 1(x0y5)
  • Step 4 : Simplify. Eliminate any powers of zero.
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Expand (2 + y)5
  • You can expand this problem using several properties of the binomial theorem.
  • Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
  • (2 + y)5 = ___(2y) + ___(2y) + ___(2y) + ___(2y) + ____(2y) + ____(2y)
  • Step 2 - Distribute the power n, notice how you start at 5 with 2, and you start at 0 with y
  • The sum of the powers of 2 and y always equal n, in this case 5
  • (2 + y)5 = ___(2y) + ___(2y) + ___(2y) + ___(2y) + ____(2y) + ____(2y)
  • (2 + y)5 = ___(25y0) + ___(24y1) + ___(23y2) + ___(22y3) + ____(21y4) + ____(20y5)
  • Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
  • (2 + y)5 = 1(25y0) + 5(24y1) + 10(23y2) + 10(22y3) + 5(21y4) + 1(20y5)
  • Step 4 : Simplify the powers of 2
  • (2 + y)5 = 1(32y0) + 5(16y1) + 10(8y2) + 10(4y3) + 5(2y4) + 1(1y5)
  • Step 5 - Simplify completely
  • (2 + y)5 = 32 + 80y + 80y2 + 40y3 + 10y4 + y5
  • or (2 + y)5 = y5 + 10y4 + 40y3 + 80y2 + 80y + 32
(2 + y)5 = y5 + 10y4 + 40y3 + 80y2 + 80y + 32
Expand (2x + 2y)5
  • You can expand this problem using several properties of the binomial theorem.
  • Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
  • (2x + 2y)5 = ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ____(( 2x )( 2y )) + ____(( 2x )( 2y ))
  • Step 2 - Distribute the power n, notice how you start at 5 with (2x), and you start at 0 with (2y)
  • The sum of the powers of (2x) and (2y) always equal n, in this case 5
  • (2x + 2y)5 = ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ____(( 2x )( 2y )) + ____(( 2x )( 2y ))
  • (2x + 2y)5 = ___(( 2x )5( 2y )0) + ___(( 2x )4( 2y )1) + ___(( 2x )3( 2y )2) + ___(( 2x )2( 2y )3) + ____(( 2x )1( 2y )4) + ____(( 2x )0( 2y )5)
  • Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
  • (2x + 2y)5 = 1(( 2x )5( 2y )0) + 5(( 2x )4( 2y )1) + 10(( 2x )3( 2y )2) + 10(( 2x )2( 2y )3) + _5(( 2x )1( 2y )4) + 1(( 2x )0( 2y )5)
  • Step 4 : Distribute the powers using properties of exponents.
  • (2x + 2y)5 = 1(( 25x5 )( 20y0 )) + 5(( 24x4 )( 2y )) + 10(( 23x3 )( 22y2 )) + 10(( 22x2 )( 23y3 )) + _5(( 2x )( 24y4 )) + 1(( 20x0 )( 25y5 ))
  • Step 5 - Simplify the powers of 2
  • (2x + 2y)5 = 1(( 32x5 )( 1y0 )) + 5(( 16x4 )( 2y )) + 10(( 8x3 )( 4y2 )) + 10(( 4x2 )( 8y3 )) + _5(( 2x )( 16y4 )) + 1(( 1x0 )( 32y5 ))
  • Step 6 - Multiply the coefficients of Pascal's Triangle, with the coefficients of x and y.
(2x + 2y)5 = 32x5 + 160x4y + 320x3y2 + 320x2y3 + 160xy4 + 32y5
Find the 6th term of (5x + 2y)7
  • Recall that the binomial theorem is defined as
  • (a + b)n = ∑k = 0n (
    n
    k
    )an − kbk; (
    n
    k
    ) = [n!/(k!(n − k)!)]
  • This particular problem would then be defined as
  • (5x + 2y)7 = ∑k = 07 (
    7
    k
    )(5x)7 − k(2y)k; where (
    7k
    ) = [7!/(k!(7 − k)!)]
  • To find any term in the binomial expansion, all you need to solve is for k.
  • The relationship between Term# and k is the following
  • Term=k+1 Term# = k + 1
  • 6 = k + 1
  • k = 5
  • Use k = 5, to find the 6th term
  • (
    7
    5
    )(5x)7 − 5(2y)5 = (
    7
    5
    )(5x)2(2y)5 = [7!/(5!(7 − 5)!)](5x)2(2y)5 = [7*6*5!/5!(2)!](5x)2(2y)5 =
  • [7*6*5!/5!(2)!](5x)2(2y)5 = [7*6*/(2)!](5x)2(2y)5 = [42/2](5x)2(2y)5 = 21(25x2)(32y5)
The sixth term of (5x + 2y)7 = 16800x2y5
Find the 30th term of (x + y)50
  • Recall that the binomial theorem is defined as
  • (a + b)n = ∑k = 0n (
    n
    k
    )an − kbk; (
    n
    k
    ) = [n!/(k!(n − k)!)]
  • This particular problem would then be defined as
  • (x + y)50 = ∑k = 050 (
    50
    k
    )x50 − kyk; where (
    50
    k
    ) = [50!/(k!(50 − k)!)]
  • To find any term in the binomial expansion, all you need to solve is for k.
  • The relationship between Term# and k is the following
  • Term=k+1
  • Term# = k + 1
  • 30 = k + 1
  • k = 29
  • Use k = 29, to find the 30th term
  • (
    50
    29
    )(x)50 − 29(y)29 = (
    50
    29
    )(x)21(y)29 = [50!/(29!(50 − 29)!)](x)21(y)29 = [50!/29!(21)!](x)21(y)29 =
  • [50*49*48...28*29!/29!(21)!](x)21(y)29 = [50*49*48...28/(21)!](x)21(y)29 = [50*49*48...28/(21)!](x)21(y)29 =
The 30th term of (x + y)50 = 67,327,446,062,800x21y29
Find the 4th term of (6x + y)7
  • Recall that the binomial theorem is defined as
  • (a + b)n = ∑k = 0n (
    n
    k
    )an − kbk; (
    n
    k
    ) = [n!/(k!(n − k)!)]
  • This particular problem would then be defined as
  • (6x + y)7 = ∑k = 07 (
    7
    k
    )( 6x )7 − kyk; where (
    7
    k
    ) = [7!/(k!(7 − k)!)]
  • To find any term in the binomial expansion, all you need to solve is for k.
  • The relationship between Term# and k is the following
  • Term# = k + 1
  • 4 = k + 1
  • k = 3
  • Use k = 3, to find the 4th term
  • (
    7
    3
    )(6x)7 − 3(y)3 = (
    7
    3
    )(6x)4(y)3 = [7!/(3!(7 − 3)!)](6x)4(y)3 = [7!/3!(4)!](6x)4(y)3 =
  • [7*6*5*4!/3!(4)!](6x)4(y)3 = [7*6*5*/3!](6x)4(y)3 = 35(6x)4y3 = 35(64x4)y3 = 35(1296x4)y3 = 45360x4y3
The 4th term of (6x + y)7 = 45360x4y3
Find the 4th term of (2x − 4y)7
  • Recall that the binomial theorem is defined as
  • (a + b)n = ∑k = 0n (
    n
    k
    )an − kbk; (
    n
    k
    ) = [n!/(k!(n − k)!)]
  • This particular problem would then be defined as
  • (2x + ( − 4y))7 = ∑k = 07 (
    7
    k
    )( 2x )7 − k( − 4y)k; where (
    7
    k
    ) = [7!/(k!(7 − k)!)]
  • To find any term in the binomial expansion, all you need to solve is for k.
  • The relationship between Term# and k is the following
  • Term# = k + 1
  • 4 = k + 1
  • k = 3
  • Use k = 3, to find the 4th term
  • (
    7
    3
    )(2x)7 − 3( − 4y)3 = (
    7
    3
    )(2x)4( − 4y)3 = [7!/(3!(7 − 3)!)](2x)4( − 4y)3 = [7!/3!(4)!](2x)4( − 4y)3 =
  • [7*6*5*4!/3!(4)!](2x)4( − 4y)3 = [7*6*5*/3!](2x)4( − 4y)3 = 35(2x)4( − 4y)3 = 35(24x4)( − 4)3y3 = 35(16x4)( − 64)y3 = − 35840x4y3
The 4th term of (2x − 4y)7 = − 35840x4y3
Find the 4th term of (2x3 − 4y3)7
  • Recall that the binomial theorem is defined as
  • (a + b)n = ∑k = 0n (
    n
    k
    )an − kbk; (
    n
    k
    ) = [n!/(k!(n − k)!)]
  • This particular problem would then be defined as
  • (2x3 + ( − 4y3))7 = ∑k = 07 (
    7
    k
    )( 2x3 )7 − k( − 4y3)k; where (
    7
    k
    ) = [7!/(k!(7 − k)!)]
  • To find any term in the binomial expansion, all you need to solve is for k.
  • The relationship between Term# and k is the following
  • Term=k+1
  • Term# = k + 1
  • 4 = k + 1
  • k = 3
  • Use k = 3, to find the 4th term
  • (
    7
    3
    )(2x3)7 − 3( − 4y)3 = (
    7
    3
    )(2x)4( − 4y)3 = [7!/(3!(7 − 3)!)](2x3)4( − 4y3)3 = [7!/3!(4)!](2x3)4( − 4y3)3 =
  • [7*6*5*4!/3!(4)!](2x3)4( − 4y3)3 = [7*6*5*/3!](2x3)4( − 4y3)3 = 35(24x12)( − 43)(y9) = 35(16x12)( − 64)(y9) = − 35840x12y9
The 4th term of (2x3 − 4y3)7 = − 35840x12y9
Find the 4th term of (x2 + y3)14
  • Recall that the binomial theorem is defined as
  • (a + b)n = ∑k = 0n (
    n
    k
    )an − kbk; (
    n
    k
    ) = [n!/(k!(n − k)!)]
  • This particular problem would then be defined as
  • (x2 + y3)14 = ∑k = 014 (
    14
    k
    )( x2 )14 − k(y3)k; where (
    14
    k
    ) = [14!/(k!(14 − k)!)]
  • To find any term in the binomial expansion, all you need to solve is for k.
  • The relationship between Term# and k is the following
  • Term# = k + 1
  • 4 = k + 1
  • k = 3
  • Use k = 3, to find the 4th term
  • (
    14
    3
    )(x2)14 − 3(y3)3 = (
    14
    3
    )(x2)11(y3)3 = [14!/(3!(14 − 3)!)](x2)11(y3)3 = [14!/3!(11)!](x22)(y9)
  • [14!/3!(11)!](x22)(y9) = [14*13*12*/3!](x22)(y9) = 364x22y9
The 4th term of (x2 + y3)14 = 364x22y9
Find the 10th term of (x2 + y3)14
  • Recall that the binomial theorem is defined as
  • (a + b)n = ∑k = 0n (
    n
    k
    )an − kbk; (
    nk
    ) = [n!/(k!(n − k)!)]
  • This particular problem would then be defined as
  • (x2 + y3)14 = ∑k = 014 (
    14
    k
    )( x2 )14 − k(y3)k; where (
    14
    k
    ) = [14!/(k!(14 − k)!)]
  • To find any term in the binomial expansion, all you need to solve is for k.
  • The relationship between Term# and k is the following
  • Term=k+1
  • Term# = k + 1
  • 10 = k + 1
  • k = 9
  • Use k = 9, to find the 10th term
  • (
    14
    9
    )(x2)14 − 9(y3)9 = (
    14
    9
    )(x2)5(y3)9 = [14!/(9!(14 − 9)!)](x2)5(y3)9 = [14!/9!(5)!](x10)(y27)
  • [14!/9!(5)!](x10)(y27) = [14*13*12*11*10*/(5)!](x22)(y9) = 2002x10y27
The 10th term of (x2 + y3)14 = 2002x10y27

*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

Binomial Theorem

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
  • Pascal's Triangle 0:06
    • Expand Binomial
    • Pascal's Triangle
  • Properties 6:52
    • Example: Properties of Binomials
  • Factorials 9:11
    • Product
    • Example: Factorial
  • Binomial Theorem 11:08
    • Example: Binomial Theorem
  • Finding a Specific Term 18:36
    • Example: Specific Term
  • Example 1: Expand 24:39
  • Example 2: Fourth Term 30:26
  • Example 3: Five Terms 36:13
  • Example 4: Three Iterates 45:07

Transcription: Binomial Theorem

Welcome to Educator.com.0000

In today's lesson, we are going to be covering the binomial theorem.0002

We are going to start out by talking about what we mean when we say we are going to expand a binomial.0007

So, if you expand a binomial, this is what happens.0013

I have (a + b)n; and I am going to expand that for n = 1, 2, 3, 4.0019

I am going to let n equal 1; and I am going to then get (a + b)1, which is just a + b.0027

If I let n equal 2, then I have(a + b)2; that is going to give me, if you will recall, a2 + 2ab + b2.0041

Let n equal 3: you get (a + b)3, which is a little bit more complicated to work out.0057

But if you figure that all out, (a + b)(a + b)(a + b), you would end up with a3 + 3a2b + 3ab2 + b3.0064

Letting n equal 4, (a + b)4 is going to give you a4 + 4a3b +0079

6a2b2 + 4ab3 + b4.0091

One more: let n equal 5: this is going to give me (a + b)5.0099

This equals a5 + 5a4b + 10a3b2 + 10a2b...0108

actually, this is going to be b3...+ 5ab4 + b5.0127

All right, before we go on to talk about Pascal's triangle, let's note a few things about this expansion.0136

And these will be summed up on the next slide.0144

A few things that you will notice are that the first term is an, and the last term is bn.0146

The second thing is that the number of terms equals n + 1.0157

You will also notice a pattern: let's look at n = 4, (a + b)4.0179

The first term is an, so it is a4; in each subsequent term, the power that a is raised to (the exponent) decreases by 1.0185

So, I started out with a4; then here it's a3; a2; a; and then a is gone.0195

So, the exponent for a decreases by 1 for each term; b does the opposite.0203

There is no b here; this would be b0...it is 1; and then in my next term, I get b to the first power, b2, b3, b4.0213

So, with each term, exponents for a are decreasing; exponents for b are increasing.0223

The next thing that you might notice is the coefficients; let's look at the coefficients.0230

The coefficient of the second term is equal to n; again, this is all summed up on the next slide, but this is just to introduce it here.0234

So, if I am looking at n = 3 (that is (a + b)3), and I expand that; then I get that the second term has a coefficient of 3.0242

The second term here has a coefficient of 4; the second term here has a coefficient of 5.0252

Pascal's triangle--we are getting to that now.0258

I can take the coefficient of these terms and use them to create an array that is called Pascal's triangle.0260

Looking at it, it actually starts out with 1; so if we let n equal 0, that is just going to give me (a + b)0, which is going to be 1.0266

Then, in my next row, I am going to have 1, 1, and 1; those are my coefficients.0277

In the next row, they are 1, 2, 1; in the next row, 1, 3, 3, 1.0289

Now, I can look up here to get the next row; but I don't even need to, because each number is the sum of the two numbers just above it.0302

So, 1 + 1 forms 2; 1 + 2 gives me 3; 2 + 1 gives me 3.0314

So, all I have to do, actually--these outside numbers are always 1; but I could just say 1 + 3...this is going to be 4.0325

3 + 3 is equal to 6; 3 + 1...that is equal to 4.0334

So, I look up here to verify that; and indeed, it is 1, 4, 6, 4 1.0340

In the next row, I would add 1, and I add 1 + 4; I am going to get 5; 4 + 6 gives me 10;0345

6 + 4 gives me 10; 4 + 1 gives me 5; and again, I have a 1 out here.0352

And that matches up with what I have for the expansion of (a + b)5.0357

So, Pascal's triangle is an interesting array that comes from the coefficients of the expansion of this binomial.0363

Something else to note is the symmetry around the middle term; you might have seen that up here, but it is even more obvious right here.0371

The middle term is 2; it is flanked by 1's; here I have two middle terms, because there is an even number of terms;0377

so the two middle terms are 3; they are flanked by 1's; now I have a middle term of 6; next to it are 4's; outside of that, 1's.0384

Two middle terms are 10; the next terms are 5; the next terms are 1.0392

And that is helpful, because if you are looking for the coefficients, you don't need to find all of them; you only need to find half of them.0397

And then, you can use this symmetry to know what the other half are.0403

Summing up these properties of the expansion of (a + b)n: there are n + 1 terms.0409

Let's use an example that we just talked about, (a + b)5, which gave me a5 +0418

5a4b + 10a3b2 + 10a2b3 +0427

5ab4 + b4, to illustrate this.0437

There are n + 1 terms; so here, n = 5; therefore, I am going to have n + 1 terms, equals 5 + 1, so I am going to have 6 terms.0442

And I do have 1, 2, 3, 4, 5, 6 terms.0455

The first term is going to be an (that is a5), and the last term is bn, which should be b5 right here.0458

In successive terms, the exponent of a decreases by 1, while that of b increases by 1.0470

I start out with a5, then a4, to the third, to the second, and then a.0475

b goes the opposite way: I start with b, b2, b3, b4, and b5.0482

I pointed out before that the second term has a coefficient that is equal to n.0490

Something I also pointed out in the last slide is that there is symmetry around the middle term or terms,0498

which tells me that, if I figure out these middle coefficients, well, I already know that the first and last coefficients are 1.0504

And I know that the second (and, by symmetry, second-to-last) coefficients are equal to n.0511

So then, all I would have to figure out is one of these; and I know that the other one is the same.0519

Something else to be aware of is that the sum of the exponents in each term is n.0523

Let's look right here: for this term, 10a3b2, if I add up 3 and 2, that is going to give me 5.0528

So, the sum of the exponents is equal to 5.0537

Or looking at this, 10a2b3, I get the same thing.0540

The sum of the exponents of each term is equal to n.0544

The next concept that we will learn requires that you remember how to work with factorials.0552

So, I am going to go ahead and review factorials: factorials are products, and the symbol that looks like an exclamation point is actually read as "factorial."0557

This is a product: n! is equal to n times n - 1 times n - 2 times n -3...and you continue on that way until you get all the way to 2, and then finally 1.0567

For example, 4 factorial: here, n = 4; this is going to be equal to 4, so here n = 4; n - 1...4 - 1 is 3, so that is 4 times 3.0586

4 - 2 gives me 2; 4 - 3 gives me 1; and this equals...4 times 3 is 12, times 2, times 1, is just 24.0599

So, 4! is just the product of 4 times 3 times 2 times 1.0613

If you are working with fractions that involve factorials in the numerator and denominator, you can often do a lot of canceling out to make things easier.0618

If you were working with 8!/6!,that gives you 8 times 7 times 6 times 5 times 4, and then on down to 1.0625

In the denominator, you have 6 times 5 times 4 times 3 times 2 times 1.0637

So, instead of multiplying all of this out and dividing, it makes a lot more sense to start canceling out all of these common terms.0644

1 through 6; 1 through 6; this just leaves me with 8 times 7, or 56.0657

Now, the reason we reviewed it is to work with this formula.0665

First, we are talking about the binomial theorem, and then the binomial formula.0668

The binomial theorem is really what we just discussed; and it is the idea that, when you expand (a + b)n,0672

you are going to get an equation with the properties we just discussed.0682

And those properties are that the first term is going to be an;0688

the second term is going to have some coefficient; and it is going to be raised to the power an - 1.0692

b is going to be to the first power; then you are going to get another term; and you are going to have a total of n terms.0699

You are going to have another term; it is going to have a coefficient that is going to be an - 2.0707

b is going b2; it is going to increase by 1.0711

Then, you are going to get another coefficient, an - 3; b is going to become b3.0714

And that is going to go on and continue on until you get abn - 1, bn, and so on.0721

Excuse me, plus...that is an - 1, times b, plus bn; the last term will be bn; the first term is an.0731

Each of these terms can be given in the general form an - kbk.0751

And we will look at some examples in a few minutes, and talk about what this means, as we relate it to the binomial formula.0760

This is the binomial formula; this is pronounced "n choose k."0769

Now, why are we even looking at this--what is this for?0781

Think about how we can find these coefficients.0785

We can use Pascal's triangle, but that could be really impractical and quite a bit of work,0788

if you have to go through that whole large array, if you have a really big expansion.0793

And if you just find the coefficient for a particular term, you are going to have to go through a lot of work to get there.0799

So, as usual, we have a formula that gets us directly there, without having to go through all that work.0804

This formula will give me the coefficients for a term that has the exponent an - kbk.0812

So, the binomial formula allows me to find this coefficient.0823

Let's use an example of (a + b)7: I can find the coefficient, using this, of a particular term;0829

or I can expand the whole thing and find all of the terms.0839

Let's look at what this expansion would look like: it would look like this.0841

I know that the first term is going to be a7; recall that the second coefficient is going to be equal to n.0845

So this one I also know; it is 7; and then, I know that a...the exponent is going to decrease by 1; and b is going to appear, raised to the first power.0851

I also know that I am going to have 8 terms, because I am going to have n + 1 terms, or 7 + 1.0866

So, I have another term here; I don't know the coefficient, but I do know that it is going to be a5b2.0872

I have another term with an unknown coefficient: a4b3.0878

I have another term, again with an unknown coefficient: a3b4.0884

Another term is a2b5; another term is ab6; plus b7.0890

I have 8 terms, so I am going to have these two middle terms the same; these two terms will be the same;0900

the second-to-last terms will be the same; so I know this is 7; and these two outer terms will be the same, 1 and 1.0907

I may be asked to find this expansion; I found this much; now, in order to finish it out, I need to find this coefficient and this coefficient.0917

If I have those, I have the rest, because I would have the first half, and then I just reflect for the second half.0927

Let's use this formula to find the coefficient for this third term.0935

In order to use this formula, I need to know two things: I need to know n, and I need to know k.0950

n is easy--we know n equals 7; k...if you look up here, k is the exponent of b for that term.0956

So, I am looking for the coefficient that goes right here, where n is 7 and k is 2.0969

Once I have that, it is a matter of using the formula: n!...this is going to be 7!/k!, which is 2!...times (7 - 2)!;0978

this equals 7! divided by 2!, times 5!.0995

I am going down here to complete this: 7 times 6 times 5 times 4 times 3, and so on.1002

In the denominator, I have 2!, 2 times 1, times 5 times 4 times 3 times 2 times 1 (factorials).1010

Again, I am going to cancel out common factors to make this much easier to work with.1023

This leaves me with 7 times 6, divided by 2; that is simply 42/2, or 21.1030

Therefore, the coefficient for the third term is actually 21.1043

Since I know that, and I know that we are going 1, 7, 21, starting from this end1052

(we are also going 1, 7, 21), I could do the same thing to find the fourth term coefficient.1057

The fourth term coefficient would have, again, an n equal to 7; but this time, k would be equal to 3.1067

Note one thing: instead of having to write all this out, to figure out what k is and everything,1077

you can just be aware that the term number equals k + 1.1082

Let's look at the fourth term: the term number is 4--that equals k + 1; k = 3.1092

For the fourth term, k = 3; so that is 1, 2, 3, 4; I know n = 7; and since this is the fourth term, I quickly know that k = 3.1101

So, you can use the binomial theorem, and specifically the binomial formula, to find a specific term of the expanded form of (a + b)n.1117

And I talked, in the last section, about how you can use that formula to find the whole expansion.1126

Or you may just need to find a certain coefficient within the expansion.1133

Recall that the binomial formula is equal to n!, divided by k!, times (n - k)!, where n is this n,1140

and k is the value of the exponent for the b in the term that you are looking for the coefficient of.1158

As an example, we are going to do something slightly more complex.1167

Instead of just a + b, we are actually going to say a - 2b; and we are going to expand that--it is going to be to the fifth.1169

So, we have to account for this -2--that the second term here is not just b; it is -2b.1178

I know that the first term is going to be a5.1188

I know that the second term is going to have a coefficient of n, so it is a coefficient of 5, and that a is going to decrease by 1.1193

I have to be very careful and not just write b; I have to write -2b.1200

Then, I have an unknown coefficient; but I do know that I am going to get a3, because the exponent that a is raised to will decrease by 1.1206

And here, I just have to say that b is really -2b; that is going to be squared; it is going to increase by 1.1216

I have another unknown coefficient, and it becomes a3 and goes down to a2; -2b squared goes up to -2b cubed.1226

Then, there is another coefficient here, and I get a3, -2b to the fourth, and then finally...1237

actually, this is just a; it goes from a2 to a; and then finally, a is gone, and I have -2b5.1247

By symmetry, I know the coefficient of these outer terms, right here, is actually 1; you can think of that as 1.1256

The two outer terms, then, have coefficients of 5.1266

All right, so back to finding a specific term: let's say that I want to find the coefficient for this third term.1270

n is going to equal 5; so in order to find the coefficient for this third term, I need to use the binomial formula; I need to have n, and I need to have k.1298

k...I could just say it is one less than the term number, so k is 2.1307

Or I could look up here and say it is whatever power this is raised to, which is 2.1311

Next, I insert these values into the formula; that is going to give me 5!, divided by 2!, times (n - 2)!.1318

This is equal to 5! divided by 2!, times 3!, equals 5 times 4 times 3 times 2 times 1, divided by 2! (2 times 1), times 3 times 2 times 1.1334

Cancel out common factors; that leaves me with 5 times 4, divided by 2 times 1, or 20 divided by 2, is 10.1354

Therefore, coefficient for the third term is equal to 10.1368

I can go back up and put this in here, but there is some simplifying that needs to be done.1374

So, the third term equals 10a3, (-2b)2; this equals 10a3...-2 times -2 is 4b2.1379

This equals 10 times 4, a3, b2, which equals 40a3b2.1398

It is a little bit more complicated than just putting the 10 in there.1408

And then, something to be careful of, as well: then I can say that by symmetry, this is also 10.1411

But I need to work with what is in here, which is actually a2(-2b)3, so it is somewhat different.1416

So, let's look at what this fourth term is: 1, 2, 3,...the fourth term.1425

Again, I am going to use that coefficient, 10, but here I have a2(-2b)3.1431

So, it is 10a2...-2 times -2 is 4, times -2 is -8; so it is times -8b3.1439

I can rewrite this as 10 times -8a2b3 = -80a2b3.1454

When I simplify that fourth term, this is what it is going to look like.1466

That was using the binomial formula to find a specific term of the expanded form.1471

In the first example, we need to expand (x + 2)4.1481

The general form is (a + b)n; here, I do the same thing, just realizing that a is equal to x, and b is equal to 2.1486

I am going to get something that looks like this: (a + 2)4; I am going to have five terms.1497

I am going to have n equal 4; so, there are going to be 5 terms.1502

The first term is going to be a2; in this case, it is x, so it is x2.1508

Excuse me, not squared; to the n--it is going to be a to the n, which is 4, so it is actually going to be x4.1517

In the next term, I have the coefficient; and the coefficient of the second term is equal to n.1525

So, that is going to be 4, times x...this exponent is going to decrease by 1...times 2.1531

The next term: I am going to have some unknown coefficient; x is going to go from cubed to squared;1543

2 is going to go from being the first power to the second power.1555

I am going to have another term, and then x2 is just going to become x; 2 is going to go from being squared to being cubed;1560

and then, finally, x drops out, and I get, for my last term, 2n, or 24.1572

By symmetry, I know that I am going to have a middle term (I have an odd number of terms--I have one middle term);1581

the two next to it are going to have the same coefficient--it is going to be 4.1587

I can simplify this later on, because I am going to have to deal with this 23 and this 2, and multiply that times the coefficient.1591

But right now, let's just leave it like this.1600

I am almost done with my expansion; for this middle term, there are two ways I could go, because this is not that large of an expansion.1603

I could use the binomial formula, and we are going to get more practice with that in a minute.1612

But this time, I am actually going to use Pascal's triangle.1615

Remember that it starts out with (a + b)0, and you end up with 1 as the only coefficient, because this equals 1.1617

Then, I get (a + b)1 = (a + b); that gives me 1 and 1 as coefficients.1632

Once I have those two, I can find the rest, because I know I have 1's out here.1639

And to find the other terms, I add the two above: 1 + 1 gives me 2.1645

So, for (a + b)2, these are my coefficients; then I want (a + b)3.1651

And what I am finally looking for, to find this coefficient, is when (a + b) is expanded to the fourth.1659

So, I have a 1 out here and a 1 out here; 1 + 2 gives me 3; 2 + 1 gives me 3;1667

so the coefficients for this expansion (that is going to have 4 terms) are going to be 1, 3, 3, 1.1674

Finally, the one I am looking for: I have a 1 and a 1, which I have up here;1680

1 + 3 is 4; 1 + 3 is 4, which I already knew; this gives me my middle coefficient of 6.1685

Now, I could have used the binomial formula to find this; but I wanted1695

to just do it a little bit differently this time, since it was a fairly small expansion, and use Pascal's triangle.1699

Now, I have the expansion, but it is not really simplified; we need to simplify these various terms.1706

So, let's take...the first term is already simplified; the second term equals (here is the second term) 4 times x3 times 2; that equals 8x3.1712

The third term equals (that is 1, 2...the third term) 6x2, 22, equals 6x2 times 4, equals 24x2.1733

That is the third term; the fourth term equals 4x times 2 cubed; that is 4x times 8; 8 times 4 is 32, so that is 32x.1750

The fifth term equals 24; 23 is 8, times 2 again gives me 16.1764

Rewriting it all here, the expansion of (x + 2)4 is1774

x4 + (second term) 8x3 + (third term) 24x2 + (fourth term) 42x + (the final term) 16.1779

Again, I was able to expand this using my properties that I am aware of for the expansion of a binomial1794

to find what these exponents are, how many terms there are going to be, and what this coefficient and this coefficient are.1802

I was left with just finding one coefficient that I could have used Pascal's triangle for.1810

That is what I did; I found the 6 (or I could have used the binomial formula).1814

Once I had these coefficients, all I did was to simplify; and this is the expansion, right down here.1818

Here, I am asked to find the fourth term of the expansion of (a + 2b)8.1829

So, I am asked to find this specific term; and we know that we can use the binomial formula, n!/k! times (n - k)!.1834

I know that I have n; n = 4; what is k?1847

Well, the term number equals k + 1; the term number is 4, so this means 4 = k + 1.1853

This gives me k = 3; so I have n = 4; k = 3; I can find the coefficient.1866

I need to find the coefficient of this fourth term; it is going to have a coefficient.1875

For the first term, it is going to be a raised to some power; for the second component of it, it is going to be 2b (not just b), raised to some power.1884

How do I know what these powers are?1902

Well, recall that it is going to be equal to (let's give ourselves some more room)...1904

when you find a term used in a binomial formula, you also know that it is in the form n - k,1910

and then the second part is raised to the k power, because that is what k is--1919

it is whatever you are raising b to; and in this case, b = 2b.1926

I also know that the sum of the exponents has to equal n; the sum of the exponents of a term equals n.1934

This makes sense, because n - k + k needs to equal n; the sum of the exponents must equal n, and that works out; so this makes sense.1945

So, the fourth term: I need to find the coefficient using the binomial formula.1954

I know what the power a is going to be raised to is, n - k; and I know that the power that 2b is going to be raised to is k.1960

Let's go ahead and find the coefficient: the coefficient of the fourth term is going to be equal to n!, which is 4!...1967

excuse me, n is actually 8--a correction on that: we are finding the fourth term, but n is right here, so n is actually 8.1981

So, it is n! (that is 8!), divided by k! (that is 3!); n (which is 8), minus 3, factorial, equals 8!/3! times 5!.1993

That gives me 8! divided by 3 times 2 times 1, times 5!.2013

I do my canceling to make things easier to work with.2026

This leaves me with 8 times 7 times 6, divided by 3 times 2 times 1.2032

6 is equal to 3 times 2, so I have more common factors and more canceling.2040

That just leaves me with 8 times 7, divided by 1, which is 56; that was the hardest part here--finding the coefficient.2045

The coefficient of this fourth term is equal to 56.2053

To write that out completely, I am going to write it out as 56an - k; that is 8 - 3;2058

times (2b) raised to the k power (k is 3); this equals 56a5(2b)3.2071

Simplifying further: 56a5...2 cubed is 8, times b3.2087

If you multiply out 56 times 8, you will get 448; a5, b3.2095

This is the fourth term of this expanded form of (a + 2b)8.2103

It is a little bit complicated: the first thing I had to do is find the coefficient.2111

And I did that by using my binomial formula and knowing that n = 8 and k = the term number, plus 1.2115

The term number plus 1 is k, so k is one less than the term number: k is 3.2128

I was able to find, then, using this formula, that the coefficient is 56.2133

The second thing I had to do is figure out what power I should raise a to, and what power I should raise 2b to.2139

Well, when I am using the binomial formula, I am finding the coefficient for a term in this form, which is an - k2bk.2145

So, I knew my n and my k; so I put my coefficient here and said, "OK, I have a8 - 3(2b)3," using this.2153

Simplify, and then multiply to get 448a5b3.2163

We are asked to expand (2x + 3y)5...this binomial to the fifth.2172

And the general form is (a + b)n.2178

So, here we actually have that a is equal to 2x, and b is equal to 3y; so that makes things a little bit more complicated at the end.2181

We have to take these into account when we simplify our final answer.2190

Since this is the fifth power, I have an n equal to 5.2197

I am going to have a total of 6 terms.2202

It is going to look something like this at the end: (2x + 3y)5 =...my first term is 2x raised to the n power,2205

which is the fifth; plus the coefficient, and then 2x is going to decrease to the fourth power, and I am going to have my 3y here.2214

Plus...some other coefficient; 2x becomes cubed, and 3y increases from an exponent of 1 to an exponent of 2.2225

Another coefficient is going to bring me to 2x23y3;2241

another coefficient...now I just have 2x, and 3y4; and then, finally, 3y5.2250

And I expect to have 6 terms--let's verify that I am not missing anything: 1, 2, 3, 4, 5, 6.2261

Down here at the bottom, I am going to start building up what our final answer is going to look like, because I have some simplifying that I need to do.2270

The first term, I can figure out: I already know that it is 2x5.2281

23 is 8, times 2 is 16, times 2 is 32; so that is 32x5.2286

The second term: recall that the coefficient for the second term is equal to n--that is 5.2293

Also, remember that symmetry is involved: since there are 6 terms (that is an even number of terms), I am going to have two middle terms.2301

These two are going to have the same coefficient; the two next to those will have the same coefficients; and the two outer have the same coefficients.2309

It is not the same value, once I simplify them; but the same coefficient.2319

So, since these two are the same, and these two are the same, this term also has a coefficient of 5.2323

Let's work this one out up here; this is the second term, and I know that it is equal to 5 times (2x)4 times 3y.2332

That is...2 times 2 is 4; 2 cubed is 8, times 2 is 16; so that is times 16x4, times 3y.2346

Multiplying 5 times 16 times 3 is going to give you 240x4y; I am going to put that in here; that is 240x4y.2358

I don't know what I have for these two middle ones, but it is going to be something here, and it is going to be x3y2.2373

Something is here, then x2y3; I can figure this one out:2384

right here, this is the third term; it equals 5 times 2x, times (3y)4.2394

5 times 2x is just 10x; 3 times 3 is 9, times 3 is 27, times 3 is going to give me 81.2406

So, that is times 81y4; so this one is easy, because it is just 81 times 10, is 810xy4.2419

So, that was...actually not the third term: correction--we don't know the third term; the first, second, third, and fourth...2436

the two middle terms we don't know; we do know the fifth term--this is the fifth term.2448

All right, so the fifth term I can fill in: 810xy4.2456

The sixth term, right here, I can also fill in: the sixth term equals (3y)5.2465

33 is 27; 34 is 81; 35, if you figure it out, is 243; 243y5.2475

Just using the properties of binomial expansions, I got most of the terms.2489

And since there is symmetry, the two middle terms will have the same coefficient inserted up here.2496

And then, I can work out the rest and finalize it.2505

So, I only need to find one coefficient, using the binomial formula.2507

The binomial formula, recall, is n!/k!, times (n - k)!.2511

I know that n = 5; for the first, second, third term, k is equal to 2; you could just say "Third term: 3 - 1 is 2";2521

or you could look up here and say, "OK, 3y is raised to the second power, so k = 2."2532

Therefore, I am going to have 5!/2!, times (5 - 2)!; this equals 5!/2! times 3!,2537

which equals 5 times 4 times 3 times 2 times 1; 2 times 1; 3 times 2 times 1.2559

1, 2, and 3 all cancel; this gives me 5 times 4, divided by 2; that is 20/2, or 10.2568

So, the missing coefficient here is 10; by symmetry, the missing coefficient right here is also 10; I am almost done.2578

So, for the first, second, third term, I have my coefficient; but I need to simplify that a bit.2588

My third term equals 10 times (2x)3 times (3y)2.2595

This is going to be equal to 10 times...2 cubed is 8x3; 3 squared is 9y2.2605

This is going to give me 80 times 9x3y2, which is just 720x3y2.2617

That is my third term, so I am going to put a 720 right here.2625

The fourth term: I also have a 10 right here.2630

What is slightly different, though: it is 10 times (2x)2(3y)3.2635

This is going to give me 10 times 4x2; 3 cubed is 27, so it is y3.2642

This is 40 times 27...let's put all of the constants together...x2y3.2651

40 times 27 actually comes out to 1080; you can multiply that out for yourself; x2y3.2663

The fourth term, therefore: I am going to put a 1080 right here.2673

So, this is quite a bit of work; but we actually only needed to find one coefficient using the binomial formula,2676

because by symmetry, these two middle ones were the same (the two middle coefficients that we will put in the blank);2682

and I knew that the second and the second-to-the-last had coefficients equal to n.2690

Once I found that a 10 went into these blanks, all I needed to do was work out what each of these terms would simplify to,2695

to end up with the expansion that I found down here.2703

We are asked to find a specific term this time, not the entire expansion.2710

I just want to find the fourth term of (x - 3y)7, using the binomial formula, n!/k!, times (n - k)!.2713

All right, n = 7; what does k equal? Well, it is the fourth term; k is going to be one less than that; 4 - 1...I am going to have k = 3.2725

This is going to give me the coefficient for a term in this form.2737

It is going to have a coefficient, and then it is going to have x raised to the n - k power, times -3y, raised to the k power.2741

This is going to become x7 - 3, (-3y)3.2757

Therefore, I am going to get some coefficient (that I am going to find in a moment), times x4, times (-3y)3.2768

Let's go ahead and find the coefficient: this missing coefficient equals n!, 7!, divided by k!, times (7 - 3)!;2775

that equals 7!/3!, times 4!, equals (let's write this out)...divided by 3 times 2 times 1, times 4 times 3 times 2 times 1.2793

Do some canceling out: 1's, 2's, 3's, and 4's are the same.2812

This leaves me with 7 times 6 times 5, divided by 3 times 2 times 1.2819

6 has common factors here of 3 and 2; I can just cancel those out.2825

So, this is 7 times 5, divided by 1; so just 7 times 5 equals 35.2832

I know that my coefficient is, then, 21 that is what is going to go in this blank.2838

So, I am going to end up with...the fourth term is going to be 35 (I am going to put 35 right here in this blank)2842

x4(-3y)3, which equals 35x4...-3 squared is 9, times -3 again is -27.2849

So, it is -27y3; so it is time to get out your calculators and figure out 35 times -27, or multiply it out, if necessary.2864

And you will find that that equals -945x4y3.2875

OK, we found the fourth term of this expansion by using the binomial formula to find the coefficient,2881

and also, an awareness that the exponent of this x is going to be n - k; the exponent for the -3y is going to be k.2887

We figured out the coefficient (it is 35), and then simplified, using our rules for working with exponents.2896

That concludes this session on the binomial theorem here at Educator.com; thanks for visiting!2905

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