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Lecture Comments (10)

2 answers

Last reply by: Tiffany Warner
Wed May 25, 2016 7:17 PM

Post by Tiffany Warner on May 25 at 06:21:53 PM

Hi Professor!

I am really struggling with one of the practice questions given.
Find all the roots of f(p) = 2 p4 ? 8p3 ? 14p2 + 44p + 48 given that f(3) = 0 and f(?2) = 0.

The first part of the problem seems straight forward. We are given two roots, therefore two factors.


In example 2 of the lecture, you show us how to tackle factoring a cubic polynomial. However, I’m lost with this beast.

I figured expanding the two factors would make it more simple and the steps did reinforce that idea.
So we have (p-3)(p+2)(_p^2+_p+_)
Which expanded becomes

They show us in the steps what to fill in those blanks with. I understand how they got -8. (-8)(-6)=48
I also understand how they got the coefficient of 2 in front of p^2.
However I have no idea how they went about filling in that middle blank in front of p. I’m clearly missing something. If you could provide some guidance, I’d be very grateful!

Thank you!

0 answers

Post by Jamal Tischler on December 21, 2014

I've seen that theorem in Horner's factoring table.

3 answers

Last reply by: Professor Selhorst-Jones
Fri Dec 12, 2014 8:38 PM

Post by abendra naidoo on December 12, 2014

These are good teaching modules.
I have 2 questions:
1)How is an equation different from a function.
2) I can correctly operate exponents and logs but still don't see why logs are necessary? Why not just use exponents - is it because log tables exist? How can I better understand this concept?

1 answer

Last reply by: Professor Selhorst-Jones
Mon Oct 20, 2014 11:59 AM

Post by Saadman Elman on October 19, 2014

In 21:49- 21:52 you meant to say X-5 is a factor. And X=5 is a zeros/roots. So X-5 is a factor. But you said x-5 is a root. Waiting for your reply.

Roots (Zeros) of Polynomials

  • The roots/zeros/x-intercepts of a polynomial are the x-values where the polynomial equals 0.
  • If you have access to the graph of a function/equation, it is very easy to see where the roots are: where the graph cuts the horizontal axis (the x-intercepts)! Why? Because that's where f(x) = 0 or y=0.
  • While you can occasionally find the roots to a polynomial by trying to isolate the variable and directly solve for it, that method often fails or is misleading.
  • To find the roots of a polynomial we need to factor the polynomial: break it into its multiplicative factors. Then we can set each factor to 0 and solve to find the roots.
  • Factoring can be quite difficult if you're trying to factor a very large or complicated polynomial. There is no procedure that will work for factoring all polynomials.
  • In general, if we have a quadratic trinomial (something in the form ax2 + bx + c), we can factor it into a pair of linear binomials as
    ax2 + bx + c  = ( x +  ) ( x +  ).
    Think about what has to go in each blank for it to be equivalent to the polynomial you started with.
  • Whenever you're factoring polynomials, make sure you check your work! Even on an easy problem, there are ample opportunities to make a mistake. That means you should always try expanding the polynomial (it's fine to do it in your head) to make sure you factored it correctly.
  • In general, factoring higher degree polynomials is similar to what we did above. Figure out how you can break down the polynomial into a structure like the above, then ask yourself how you can fill in the blanks.
  • If you already know a root to a polynomial, it must be one of its factors. For example, if we know x=a is a root, then the polynomial must have a factor of (x−a). This makes factoring the polynomial that much easier.
  • Not all polynomials can be factored. Sometimes it is impossible to reduce it to smaller factors. In such a case, we call the polynomial irreducible. [Later on, we'll discuss a a hidden type of number we haven't previously explored when we learn about the complex numbers. These will allow us to factor these supposedly irreducible polynomials. However, for the most part, we won't work with complex numbers so such polynomials will stay irreducible.]
  • There is a limit to how many roots/factors a polynomial can have. A polynomial of degree n can have, at most, n roots/factors.
  • We also get information about the possible shape of a polynomial's graph from its degree. A polynomial of degree n can have, at most, n−1 peaks and valleys (formally speaking, relative maximums and minimums).

Roots (Zeros) of Polynomials

Find the zeros of f(x) = x2+2x − 15.
  • The zeros of f(x) are all those x where f(x)=0. That means we want to find all the x such that
    0 = x2+2x − 15.
  • We do this by factoring the polynomial. Break the polynomial into factors. Because we have a quadratic, we know it will probably break into factors of the form
    ( x +  ) ( x +  ) .
  • We know that the coefficients in the front blanks for each parenthetical must multiply to make 1, because that is the coefficient in front of x2. We know the numbers in the other blanks must multiply to make −15. Finally, we know that when we add those numbers together, they must make 2 because we have 2x.
  • Working through this, we find (x−3)(x+5) works, which we can check. If we expand those factors, they do indeed come out to be x2 + 2x −15, so we know the factoring is correct.
  • Once we have 0=(x−3)(x+5), we can find the values for x. We set each of the factors equal to 0. [We can do this by the same logic that 0 = a·b means a and/or b must be 0.]
    0 = x−3        0 = x+5
x = −5,  3
Find the roots of g(t) = t2 − 7t − 18.
  • The roots of g(t) mean the same thing as the zeros: all those t where g(t)=0. That means we want to find all the t such that
    0 = t2 − 7t − 18.
  • We do this by factoring the polynomial. Break the polynomial into factors. Because we have a quadratic, we know it will probably break into factors of the form
    ( t +  ) ( t +  ) .
  • We find (t+2)(t−9) works, which we can check. If we expand those factors, they do indeed come out to be t2 − 7t − 18, so we know the factoring is correct.
  • Once we have 0=(t+2)(t−9), we can find the values for t. We set each of the factors equal to 0. [We can do this by the same logic that 0 = a·b means a and/or b must be 0.]
    0 = t+2        0 = t−9
t = −2,  9
Solve for x:    2x2 + 11x + 30 = x2 −4x −20.
  • Because we're trying to solve something involving polynomials, we will need to use factoring. However, before factoring can be useful, we need to get one side equal to 0. [This is because we want to eventually have something of the form 0 = a·b and then say 0=a and 0=b.]
  • Use algebra to get one side of the equation equal to 0. We obtain
    x2 + 15x + 50 = 0.
  • Factor the left side to get
    (x+10)(x+5) = 0.
  • Now that we have it in the form of 0 equals things multiplied together, we can set each of those things equal to 0:
    0 = x+10        0 = x+5
x = −10, −5
Find the zeros of h(x) = x3 − 9x2 + 23x − 15.
  • The zeros of h(x) are all those x where h(x)=0. That means we want to find all the x such that
    0 = x3 − 9x2 + 23x − 15.
  • We do this by factoring the polynomial. Break the polynomial into factors. Start off easy by trying to break it into something of the form
    ( x +  )  ( x2 +  x +  ).
  • While there are other possibilities (depending on which factor we pull out first), we might get
    [With something difficult like this, it's even more important to check your work and expand the polynomial in your head to be sure you didn't make a mistake.]
  • Now factor the remaining x2−8x+15 to obtain
  • We now have 0 = (x−1)(x−3)(x−5). Set each of the three factors to 0 and solve:
    0 = x−1        0 = x−3        0 = x−5
x=1,  3,  5
Find the other roots to y = x3 + 4x2 − 7x −10 if one of the roots is x=2.
  • If you know a root, that automatically implies a factor. A root of x=a implies a factor of (x−a). Thus, since we know x=2 is a root of the polynomial, we know (x−2) is a factor, which will help us in our factoring.
  • Pulling out (x−2), we know we'll have something of the form
    (x−2)  ( x2 +  x +  ),
    now we just need to appropriately fill in the blanks.
  • Begin by filling in the blanks that are easiest. You know the coefficient on x2 must be 1 because in the fully expanded polynomial it is x3. Furthermore, you know the blank for the constant must be 5 because in the fully expanded polynomial you have −10. Then use that information to figure out the middle blank. Eventually you obtain
    (x−2)(x2 + 6x +5).
  • Factor the right parenthetical to finish factoring the entire polynomial:
  • Thus, to find the roots, we have 0 = (x−2)(x+1)(x+5). Set each of these equal to 0, then solve:
    0 = x−2       0 = x+1        0 = x+5
The additional roots are x=−5, −1 (and there is still the root that was given to us in the problem: x=2).
Find all the roots of f(p) = 2 p4 − 8p3 − 14p2 + 44p + 48 given that f(3) = 0 and f(−2) = 0.
  • If you know a root, you automatically know a factor. Remember, a root/zero is where the function equals 0. Thus, since we know f(3) = 0 and f(−2) = 0, we know p=3 and p=−2 are roots. A root of p=a implies a factor of (p−a). Therefore we already know two roots: (p−3) and (p+2).
  • If we pull out those two factors, we'll have something of the form
    (p−3)(p+2) ( p2 +  p +  ).
  • As it stands, it's a little difficult to fill in the blanks. To make it easier to see how the blanks should be filled in, expand the left side:
    (p2 − p − 6)  ( p2 +  p+  )
  • Working it out, we obtain
    (p2 − p − 6)  (2p2−6p−8).
  • Finish factoring the above:
    [There are actually other ways to factor the right side. You could also obtain
    (p−3)(p+2)(p−4)(2p+2)     or     2·(p−3)(p+2)(p−4)(p+1)
    Still, whatever way you factor it, you will wind up getting the same answer for the roots.]
  • We now are looking to solve 0 = (p−3)(p+2)(2p−8)(p+1). Set each factor to 0:
    0 = p−3        0 = p+2        0 = 2p−8        0 = p+1
p=−2, −1, 3,  4
Give a polynomial that has degree 3 and roots at x=−4, 3,  9.
  • Each root implies a factor where the constant has the opposite sign:
    x=−4, 3,  9     ⇒     (x+4),    (x−3),    (x−9)
  • If we multiply them all together, we have a polynomial with degree 3:
  • OPTIONAL: You could expand the polynomial if you wanted (or if a problem required it).
  • OPTIONAL: If you wanted, you could multiply the entire polynomial by a constant number because it has no effect on the roots. Notice that a·(x+4)(x−3)(x−9) has the exact same roots, because the same values for x will cause the expression to become 0.
(x+4)(x−3)(x−9) or, equivalently, x3−8x2−21x+108
[Additionally, it would also be correct to have a constant multiple of the above, such as −2·(x+4)(x−3)(x−9) = −2x3+16x2 +42x −216.]
Give a polynomial that has degree 4 and roots at x=2, 5.
  • Each root implies a factor where the constant has the opposite sign:
    x=2,  5     ⇒     (x−2),    (x−5)
  • However, if we just multiply them all together, we only have a polynomial with degree 2:
  • We need to increase the degree of the polynomial, but without adding any additional roots. We can do this by adding multiplicity to a single root. Have the same root show up multiple times in the expansion, such as
    In such a way, we now have a polynomial of degree 4.
  • OPTIONAL: You could expand the polynomial if you wanted (or if a problem required it).
  • OPTIONAL: If you wanted, you could multiply the entire polynomial by a constant number because it has no effect on the roots.
  • OPTIONAL: Instead of just using multiplicity and having roots appear more than once, you could also use a polynomial factor that is irreducible, such as (x2 +1). Because we cannot solve 0=x2+1 in the real numbers, it will not add any additional roots. For example, we could have a polynomial like
There are multiple possible answers: any polynomial that contains factors of (x−2) and (x−5) (but no other factors [unless irreducible]) where the total number of factors is 4. Some examples:
(x−2)3 (x−5)           (x−2)2 (x−5)2
[There are some other optional ways to answer this problem. See the steps above if you're curious.]
What is the maximum number of roots the below polynomial can have?
x5 − 8x4 + 2x3 + 34x2 + x + 42
  • The maximum number of roots/factors a polynomial can have is based on its degree. A polynomial of degree n can have, at most, n roots/factors. Notice that this is a maximum: there is no guarantee it will have that many, we can only be sure it will not have more than that.
  • The degree of the polynomial in this problem is 5, so the maximum number of roots is 5.
  • In actuality, if you factor this polynomial, it does not have 5 roots. It only has 3 roots because it factors to
    While (x−3), (x−7), and (x+2) all produce roots [x=3, 7, −2], the factor (x2+1) does not produce any. Why? Because x2+1=0 cannot be solved with any real numbers, so it is irreducible. We will learn more about this idea in later lessons.
The polynomial can not have more than 5 roots.
[If you're curious about the precise number of roots that it has, check out the final step for this problem.]
The area of a rectangle is 66 m2. If the length of the rectangle is 5 m longer than the width, what are the dimensions of the rectangle?
  • Begin by setting up the varaibles:
    w = width       l = length
  • Create equations from the information given:
    w ·l = 66        w+5 = l
  • Notice that we can plug in w+5 for l in the area equation to produce:
    w (w+5) = 66
  • This is a quadratic equation, so we can solve it by factoring. Begin by getting 0 on one side of the equation, then factor it:
    w (w+5) = 66     ⇒     w2 +5w − 66=0     ⇒     (w+11)(w−6) = 0
  • We have two possibilities for w: w = −11 and w=6. Notice that a negative value for length makes no sense, so we throw out that extraneous solution. This leaves us with w=6.
  • Once we know the width, we can find the length. Don't forget to put units in your answer.
The width is 6 m and the length is 11  m.

*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.


Roots (Zeros) of Polynomials

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

Transcription: Roots (Zeros) of Polynomials

Welcome back to

Today, we are going to talk about roots (also called zeroes) of polynomials.0002

We briefly went over what a root is when we talked about the properties of functions.0006

But let's remind ourselves: the zeroes of a function, the roots of an equation, and the x-intercepts of a graph are all the same thing.0009

They are inputs (which are just x-values) where the output is 0; it is where it comes out to be 0--things that will make our expression give out 0.0017

The roots, zeroes, x-intercepts of f(x) = x2 - 1 and y = x2 - 1 are x = -1 and x = +1,0026

because those two values, f(x) and y, are equal to 0.0036

If we plug in -1, (-1)2 will become positive 1, minus 1 is 0.0040

If we plug in positive 1, 12 is 1, minus 1 is 0.0045

Those two things cause it to give out 0, and the roots of the polynomial are the x-values where the polynomial equals 0.0049

The roots of the polynomial x2 - 1 are -1 and positive 1.0055

Being able to find roots in functions is important for many reasons; and it will come up very often when you are working with polynomials.0062

If you continue on to calculus, you will see how roots can be useful for finding lots of information about a function.0068

So, it is very important to have a grasp of what is going on there and be able to find roots.0073

Roots in graphs: if you have access to the graph of a function/equation, it is very easy to see where the roots are.0078

Of course, you might not see precisely, because it is a graph, after all, and it might be off by a little bit.0083

But you can get a very good sense of where they are: it is where the graph cuts the horizontal axis--the x-intercepts.0088

Why? Because here we have f(x) = 0 or y = 0, depending on if it is a function or an equation.0094

Since that means our height is at 0 there, then every place where we cross the x-axis must be a root--it's as simple as that.0101

This also gives us a nice mnemonic to remember what the word "root" means--0111

it can be a little hard to remember the word "root," since we aren't that used to using it.0115

But we could remember it as where the equation or the function grows out of the x-axis--where it is 0.0118

It is like it is the ground; think of it as a plant rooted in the ground.0125

A function or equation has its roots in the x-axis; a tree has its roots in the earth, and a function has its roots in a height of 0.0129

So, a root is where it is growing up and down; it is where it is held in our plane, held in our axes.0140

And that is one way to remember what a root is.0150

How do we find the roots of a polynomial? Well, at first we might try a naive approach and attempt to solve the way we are used to.0153

Naive is just what you have done before--what seems to make sense--without ever really having had a whole lot of experience about it.0160

So, the naive attempt would be probably to just isolate the variable on one side.0166

That is what we did with a bunch of other equations before, so let's do it again.0170

Now, in some cases, this will actually work, and we will find all of the real solutions.0173

For example, if we have f(x) = x - 3, then we set it equal to 0, because we are looking for the roots; we are looking for when 0 = x - 3.0176

We move that over, and we get x = 3; great.0184

Or if we had y = x3 + 1, we set that to 0, and we have 0 = x3 + 1,0187

because we are looking for what x's cause us to have 0 = x3 + 1.0192

So, we have x3 = -1; we take the cube root of both sides, and we get x = 3√-1, which is also just -1.0197

So, in both of these cases, the naive method of isolating for variables worked just fine.0204

But that is definitely not going to be the case for all situations.0208

The naive method of isolation will fail us quite quickly, even when used on simple quadratic polynomials.0212

Consider f(x) = x2 - x - 2--that is not a very difficult one.0217

But this method of trying to isolate will just fail us utterly if we use it here.0223

So, 0 = x2 - x - 2...we might say, "Well, let's get the numbers off on one side."0227

We have 2 = x2 - x; but then, we don't have just x; so let's pull out an x; we get 2 = x(x - 1).0232

And well, we are not really sure what to do now; so let's try another way.0240

0 = x2 - x - 2; let's move x over, because we are used to trying to get just x alone.0243

So, we have x here; but then we also have x2 here; so let's divide by x.0248

We get 1 = (x - 2)/x; once again, we are not really sure what to do.0252

Let's try again: 0 = x2 - x - 2, so let's move over everything but the x2.0256

Maybe x2 is the problem; so we will get x + 2 = x2.0262

We take the square root of both sides; we remember to put in our ± signs: ±√(x + 2) = x.0265

I don't really know how to figure out what x's go in there to make that true.0271

So, in all three of these cases, it is really hard to figure out what is going on next.0274

If we are going to try to isolate, we are going to get these really weird things.0280

This method of isolation that we are used to isn't going to work here, because we can't get x alone; we can't get the variable alone on one side.0283

It is not going to let us find the roots of polynomials if we try to isolate.0290

But at least we could trust it in those previous examples; we saw that we can trust it when it does work.0293

So, we might as well try it first--no, it is even worse: the naive method of isolation0298

can make us miss answers entirely, even though we think we will know them all.0303

So, we will think we have found the answers; but in reality, we will only have found part of the answers.0307

Consider these two ones: if we have 0 = x2 - 1, we move the 1 over; and then we take the square root of both sides.0311

The square root of 1 is 1; the square root of x2 is x; great.0317

For this one over here, we have 0 = p2 + 3p, so we realize that we can divide both sides by p.0320

And since 0/p is just 0, we have 0 = p2/p (becomes p); 3p/p becomes 3; so we have 0 = p + 3.0327

We move the 3 over; we get p = -3; great--we found the answers.0336

Not quite: those above things are solutions, but in each case, we have missed something.0340

We have been tricked into missing answers by trying to follow this naive method.0345

The other solutions for this would be x = -1 and p = 0, respectively.0349

The mistakes that we forgot were a ± symbol over on this one; we forgot to put a ± symbol when we took the square root;0353

and then, the other one was dividing by 0, because when we divided, we inherently forgot0361

about the possibility that, if we were actually dividing by 0, we couldn't divide by 0.0366

So, those are the two mistakes; but even if you personally wouldn't have made those same mistakes,0371

this example shows how it is easy to forget those things in the heat of actually trying to do the math.0375

You might forget about that; you might accidentally make one of these mistakes; so it is risky to try this method of isolation.0381

We need something that works better.0387

We find the roots of the polynomial by factoring; we break it into its multiplicative factors.0389

Let's look at how this works on the example that we couldn't solve with naive isolation.0395

We have f(x) = x2 - x - 2; we have 0 = x2 - x - 2, because we are looking for when f(x) is equal to 0.0399

And then, we say, "Let's factor it; let's break it into two things."0407

So, we have (x - 2) and (x + 1); and if we check, that does become it: x times x becomes x2; x times 1 becomes + x;0409

-2 times x becomes - 2x; -2 times 1 becomes -2; so yes, that checks out to be the same thing as x2 - x - 2.0417

0 = x - 2 and 0 = x + 1 is how we now set these two things equal to 0; we have 0 = x - 2 and 0 = x + 1.0425

And we get x = 2 and x = -1, and we have found all of the solutions for this polynomial.0435

Why does this work, though? We haven't really thought about why it works.0441

And we don't want to just take things down and automatically say, "Well, my teacher told me that, so that must be the right thing."0444

You want to understand why it is the right thing.0449

Teachers can be wrong sometimes; so you want to be able to verify this stuff and say, "Yes, that makes sense,"0451

or at least have them explaining and saying, "Well, we don't understand quite enough yet;0456

but later on you will be able to see the proof for this"; you really want to be able to believe these things,0460

beyond just having someone tell you by word of mouth.0463

So, to figure out why this has to be the case, we will consider 0 = ab.0466

The equation is only true if a or b, or both of them, is equal to 0; if neither a nor b is equal to 0, then the equation cannot be true.0473

If a = 2 and b = 5, then we get 10, which is not equal to 0.0482

As long as a or b is 0, it will be true, because it will cancel out the other one.0489

But if both of them are not 0, then it fails, and it is not going to be 0.0494

It is the exact same thing happening with x2 - x - 2; we have 0 = x2 - x - 2,0499

which we then factor into (x - 2) and (x + 1); so let's use two different colors, so we can see where this matches up.0505

We are pairing this to the same idea of the a times b equals 0; it is (x - 2) (x + 1) = 0.0513

The only way that this equation can be true is if x - 2 = 0 or x + 1 = 0.0520

Just as we showed up here, it has to be the case that a or b equals 0 for that to be true.0528

So, it must be the case that either (x - 2) or (x + 1) equals 0, if this is going to be true.0534

So, our solutions are when either of the two possibilities is true--if the possibility is true, if one of them is true, then the whole thing comes out.0539

So, either case being true makes it acceptable; that gets us 2 (keeping with our color coding) and -1 as the two possibilities.0546

So, by breaking x2 - x - 2 into its factors, we can find its roots.0557

So, this is how we find polynomial roots, in general: the first thing we do is set the whole thing as 0 = polynomial.0562

We have to have some polynomial, and then it is 0 equals that polynomial.0569

The next thing we do is factor it into the smallest possible factors; we break it down into multiplicative factors.0573

And then finally, we set each factor equal to 0, and we solve for each of them.0580

So, in step number 2, we are going to get things like 0 = (x + a)(x + b)(x + c)(x +d) and so on, and so on, and so on.0585

And then, in step 3, we set each factor to 0; so we get things like x + a = 0, at which point we can solve and say, "Oh, x = -a."0595

That is one of our possible solutions; and from there, you can work out all of the roots of the polynomial.0604

Caution--this is very important: notice that it is extremely, extremely important to begin by setting the equation as 0 = polynomial.0610

I have seen lots of mistakes where people forgot to set it as 0 = polynomial.0618

If it isn't, if it was something like 5 = (x - 2)(x + 1), we can't solve for the solutions from those factors.0621

Those factors are now meaningless; they aren't going to help us.0628

We need the special property that 0 turns everything it multiplies into 0.0631

Without that special property, this method just won't work.0635

Consider if we had something like 5 = ab; there is no way that we could just figure out what the answers are here.0638

It is not just simply that a has to be 0 or b has to be 0, because a could be 5 and b could be 1.0644

Or b could be 5 and a could be 1, or a could be 25 and b could be 1/5; or a could be 100 and b could be 1/20.0649

We have lots of different possibilities--a whole spectrum of things; there are way too many possible solutions.0658

We need that special property of 0 = ab to be able to really say, "That thing is 0, or that thing is 0"; that is what we know for sure.0663

That is how we get useful information out of it.0671

That is why it has to be 0 = polynomial; if you don't set it up as that before you try to factor it,0674

before you try to do the other steps, you are just not going to be able to get the answer,0679

because we need that special property that 0 has when it multiplies other things.0682

0 multiplying something automatically turns it to 0; if we don't have that special property, things just won't work.0686

Factoring is not necessarily easy: say we have something like0693

x5 + 6.5x4 - 17x3 - 41x2 + 24x, and we want to know what the zeroes are.0697

Well, if we knew its factors, we would be able to break it into (x + 8)(x + 2) times x times (x - 0.5) times (x - 3).0703

And it would be really easy to figure out what the polynomial's roots are.0710

At this point, we say, "Well, great; x + 8 becomes x = -8; x + 2 becomes x = -2; x + 0 becomes x = 0;0712

x - 0.5 becomes x = positive .5; and x - 3 becomes x = positive 3--great; I found it; it is really easy to find its roots."0722

But how do you factor a monster like that?0732

Ah, there is the problem: factoring can be quite difficult.0734

Luckily, by this point, you have been certainly practicing how to factor for years in your algebra classes.0738

By now, you have done lots of factoring; you are used to this; you have played with polynomials a bunch in previous math classes.0743

And all that work has a use, and it is here, finding roots; we can break things down into their factors and find roots.0748

But there is no simple procedure for factoring polynomials.0754

Once again, remember: if you were confronted by something like this, you would probably have a really difficult time0757

figuring out what its factors were--figuring out how you can break that down into factors.0762

There is no simple procedure for factoring polynomials.0768

High-degree polynomials can just be very difficult to factor; happily, we are not going to really see such polynomials.0770

Most courses on this sort of thing don't end up giving you very difficult things to factor at this stage.0777

So, we won't really have to worry about factoring really difficult polynomials.0782

We will be able to stick to the smaller things.0787

So, let's have a quick review of how you factor small things like quadratics.0790

We are going to see a bunch of quadratics; they are very important--they pop up all the time in science.0795

So, let's look at a brief review of how to factor a quadratic polynomial.0799

Remember, quadratic is a degree 2, and a trinomial just means 3 terms.0801

So, if we have a quadratic trinomial in its normal form, then we have ax2 + bx + c.0806

If we want to factor that, we will turn it into a pair of linear binomials: degree 1 and 2 terms (linear and binomial).0811

We would want to break this into (_x + _) (_x + _).0819

Great; to be equivalent to the above, the coefficients of x must give a product of a.0824

So, a has to come out with the red dot here, times the red dot here.0830

And the constants have to give a product of c; so the blue dot here times the blue dot here has to come out to be c.0835

Also, b has to add up from the products of the outer blanks and the inner blanks.0843

So, it has to be that the red dot here times the blue dot here, plus the blue dot here times the red dot here, comes up to be this b here.0848

So, it is mixed out of the two of them.0863

Don't worry if that is a little bit confusing right now; you will be able to see it as we work through examples.0865

The b has to add up to the products of the outer blanks and the inner blanks.0868

The a has to come from the first blanks, and the c has to come from the last blanks.0872

Don't worry about memorizing this, though--it is just a sense of what is going on in practice.0876

Let's look at an example: we want to factor 2x2 - 5x - 12.0880

So, we know, right away, that we want to break it into the form (_x + _)(_x + _).0884

The first thing we notice is that we have this 2 at the front; and 2 only factors into 2 times 1.0890

We can't break it up into anything else really easily; so let's put 2 times 1 down as 2x times x.0895

We have to put the 2 somewhere; so it is either going to be (2x + _)(1x + _), or it is going to be the 2 over here and the 1 here.0901

It doesn't really matter what order we put it in; so we will put the 2 at the front.0910

We have (2x + _)(x + _); now, what is going to go into those other blanks?0913

Now, we need to factor the -12; so let's factor 12 first: we notice that 12 can break into 1 times 12, 2 times 6, or 3 times 4.0917

And one of the factors has to be a negative, because we have a negative in front of the 12.0926

So, they have to be able to multiply to make a -12; so there is going to have to be a negative on either the 1 or the 12.0931

One of those two will have to have a negative on it; we don't know which one, but it is going to be one of them.0938

Or it is going to be -2, or -6; and then finally, for the last pair, it would be -3 or -4.0943

I am not going to use all of these at once; we will have to figure out which one is right.0948

But one of them will have to be negative, because of this negative sign up here.0950

We start working through this; and we know that there is this 2x here at the front.0955

We have this 2x at the front; so it is going to multiply this one, and it is going to effectively double whatever we put here.0960

So, the difference between one of the numbers doubled and its sibling (the one here times this one in front of it) must be -5, because we get -5x.0967

We notice that 3 - 2(4)...2 times 4 is 8; 3 is 3; so the difference between those is 5.0979

So, we can set it up as 3 - 2(4) = -5; and we get (2x + 3)(x - 4); and we have factored this out; we have been able to work it out.0986

Now, there are various ways, various tricks that have been taught to you in previous classes.0998

But the important thing is just to set up and have an expectation of what form you are trying to get.1005

And then, plug things in and say, "Yes, that would work; that would get me what I am looking for" or1009

"No, if I plug that in, that won't work; that won't get me what I am looking for."1013

As long as you work through that sort of thing, you will be able to find the answer eventually.1016

It is always a good idea, though, to check your work; you will find the answer, but it is really easy to make mistakes.1020

So, even on the easiest of problems (like the one we were just working on), there are lots of chances to make mistakes; trust me.1026

I make mistakes; everybody makes mistakes; the important thing is to catch your mistakes before they end up causing problems.1032

So, it means that you should always try expanding the polynomial after you have factored it, to make sure you factored correctly.1038

And it is OK to do this in your head; once you get comfortable with doing this sort of thing1044

(and by now, honestly, you probably have had enough experience with this that you can just do this in your head reasonably quickly),1048

it is OK to do it in your head; the important part is that you want to have some step where you are checking back on what you are doing.1053

So, either do it on the paper (if it is a long one) or do it in your head (if it is something that is short enough, easy enough, for you to check).1058

But you want to make sure that you are checking your work.1063

For example, if we have 2x2 - 5x - 12, we figured out that that breaks into (2x + 3) and (x - 4).1065

But we want to check and make sure that it is right.1071

So, we check and make sure: 2x + 3 times x - 4...we get 2x2, and then 2x(-4), 3x...we get -8x + 3x...1073

3 times -4...we get -12; we combine our like terms of -8x + 3x, and we get 2x2 - 5x - 12.1083

Sure enough, it checks out; we have what we started with, so we know that our factoring was correct; we did a good job.1092

Factoring higher-degree polynomials: in general, factoring polynomials of any degree is going to be similar to what we just did on these previous few slides.1100

The only difference is that it will become more complex as they become longer, as we get to higher and higher degrees.1107

So, for example, if we had something like a cubic--if we had ax3 + bx2 + cx + d--1112

we would probably want to set it up as (_x + _)(_x2 + _x + _).1118

If we are going to be able to break it up and factor it, it is going to have to factor into these two things, (_x + _)(_x2 + _x + _).1123

Notice also that this is a degree 1, and this is a degree 2; and when you multiply these two together, you will get back to a degree 3 over here.1131

Adding up the degrees on the right side has to be what we had on the left side.1141

We could also work on quartic, a degree 4; and we might try one of these two templates.1145

If we have ax4 + bx3 + cx2 + dx + e, we might break it1149

into (_x + _)(_x3 + _x2 + _x + _); or we might break it into (_x2 + _x + _)(_x2 + _x + _).1153

And so on and so forth...clearly, this is going to become more difficult as we get to higher and higher degrees.1163

The higher the degree of a polynomial, the more complicated our template is going to have to be for where we are going to fit things in.1168

The more choices we are going to have, the more difficult it is going to be to do this.1173

Luckily, we are only occasionally going to need to factor cubics, things like this where it is degree 3.1177

And we are very, very seldom going to see anything of higher degree.1182

So, don't worry too much about having really difficult ones.1186

But just be aware that factoring really large, high-degree polynomials can actually be pretty difficult to do.1188

Roots imply factors: we have this useful trick if we want to break out these higher polynomials.1196

If we know one of a polynomial's roots, we automatically know one of its factors.1201

Remember: one use of finding the factors of a polynomial is to find its roots.1206

If you find a factor of the form (x - a), then you set that equal to 0, and you know that that is going to be x = a.1209

It turns out that the exact opposite is true; if we know a polynomial has a root at x = a, then it also means we have a factor of x - a.1216

So, if we have a root x = a, then that turns into a factor, x - a, just because of this equation here,1225

where we are setting that factor equal to 0, which gives us the root.1232

We won't prove this; it requires a little bit of difficult mathematics, and some things that we actually haven't covered in this course yet.1236

But we can see it as a theorem: Let p(x) be a polynomial of degree n; then if there is some number a,1242

such that p(a) = 0 (that is to say that a is a root of our polynomial, p--if we plug a in, we get 0),1250

then there is some way to break it up so that it is p(x) = (x - a), our factor (x - a) that we know1257

from our root at x = a, times q(x), where q is some other polynomial of degree n - 1,1264

because this here is degree 1; so when we multiply it by a degree n - 1, we will be back up to our degree n polynomial that we originally had.1272

If we manage to find one or more roots, sometimes the problem will give them to use; other times, we will get lucky, and we might just guess one.1285

This theory means we automatically know that many factors of the polynomial.1291

For example, say we know that p(x) = x3 - 2x2 - 13x - 10.1295

And we are told that p(5) = 0; we know that 5 is a root.1300

Then, we automatically know that at x = 5 we have a zero; so (x - 5) is our root.1305

We plug that in; we know it is going to be (x - 5)(_x2 + _x + _).1312

Now, we don't know what is in these blanks yet; we still don't know what is going to be in there.1317

But we are one step closer to figuring out what those factors are, for being able to figure out what has to go in those blanks.1325

Later on, in another lesson, we will use this fact to great advantage--1334

this fact that knowing the root automatically means you know a factor--1336

When we learn about the intermediate value theorem to help us find roots,1340

and the polynomial division using those roots to break down large polynomials into smaller, more manageable factors.1344

Not all polynomials can be factored, though; even with all of this talk of factoring polynomials, there are some that cannot be factors,1352

not because it is difficult or really hard to do, but because it is just simply impossible.1358

Consider this polynomial: f(x) = x2 + 1; if we try to find its roots, then we have 0 = x2 + 1.1363

So, we have x2 = -1; but there is no number that exists that can be squared to become a negative number.1370

No number can be squared to become a negative number; why?1377

Consider: if we have (-2)2, that becomes positive 4; if we square any negative number, it becomes a positive number.1380

If we square any positive number, it stays positive; if we square 0, it stays 0.1386

So, there is no number that we have, that we can square, and get a negative number out of it.1391

Thus, the polynomial has no roots; and since it has no roots from that theorem we just saw, it can't have any factors.1395

So, it has no roots; therefore, it cannot be reduced into smaller factors.1401

And something that cannot be reduced, we call irreducible; it is not reducible.1407

Now, I will be honest: what I just told you isn't really the whole story.1413

More accurately, we can't factor all polynomials yet.1418

The previous slide that we just saw is perfectly true, but only if you are working with just the real numbers (which have the symbol ℝ like that).1422

Now, that is what we are normally working with; so it is kind of reasonable to say this.1432

But it turns out that there is a hidden type of number that we haven't previously explored.1435

You might have seen this in previous math classes, even.1439

We will learn about the complex numbers later on; complex numbers can give us a way to factor these supposedly irreducible polynomials.1441

So, they are irreducible for real numbers; but they are not irreducible for complex numbers.1449

Now, we will learn about them in the lesson that is named after these numbers--our lesson on complex numbers.1454

But for now, we are just working with real numbers; and in general, we will just be working with real numbers in this course.1460

Real numbers are really useful; you can do a lot of stuff with them.1464

So, it is enough for us to be working with real numbers, generally.1467

That means, for us right now, at least some polynomials are simply irreducible.1470

And we can't always find roots for everything in a polynomial, because we can't break it down,1474

because there are things that just don't have roots, based on how real numbers work.1480

Now, we will talk about complex numbers later on; but I just wanted to point this out--1484

that I am not telling you the whole story right now, because we don't want to get confused with complex numbers.1488

But for our purposes right now, with real numbers, there are some things that are simply irreducible.1492

There is a limit to how many roots or factors a polynomial can have.1498

Now, roots and factors are basically two sides of the same thing.1501

Since x = a as a root is the same thing as knowing (x - a) as a factor, they are just two sides of the same thing.1505

So, we will consider them as roots/factors of polynomials.1516

A polynomial of degree n can have, at most, n roots/factors; so we can have a maximum of n of these.1518

Why is this the case? Well, consider this: every factor comes in the form (x + _), or even larger if the factor is irreducible.1527

It might be (_x2 + _x + _); but the very smallest has to be (x + _).1535

If we break a polynomial into its factors, we are going to get (x + _) (x + _) on and so on, up until (x + _).1540

Now, if we had more than n factors, then that would mean that we have (x + _) multiplying by itself more than n times.1547

So, if we have x multiplying more than n times, then it has to have a degree larger than n; its exponent is going to have to be greater than that.1555

If we wanted to max out at x2, but we had (x + _)(x + _)(x + _)...well, that is going to become x31564

plus stuff after it, which is not going to be x2...not going to be degree 2.1576

So, if we have a degree n polynomial, the most factors we can possibly have are n factors, n roots,1581

because otherwise, we would have too many factors, and we would blow out the degree of the polynomial.1589

Thus, the most roots/factors a polynomial can have is equal to its degree.1595

We also can get information about the possible shape of a polynomial's graph from its degree.1601

A polynomial of degree n can have, at most, n - 1 peaks and valleys.1605

Formally speaking, that is relative maximums and minimums.1610

For example, if we have x4, then that means we have n = 4 (our degree).1613

So, n - 1 = 3; so we look over here, and we have one valley here, one peak here, one valley here.1618

This is also a relative minimum, a relative minimum (that is what we mean by "valley), and a relative maximum (that is what we mean by "peak").1630

So, n - 1 tells us the most bottoms and tops we can have, before they either go off to positive infinity or go off to negative infinity.1642

Now, we can't prove this here, because it requires calculus; but it is connected with the maximum number of roots in a polynomial.1650

And if you go on to take calculus (which I heartily recommend), you will very clearly, very quickly see it.1656

It becomes very clear in calculus; it is one of the important points of what you do in calculus.1661

So, you will think, "Oh, that makes a lot of sense," because the possible peaks and valleys1665

are connected to a polynomial that has a degree that is 1 less; and that is why it is connected.1670

Don't worry about it too much right now; but it is very interesting, and very obvious, if you go on to take calculus.1675

Notice that, in both of the previous properties, it was described as "at most."1681

Just because a polynomial has degree n does not mean it will have n distinct roots or n - 1 peaks and valleys.1687

We aren't necessarily going to have to have that many; it is just that we can have up to that many.1693

Consider f(x) = x5 + 1; this graphs like this, but from this graph, we can see clearly:1697

we only have one root, and we have no peaks or valleys.1704

The degree gives an upper limit on how many there can be, but it doesn't tell us how many there will be.1708

It just that the maximum is this; but you could definitely have fewer.1713

All right, we are ready for some examples.1719

The first one: we want to find the zeroes of f(x) = 3x2 - 23x + 14.1721

So, this is a textbook example--literally a textbook example, since this is effectively a textbook.1726

So, you plug in 0, because we are looking for when f(x) is equal to 0.1731

0 = 3x2 - 23x + 14; we know we are going to be looking for 0 =...something where it is going to be (_x + _)(_x + _).1735

What are we going to slot in there? Well, we notice that here is 3; the only way we can break up 3 is 3 times 1.1751

There are no other choices; so we either have to have 3 go for the first x or 3 go to the second x.1758

So, let's set it as 0 = (3x + _); and we will have 1x, so just x, plus _.1762

Great; now, at this point, we also say, "We have 14 over here; how can 14 break up?"1772

Well, we can have 1 times 14, or we can have 2 times 7; those are the only choices.1777

So, we are going to have to plug in either 1 times 14 or 2 times 7.1783

But now, we also have to take this -23 into consideration.1786

If we have -23, then we are going to have at least one negative over here.1789

And since it comes up as a positive, it is going to have to be that they are both negatives.1793

So, one of them is negative, so they are going to both be negatives; so it will be -1(-14) or -2(-7).1796

So, -1 times -14...we will notice that, either way we put that in, that won't work out.1802

But we can plug in -2 times -7, and we can amp up this -7; so + -7 here...put in the -2 here...and we get 0 = (3x - 2)(x -7).1807

We have managed to factor it; let's really quickly check what we have here--does this work out?1823

Check (3x - 2)(x - 7); we would get 3x2 - 21x - 2x + 14, so 3x2 - 23x + 14; it checks out; sure enough, it is good.1828

So, at this point, we break this down into two different possibilities: either 3x - 2 = 0, or x - 7 = 0.1847

So, 3x - 2 = 0 or x - 7 = 0 are the two different worlds where this will be true,1855

where we will have found a root where the whole expression will be equal to 0.1861

So, 3x - 2 = 0: we get 3x = 2; x = 2/3; over here, x - 7 = 0; we have x = 7; so our answers are x = 2/3 and x = 7; those are the roots for this polynomial.1864

Great; if f(2) = 0, factor f(x) = x3 - 7x + 6.1881

Remember: if we know that at x = 2 we have a zero (at x = 2 there is a root), then that means there is a factor in that polynomial of (x - 2).1888

How do we figure that out? Well, we notice that x = 2; then it is x - 2 = 0, so that implies that it has to be a factor of (x - 2) in there.1904

We can use that piece of information; we know that f(x) is going to have to break down with an (x - 2) in there.1912

So, let's set it up like normal: 0 = x3 - 7x + 6; but what we just figured out here...we know that there is a factor of (x - 2).1917

So, we can also write this as 0 = (x - 2)(_x2 + _x + _); what are going to go into those blanks?1924

Well, at this point, we just use a little bit of logic and ingenuity, and we can figure this out.1935

Well, we know that...what is in front of this x3? It is effectively a 1.1941

So, if there is a 1 in front, we have x times x2; whatever goes into this blank is going to determine what coefficient is in front of it.1944

So, since we want a 1, it has to be that there is a 1 here, as well.1952

What about the very end? Well, the only thing that is going to create the ending constant is going to be the other constants.1957

So, the constants that we have here are -2 and whatever goes into that blank; so it must be that -2 times _ here becomes 6.1968

-2 times -3 becomes 6; so we have a -3 here.1975

And finally, what is going to go into this blank here?1980

We think about this one, and we know that we want to have 0x2 come out of this; there are no x2's up here.1984

So, we have + 0x2; so whatever we put into this blank must somehow get us a 0x2 to show up.1992

So, x times x2 is x3; so we are not going to worry about that.2001

But x times _x is going to be x2; let's do a little sidebar for this.2005

x times _x will become _x2; and -2 times...we already filled in that blank...1x2 is going to be -2x2.2010

Now, we want the 0x2 out of it; so it must be that, when we add these two things together, it comes out to be 0x2.2023

What does this have to be? It has to be positive 2x2.2031

We know that positive 2x2, minus 2x2, comes out to be 0x2; so it must be that this is a positive 2x.2035

So, we write this whole thing out: 0 = (x - 2)(x2 + 2x - 3); and we have been able to figure out that that works.2042

We check this out and do a really quick check; so x3 + 2x2 - 3x - 2x2 - 4x - 6...2053

x3 checks out; 2x2 - 2x2 cancel each other; that checks out.2067

-3x - 4x; that becomes -7x, so that checks out; -6...that checks out as well.2072

Great; we have a correct thing, so 0 = (x - 2)(x2 + 2x - 3) is correct.2078

We factored it properly; so at this point, the only thing that we have left to factor is the x2 + 2x -3.2087

0 = (x - 2)(_x + _)(_x + _); what goes in those first blanks?2095

Well, we just have a 1 in front of that; so it is 1 and 1...we don't have to worry about it that much.2103

What else is going to go in there? Well, -3 is at the very end; we have +2x, so it must be that the negative amount is smaller than the positive amount.2108

So, it is going to be + -1 and + 3; -1 times 3 gets us -3, and everything else checks out.2117

x times x is x2; plus 3x minus x...that gets + 2x; and minus 1 times 3 gets us -3; so that checks out.2128

We did another check in our heads really quickly.2136

So finally, we have 0 = (x - 2)(x - 1)(x + 3); we break this up into three different worlds;2138

set each world equal to 0: x - 2 = 0; x - 1 = 0; x + 3 = 0; so we have x = 2; x = 1; x = -3.2152

Those are all of the roots for this polynomial.2167

All right, the next example: give a polynomial with roots at the indicated locations and the given degree.2172

Now remember, a root can become a factor; so if we know that we have a root at -3, then that becomes a factor of...2177

if it was x = -3, then it becomes (x + 3); if we had x = 8, then that would become (x - 47)...2186

I'm sorry; I accidentally read the wrong one--read forward one; that is (x - 8).2196

And then finally, if we had x = 47, we would have (x - 47) as our factor.2201

So, those three things together...we have (x + 3)(x - 8)(x - 47), and that right there is a polynomial.2207

We know it has degree 3, because we have x times x times x; that is going to be the largest possible exponent we can get on our variable.2217

That will come out to be x3, so we have a degree 3; and we know it has roots in all of the appropriate places.2226

And we are done--that is it; we could expand this, and we could simplify, if we wanted.2230

We weren't absolutely required to by the problem, and this is a correct answer.2234

It is a polynomial; it is not in that general, standard form that we are used of _x to the exponent, _x to the exponent, _x to the exponent.2239

But it is still a polynomial, so it is a pretty good answer; we will leave it like that.2248

The next one: we have -2 and positive 2, so we have (x + 2) for the -2 and (x - 2) for that.2252

Why? x = -2; we move that over; we get x + 2 = 0; x = +2...we move that over and we get x - 2 = 0.2261

So, we get (x + 2)(x - 2); but if we multiply those two together, we just have a degree of 2, and we want a degree of 4.2268

So, we need to somehow get the degree up on this thing, but have the same roots--not have to accidentally introduce any more roots.2275

So, if we introduced multiplying just by x twice, we would have introduced a root at x = 0.2284

So, we can't just do that; but we do realize that if we just increase this to square it on both of them, they will still have the same roots.2289

It is just duplicate roots showing up; so (x + 2)2 + (x - 2)2...we have hit that degree of 4,2296

because each one of these will now have a degree of 2.2302

Alternately, we could have done this as (x + 2) to the 1, (x - 2) cubed, or (x + 2)3(x - 2)1.2305

Any one of these would have degree 4, and have our roots at the appropriate place.2315

Great; the final example: What is the maximum possible number of roots and peaks/valleys for each of the following polynomials?2320

So, for our first one, f(x), we notice that this has a degree of 3; so n = 3 means the maximum number of roots it can have is 3,2327

and the maximum peaks/valleys is one less; n - 1 is 3 - 1, is 2.2339

So, the maximum number of roots is 3; the maximum peaks/valleys is 2.2350

We don't necessarily know it will have that many; all we know is that that is the maximum it could possibly have.2354

The next one: we notice that the degree for this one is 47; so if n = 47, then the maximum roots are going to be equal to that degree.2359

The maximum peaks/valleys are going to be one less than that degree, so we will get 46, one less than that.2369

The final one: for this one, we think, "Oh, 103, so it is 3!"--no, we have to remember that this is not a variable.2379

This here is a variable; so it is x1, so its degree is just 1.2388

For that one, degree 1...we will change over to the color green...n = 1 means the maximum roots are just 1;2394

and the maximum peaks/valleys are going to be one less than 1, so 1 - 1 = 0.2405

Now, why is that the case? Well, think about it: 103 - 5...10 cubed is just some constant;2415

it happens to be 1,000, but that is not really the point; so 1000x - 5 is just going to be a very steep line.2420

x - will intersect here; but does it ever go up and down--does it ever undulate in weird ways?2429

No, it never does anything; we just have a nice, straight line, since it is a linear thing (linear like a line).2435

So, since it is a linear expression, it never undulates--never has any peaks or valleys.2441

So, it never has any relative maximums, and no relative minimums; and that is why we have 0 there--it makes sense.2445

All right, I hope everything there made sense; I hope you got a really good understanding of roots,2451

because roots will come up in all sorts of places; they are really important to understand.2454

It is really important to understand this general idea, because you will see it in other things, being changed around.2457

But if you understand this general idea, you will be able to understand what is going on in later things and different courses.2461

All right, see you at later--goodbye!2466