For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

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### 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:

Once you know a factor, it becomes that much easier to find the other factors.f(k) = 0 ⇔ (x−k) is a factor of f(x). - 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

Alternatively, we can write this asf(x) = d(x) · q(x) + r(x) . f(x) d(x)

= q(x) + r(x) d(x)

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

1. Divide the__first__term of the dividend (thing being divided) by the__first__term of the divisor (thing doing the division). Write the result above.

2. Multiply the entire divisor by the result, then subtract that from the dividend (so that the first term of one lines up with first term of the other).

3. Bring down the next term from the dividend.

4. Repeat the process (divide first terms, multiply, subtract, bring down) until finished with the entire dividend. - There is also a shortcut method that goes a bit faster if the divisor is in the form (x−k). This is called
*synthetic division*. [Again, this is a rather difficult technique to describe with words, but much easier to see in action. Check out the video if you haven't already to understand how to do this.] The process for synthetic division goes like this:

1. Write out the coefficients of the polynomial in order.

2. Write k in the top-left corner [from the factor you're dividing by, (x−k)].

3. Start on the left of the coefficients and work towards the right. Every step will have a "vertical" part and a "diagonal" part.

4. On the "vertical" part, add the two numbers above and below each other to produce another number. On the "diagonal" part, multiply the result from the "vertical" by k.

5. The final number produced by the process is the remainder. The other numbers are the resulting coefficients of the quotient. - If you need to do division on a polynomial that is "missing" a variable raised to a certain exponent, make sure to fill that space in with a 0 multiplying the appropriate variable with exponent. For example, if you wanted to divide x
^{4}+ 5x + 7, notice that it's "missing" x^{3}and x^{2}. 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:x ^{4}+ 5x + 7 ⇒ x^{4}+ 0x^{3}+ 0 x^{2}+ 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
- Introduction
- Reminder: Roots Imply Factors
- The Intermediate Value Theorem
- Intermediate Value Theorem, Proof Sketch
- Finding Roots with the Intermediate Value Theorem
- Dividing a Polynomial
- Long Division Refresher
- The Division Algorithm
- Polynomial Long Division
- Synthetic Division
- Synthetic Division, Example
- Which Method Should We Use
- Example 1
- Example 2
- Example 3
- Example 4

- 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

### Math Analysis Online

### Transcription: Intermediate Value Theorem and Polynomial Division

*Hi--welcome back to Educator.com.*0000

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*For example, if we have g(x) = x ^{3} - 3x^{2} - 4x + 12,*0123

*and we happen to realize that when we plug in a 2, that all turns into a 0*0128

*(and it does), then we would know that g(x) is equal to (x - 2), this thing becoming (x - 2), times (_x ^{2} + _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 heights*0305

*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 _x ^{3} + _x^{2} + _x + _--*0527

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

*Otherwise, it would not have divided out cleanly.*0534

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

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

*How do we do polynomial division?*0545

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

*So, long ago, in grade school and primary school, we were used to doing problems like 1456 divided by 3.*0553

*Let's break out long division: we have 1456, so the first thing we do is see how many times 3 goes into 1.*0559

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

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

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

*3 times 4 gets us 12; we subtract by 12: minus 12; so minus 12...we get 2.*0586

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

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

*It goes in 8 times...16, 24...so now we subtract by 24; 25 - 24 gets us 1.*0607

*We bring down the 6; we have 16; how many times does 3 go into 16? It goes in 5; we get 15.*0617

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

*There is nothing else, so that means we are left with a remainder of 1.*0633

*We have whatever our very last thing was, once we ran out of stuff; that becomes our remainder: 485 with a remainder of 1.*0640

*If we wanted to express this, we could say 1456 divided by 3; another way of thinking of that is 1456 is equal to 3(485) + 1.*0647

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

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

*So, it doesn't come out cleanly; so it comes out as 1/3.*0675

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

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

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

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

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

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

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

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

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

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

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

*We found that 1456 divided by 3 became 3 times 485 plus 1.*0747

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

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

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

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

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

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

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

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

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

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

*The thing doing the dividing is the divisor.*0808

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

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

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

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

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

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

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

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

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

*So, basically the same thing is happening over here, so this would not be 3 times 45; it should be 1/3.*0906

*1456/3 is equal to 485 + 1/3, because that is 1456/3.*0913

*We have a real connection between these two things.*0920

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

*we can break it into the divisor, times the quotient, plus some remainder, which, alternatively, we can express as*0928

*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 x ^{4} - 5x^{3} - 7x^{2} + 29x + 30, divided by x - 3.*0989

*So, we are dividing x - 3; it is dividing that polynomial: x ^{4} - 5x^{3} - 7x^{2} + 29x + 30.*0996

*OK, the first thing we do is ask, "All right, how many times does x - 3 go into x ^{4} - 5x^{3}?"*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 x ^{4}?*1025

*Well, x ^{4} divided by x would be x^{3}; so the x^{3} 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, x ^{3} is what we get out of x^{4} divided by x.*1078

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

*x ^{3} times (x - 3) becomes x^{4} - 3x^{3}.*1087

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

*Let's put that subtraction over it; minus, minus, 2 minus's become a plus...so we have -x ^{4} attacking x^{4}, so we have 0 here.*1097

*And 3x ^{3} + -5x^{3} becomes -2x^{3}.*1105

*And then, the next thing we do is bring down the -7x ^{2}.*1115

*So, -7x ^{2}: now we ask ourselves, "How many times does x go into -2x^{3}?"*1119

*Well, that is going to go in -2x ^{2}, so we get -2x^{2}.*1127

*-2x ^{2} times x - 3 becomes -2x^{3}; -2x^{2} times -3 becomes + 6x^{2}.*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 -2x ^{3} + 2x^{3} becomes 0 once again.*1149

*-7x ^{2} - 6x^{2} becomes -13x^{2}.*1154

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

*How many times does x go into -13x ^{2}? That goes in -13x.*1164

*The -13x times x - 3 gets us -13x ^{2} + 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 -13x ^{2} + 13x^{2}; 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 x ^{3} - 2x^{2} - 13x - 10 as what is left over after we divide out x - 3.*1219

*So, we know our original x ^{4} - 5x^{3} - 7x^{2} + 29x + 30 factors as...*1230

*let's write this in blue, just so we don't get it confused...(x - 3)(x ^{3} - 2x^{2} - 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 x ^{4}; the first term of our divisor,*1260

*the thing doing the dividing, is x; so how many times does x go into x ^{4}?*1266

*Well, x ^{4} 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 x ^{3} now, x times x times x; great.*1279

*That is why the x ^{3} goes here; it is the very first thing that happens.*1287

*The next thing: we take x ^{3}, and we multiply it onto (x - 3).*1291

*So, x ^{3}(x - 3) becomes x^{4} - 3x^{3}.*1298

*You multiply the entire divisor (this right here) by the result, our x ^{3}.*1304

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

*x ^{4} - 3x^{3}: we subtract that from x^{4} - 5x^{3}.*1315

*This gets distributed, so we get a negative here, a plus here; and so that becomes -2x ^{3}.*1319

*The next thing we do is bring down the next term; so our next term to deal with is this 7x ^{2}.*1325

*It gets brought down, and we have -2x ^{3} - 7x^{2}.*1331

*And then, once again, we do the same thing: how many times does x go into -2x ^{3}?*1337

*It goes in -2x ^{2}; so then, it is -2x^{2}(x - 3); and we get -2x^{3} + 6x^{2}.*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 x ^{2} + 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 ax ^{4} + bx^{3} + cx^{2} + 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 x ^{2}'s coefficient; this will be x^{3}'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 x ^{4} - 5x^{3} - 7x^{2} + 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 x ^{2}; this is our x^{3}.*1571

*So, we get x ^{3} - 2x^{2} - 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 form*1614

*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 understand*1699

*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) = 2x ^{3} + 4x^{2} - 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 poly _{up}, if it is coming from above and then going down, or the polynomial that starts below, so poly_{down}.*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) = x ^{5} - 3x^{4} + x^{3} - 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: x ^{5}, x^{4}, x^{3}...there is no x^{2}!*2195

*So, we need to figure out what x ^{2} is, so we can effectively put it in as 0x^{2}.*2202

*Remember: the coefficient that must be on the x ^{2} to keep it from appearing, to make it disappear, is a 0.*2207

*So, what we really have is the secret 0x ^{2} + 60, because we have to have coefficients*2213

*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 x ^{4};*2235

*we have a hidden 1 in front of the x ^{3}; we have a 0 on our completely-hidden x^{2}.*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 x ^{2} coefficient, our x^{3} coefficient, and our x^{4} 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 x ^{4} + 0x^{3}, so we will just omit that;*2316

*plus 1x ^{2} + 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 x ^{4}, we get x^{5}; minus 3x^{4}...x^{2}...*2362

*+ x ^{3} - 3x^{2} + 3x(x)...+ 3x^{2}...+ 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 x ^{5}; there are no other x^{5}'s, so we have just x^{5} that comes down.*2404

*3x ^{4}...do we have any other x^{4}'s?...no, no other x^{4}'s, so it is - 3x^{4}.*2408

*x ^{3}: do we have any other x^{3}'s?...no, no other x^{3}'s, so it is + x^{3}.*2414

*- 3x ^{2}: are they any others?...yes, they cancel each other out, so it is 0x^{2}.*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 x ^{4} - 2x^{3} - 11x^{2} - 8x - 60*2456

*by using the fact that x ^{2} + 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 x ^{2} + 4.*2476

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

*We have this x ^{2}, so we have to use polynomial long division.*2486

*So, x ^{2} + 4: we plug in x^{4} - 2x^{3} - 11x^{2} - 8x - 60.*2490

*Great; x ^{2} + 4 goes into x^{4}...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 x ^{2} + 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 x ^{2} go into x^{4}? It goes in x^{2}.*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 x ^{2} here.*2534

*x ^{2} times (x^{2} + 0x + 4): we get x^{4}; this is just blank still; + 4x^{2}.*2540

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

*x ^{4} - x^{4} becomes 0; -11x^{2} - 4x^{2} becomes -15x^{2}.*2554

*We bring down our -2x ^{3}; we bring down our -8x; so we have everything.*2560

*-2x ^{3} - 15x^{2} - 8x: once again, we just ask,*2566

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

*-2x ^{3} divided by x^{2} gets us just -2x^{2}.*2575

*Oops, I'm sorry: we are dividing by x ^{2}, because -2 is just x^{1}.*2582

*So, -2 times x ^{2} gets us -2x^{3}; 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: -2x ^{3} + 2x^{3} becomes 0.*2599

*-8x + 8x becomes nothing; and now we bring down the thing that didn't get touched, the -15x ^{2} and the -60.*2603

*And we have -15x ^{2} - 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

*x ^{2} + 0x + 4; how many times does that fit into -15x^{2} - 60?*2624

*Once again, we just look at the first part: -15x ^{2} divided by x^{2} becomes just -15.*2628

*-15x ^{2}...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 x ^{2} - 2x - 15.*2650

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

*We could also say this as (x ^{2} + 4)(x^{2} - 2x - 15).*2658

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

*x ^{2} + 4...can we factor that any more?...no, we can't; it is irreducible.*2670

*If we were to try to set that to 0, we would have to have x ^{2}, 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

*x ^{2} - 2x - 09 how can we factor this?*2685

*Just 1 is in front of the x ^{2}, so that part is easy; it is going to be x and x, and then... what about the next part, -15?*2688

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

*we factor it into 5 and 3; 5 and 3 have a difference of 2, so let's make it -5 and +3.*2697

*We check that: x ^{2} + 3x - 5x...-2x; -5 times 3 is -15; great.*2704

*So, at this point, we set everything to 0; x ^{2} + 4 will provide no answers.*2710

*x ^{2} + 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 polynomial*2737

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

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

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

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

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

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

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

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?