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Lecture Comments (4)

1 answer

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

Post by Fadumo Kediye on October 18, 2015

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

1 answer

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

Post by Humberto Coello on August 23, 2013

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

Intermediate Value Theorem and Polynomial Division

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


    = q(x) + r(x)


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

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

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

Intermediate Value Theorem and Polynomial Division

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

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

Transcription: Intermediate Value Theorem and Polynomial Division

Hi--welcome back to

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

that is the location that intersects that intermediate value.0388

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

How do we divide a polynomial by a factor?0516

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

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

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

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

Otherwise, it would not have divided out cleanly.0534

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

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

How do we do polynomial division?0545

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The thing doing the dividing is the divisor.0808

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

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

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

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

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

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

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

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

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

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

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

We have a real connection between these two things.0920

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

How many times does x go into x4?1025

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

And the very last thing is our remainder.1456

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Great; there is our answer.1886

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

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

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

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

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

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

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

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

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

Now, how can we see what this is?1935

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

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

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

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

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

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

So, how is this going to work?1972

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

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

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

So, now we have another two possibilities.1984

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

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

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

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

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

Those are our two possibilities; OK.2016

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

So, we have x5; there are no other x5's, so we have just x5 that comes down.2404 we have any other x4's?, no other x4's, so it is - 3x4.2408

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

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

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

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

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

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

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

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

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

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

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

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

If we are going to pull it out, can we use synthetic division?, because it is not in the form x - k.2481

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Subtraction distributes and cancels those into plus signs.2640

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

So, we get x2 - 2x - 15.2650

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

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

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

x2 + 4...can we factor that any more?, we can't; it is irreducible.2670

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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