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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### Roots (Zeros) of Polynomials

- The
*roots/zeros/x-intercepts*of a polynomial are the x-values where the polynomial equals 0. - If you have access to the graph of a function/equation, it is very easy to see where the roots are: where the graph cuts the horizontal axis (the x-intercepts)! Why? Because that's where f(x) = 0 or y=0.
- While you can occasionally find the roots to a polynomial by trying to isolate the variable and directly solve for it, that method often fails or is misleading.
- To find the roots of a polynomial we need to
*factor*the polynomial: break it into its multiplicative factors. Then we can set each factor to 0 and solve to find the roots. - Factoring can be quite difficult if you're trying to factor a very large or complicated polynomial. There is no procedure that will work for factoring all polynomials.
- In general, if we have a quadratic trinomial (something in the form ax
^{2}+ bx + c), we can factor it into a pair of linear binomials as

Think about what has to go in each blank for it to be equivalent to the polynomial you started with.ax ^{2}+ bx + c = ( x + ) ( x + ). - Whenever you're factoring polynomials, make sure you
__check your work__! Even on an easy problem, there are ample opportunities to make a mistake. That means you should always try expanding the polynomial (it's fine to do it in your head) to make sure you factored it correctly. - In general, factoring higher degree polynomials is similar to what we did above. Figure out how you can break down the polynomial into a structure like the above, then ask yourself how you can fill in the blanks.
- If you already know a root to a polynomial, it must be one of its factors. For example, if we know x=a is a root, then the polynomial must have a factor of (x−a). This makes factoring the polynomial that much easier.
- Not all polynomials can be factored. Sometimes it is impossible to reduce it to smaller factors. In such a case, we call the polynomial
*irreducible*. [Later on, we'll discuss a a hidden type of number we haven't previously explored when we learn about the*complex numbers*. These will allow us to factor these supposedly irreducible polynomials. However, for the most part, we won't work with complex numbers so such polynomials will stay irreducible.] - There is a limit to how many roots/factors a polynomial can have. A polynomial of degree n can have,
__at most__, n roots/factors. - We also get information about the possible shape of a polynomial's graph from its degree. A polynomial of degree n can have,
__at most__, n−1 peaks and valleys (formally speaking, relative maximums and minimums).

### Roots (Zeros) of Polynomials

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
- Roots in Graphs
- Naïve Attempts
- Factoring: How to Find Roots
- Factoring: How to Find Roots CAUTION
- Factoring is Not Easy
- Factoring Quadratics
- Factoring Quadratics, Check Your Work
- Factoring Higher Degree Polynomials
- Factoring: Roots Imply Factors
- Not all Polynomials Can be Factored
- Max Number of Roots/Factors
- Max Number of Peaks/Valleys
- Max, But Not Required
- Example 1
- Example 2
- Example 3
- Example 4

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

### Math Analysis Online

### Transcription: Roots (Zeros) of Polynomials

*Welcome back to Educator.com.*0000

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

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

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

*They are inputs (which are just x-values) where the output is 0; it is where it comes out to be 0--things that will make our expression give out 0.*0017

*The roots, zeroes, x-intercepts of f(x) = x ^{2} - 1 and y = x^{2} - 1 are x = -1 and x = +1,*0026

*because those two values, f(x) and y, are equal to 0.*0036

*If we plug in -1, (-1) ^{2} will become positive 1, minus 1 is 0.*0040

*If we plug in positive 1, 1 ^{2} is 1, minus 1 is 0.*0045

*Those two things cause it to give out 0, and the roots of the polynomial are the x-values where the polynomial equals 0.*0049

*The roots of the polynomial x ^{2} - 1 are -1 and positive 1.*0055

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

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

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

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

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

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

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

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

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

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

*But we could remember it as where the equation or the function grows out of the x-axis--where it is 0.*0118

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

*A function or equation has its roots in the x-axis; a tree has its roots in the earth, and a function has its roots in a height of 0.*0129

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

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

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

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

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

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

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

*For example, if we have f(x) = x - 3, then we set it equal to 0, because we are looking for the roots; we are looking for when 0 = x - 3.*0176

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

*Or if we had y = x ^{3} + 1, we set that to 0, and we have 0 = x^{3} + 1,*0187

*because we are looking for what x's cause us to have 0 = x ^{3} + 1.*0192

*So, we have x ^{3} = -1; we take the cube root of both sides, and we get x = ^{3}√-1, which is also just -1.*0197

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

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

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

*Consider f(x) = x ^{2} - x - 2--that is not a very difficult one.*0217

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

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

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

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

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

*So, we have x here; but then we also have x ^{2} here; so let's divide by x.*0248

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

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

*Maybe x ^{2} is the problem; so we will get x + 2 = x^{2}.*0262

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

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

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

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

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

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

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

*So, we might as well try it first--no, it is even worse: the naive method of isolation*0298

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

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

*Consider these two ones: if we have 0 = x ^{2} - 1, we move the 1 over; and then we take the square root of both sides.*0311

*The square root of 1 is 1; the square root of x ^{2} is x; great.*0317

*For this one over here, we have 0 = p ^{2} + 3p, so we realize that we can divide both sides by p.*0320

*And since 0/p is just 0, we have 0 = p ^{2}/p (becomes p); 3p/p becomes 3; so we have 0 = p + 3.*0327

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

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

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

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

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

*and then, the other one was dividing by 0, because when we divided, we inherently forgot*0361

*about the possibility that, if we were actually dividing by 0, we couldn't divide by 0.*0366

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

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

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

*We need something that works better.*0387

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

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

*We have f(x) = x ^{2} - x - 2; we have 0 = x^{2} - x - 2, because we are looking for when f(x) is equal to 0.*0399

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

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

*-2 times x becomes - 2x; -2 times 1 becomes -2; so yes, that checks out to be the same thing as x ^{2} - x - 2.*0417

*0 = x - 2 and 0 = x + 1 is how we now set these two things equal to 0; we have 0 = x - 2 and 0 = x + 1.*0425

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

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

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

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

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

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

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

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

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

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

*If a = 2 and b = 5, then we get 10, which is not equal to 0.*0482

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

*But if both of them are not 0, then it fails, and it is not going to be 0.*0494

*It is the exact same thing happening with x ^{2} - x - 2; we have 0 = x^{2} - x - 2,*0499

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

*We are pairing this to the same idea of the a times b equals 0; it is (x - 2) (x + 1) = 0.*0513

*The only way that this equation can be true is if x - 2 = 0 or x + 1 = 0.*0520

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

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

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

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

*So, by breaking x ^{2} - x - 2 into its factors, we can find its roots.*0557

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

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

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

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

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

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

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

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

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

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

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

*We need the special property that 0 turns everything it multiplies into 0.*0631

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

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

*It is not just simply that a has to be 0 or b has to be 0, because a could be 5 and b could be 1.*0644

*Or b could be 5 and a could be 1, or a could be 25 and b could be 1/5; or a could be 100 and b could be 1/20.*0649

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

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

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

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

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

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

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

*Factoring is not necessarily easy: say we have something like*0693

*x ^{5} + 6.5x^{4} - 17x^{3} - 41x^{2} + 24x, and we want to know what the zeroes are.*0697

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

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

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

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

*But how do you factor a monster like that?*0732

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

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

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

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

*But there is no simple procedure for factoring polynomials.*0754

*Once again, remember: if you were confronted by something like this, you would probably have a really difficult time*0757

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

*There is no simple procedure for factoring polynomials.*0768

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

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

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

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

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

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

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

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

*So, if we have a quadratic trinomial in its normal form, then we have ax ^{2} + bx + c.*0806

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

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

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

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

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

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

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

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

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

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

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

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

*Let's look at an example: we want to factor 2x ^{2} - 5x - 12.*0880

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

*The first thing we notice is that we have this 2 at the front; and 2 only factors into 2 times 1.*0890

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

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

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

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

*Now, we need to factor the -12; so let's factor 12 first: we notice that 12 can break into 1 times 12, 2 times 6, or 3 times 4.*0917

*And one of the factors has to be a negative, because we have a negative in front of the 12.*0926

*So, they have to be able to multiply to make a -12; so there is going to have to be a negative on either the 1 or the 12.*0931

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

*Or it is going to be -2, or -6; and then finally, for the last pair, it would be -3 or -4.*0943

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

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

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

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

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

*We notice that 3 - 2(4)...2 times 4 is 8; 3 is 3; so the difference between those is 5.*0979

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

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

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

*And then, plug things in and say, "Yes, that would work; that would get me what I am looking for" or*1009

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

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

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

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

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

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

*And it is OK to do this in your head; once you get comfortable with doing this sort of thing*1044

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

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

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

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

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

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

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

*3 times -4...we get -12; we combine our like terms of -8x + 3x, and we get 2x ^{2} - 5x - 12.*1083

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

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

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

*So, for example, if we had something like a cubic--if we had ax ^{3} + bx^{2} + cx + d--*1112

*we would probably want to set it up as (_x + _)(_x ^{2} + _x + _).*1118

*If we are going to be able to break it up and factor it, it is going to have to factor into these two things, (_x + _)(_x ^{2} + _x + _).*1123

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

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

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

*If we have ax ^{4} + bx^{3} + cx^{2} + dx + e, we might break it*1149

*into (_x + _)(_x ^{3} + _x^{2} + _x + _); or we might break it into (_x^{2} + _x + _)(_x^{2} + _x + _).*1153

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

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

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

*Luckily, we are only occasionally going to need to factor cubics, things like this where it is degree 3.*1177

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

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

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

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

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

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

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

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

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

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

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

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

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

*then there is some way to break it up so that it is p(x) = (x - a), our factor (x - a) that we know*1257

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

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

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

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

*For example, say we know that p(x) = x ^{3} - 2x^{2} - 13x - 10.*1295

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

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

*We plug that in; we know it is going to be (x - 5)(_x ^{2} + _x + _).*1312

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

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

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

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

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

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

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

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

*Consider this polynomial: f(x) = x ^{2} + 1; if we try to find its roots, then we have 0 = x^{2} + 1.*1363

*So, we have x ^{2} = -1; but there is no number that exists that can be squared to become a negative number.*1370

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

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

*If we square any positive number, it stays positive; if we square 0, it stays 0.*1386

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*It might be (_x ^{2} + _x + _); but the very smallest has to be (x + _).*1535

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

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

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

*If we wanted to max out at x ^{2}, but we had (x + _)(x + _)(x + _)...well, that is going to become x^{3}*1564

*plus stuff after it, which is not going to be x ^{2}...not going to be degree 2.*1576

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

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

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

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

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

*Formally speaking, that is relative maximums and minimums.*1610

*For example, if we have x ^{4}, then that means we have n = 4 (our degree).*1613

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

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

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

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

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

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

*So, you will think, "Oh, that makes a lot of sense," because the possible peaks and valleys*1665

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

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

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

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

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

*Consider f(x) = x ^{5} + 1; this graphs like this, but from this graph, we can see clearly:*1697

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

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

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

*All right, we are ready for some examples.*1719

*The first one: we want to find the zeroes of f(x) = 3x ^{2} - 23x + 14.*1721

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

*So, you plug in 0, because we are looking for when f(x) is equal to 0.*1731

*0 = 3x ^{2} - 23x + 14; we know we are going to be looking for 0 =...something where it is going to be (_x + _)(_x + _).*1735

*What are we going to slot in there? Well, we notice that here is 3; the only way we can break up 3 is 3 times 1.*1751

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

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

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

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

*So, we are going to have to plug in either 1 times 14 or 2 times 7.*1783

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

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

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

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

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

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

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

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

*So, at this point, we break this down into two different possibilities: either 3x - 2 = 0, or x - 7 = 0.*1847

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

*where we will have found a root where the whole expression will be equal to 0.*1861

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

*Great; if f(2) = 0, factor f(x) = x ^{3} - 7x + 6.*1881

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

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

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

*So, let's set it up like normal: 0 = x ^{3} - 7x + 6; but what we just figured out here...we know that there is a factor of (x - 2).*1917

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

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

*Well, we know that...what is in front of this x ^{3}? It is effectively a 1.*1941

*So, if there is a 1 in front, we have x times x ^{2}; whatever goes into this blank is going to determine what coefficient is in front of it.*1944

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

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

*So, the constants that we have here are -2 and whatever goes into that blank; so it must be that -2 times _ here becomes 6.*1968

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

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

*We think about this one, and we know that we want to have 0x ^{2} come out of this; there are no x^{2}'s up here.*1984

*So, we have + 0x ^{2}; so whatever we put into this blank must somehow get us a 0x^{2} to show up.*1992

*So, x times x ^{2} is x^{3}; so we are not going to worry about that.*2001

*But x times _x is going to be x ^{2}; let's do a little sidebar for this.*2005

*x times _x will become _x ^{2}; and -2 times...we already filled in that blank...1x^{2} is going to be -2x^{2}.*2010

*Now, we want the 0x ^{2} out of it; so it must be that, when we add these two things together, it comes out to be 0x^{2}.*2023

*What does this have to be? It has to be positive 2x ^{2}.*2031

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

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

*We check this out and do a really quick check; so x ^{3} + 2x^{2} - 3x - 2x^{2} - 4x - 6...*2053

*x ^{3} checks out; 2x^{2} - 2x^{2} cancel each other; that checks out.*2067

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

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

*We factored it properly; so at this point, the only thing that we have left to factor is the x ^{2} + 2x -3.*2087

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

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

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

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

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

*We did another check in our heads really quickly.*2136

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

*set each world equal to 0: x - 2 = 0; x - 1 = 0; x + 3 = 0; so we have x = 2; x = 1; x = -3.*2152

*Those are all of the roots for this polynomial.*2167

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

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

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

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

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

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

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

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

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

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

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

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

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

*Why? x = -2; we move that over; we get x + 2 = 0; x = +2...we move that over and we get x - 2 = 0.*2261

*So, we get (x + 2)(x - 2); but if we multiply those two together, we just have a degree of 2, and we want a degree of 4.*2268

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

*So, if we introduced multiplying just by x twice, we would have introduced a root at x = 0.*2284

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

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

*because each one of these will now have a degree of 2.*2302

*Alternately, we could have done this as (x + 2) to the 1, (x - 2) cubed, or (x + 2) ^{3}(x - 2)^{1}.*2305

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

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

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

*and the maximum peaks/valleys is one less; n - 1 is 3 - 1, is 2.*2339

*So, the maximum number of roots is 3; the maximum peaks/valleys is 2.*2350

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

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

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

*The final one: for this one, we think, "Oh, 10 ^{3}, so it is 3!"--no, we have to remember that this is not a variable.*2379

*This here is a variable; so it is x ^{1}, so its degree is just 1.*2388

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

*and the maximum peaks/valleys are going to be one less than 1, so 1 - 1 = 0.*2405

*Now, why is that the case? Well, think about it: 10 ^{3} - 5...10 cubed is just some constant;*2415

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

*x - 5...it will intersect here; but does it ever go up and down--does it ever undulate in weird ways?*2429

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

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

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

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

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

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

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

*All right, see you at Educator.com later--goodbye!*2466

1 answer

Last reply by: John Stedge

Mon Aug 20, 2018 4:56 PM

Post by John Stedge on August 20 at 04:55:37 PM

i^2=-1

2 answers

Last reply by: Tiffany Warner

Wed May 25, 2016 7:17 PM

Post by Tiffany Warner on May 25, 2016

Hi Professor!

I am really struggling with one of the practice questions given.

Find all the roots of f(p) = 2 p4 ? 8p3 ? 14p2 + 44p + 48 given that f(3) = 0 and f(?2) = 0.

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

(p-3)(p+2)

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

I figured expanding the two factors would make it more simple and the steps did reinforce that idea.

So we have (p-3)(p+2)(_p^2+_p+_)

Which expanded becomes

(p^2-p-6)(_p^2+_p+_)

They show us in the steps what to fill in those blanks with. I understand how they got -8. (-8)(-6)=48

I also understand how they got the coefficient of 2 in front of p^2.

However I have no idea how they went about filling in that middle blank in front of p. I’m clearly missing something. If you could provide some guidance, I’d be very grateful!

Thank you!

0 answers

Post by Jamal Tischler on December 21, 2014

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

3 answers

Last reply by: Professor Selhorst-Jones

Fri Dec 12, 2014 8:38 PM

Post by abendra naidoo on December 12, 2014

Hello,

These are good teaching modules.

I have 2 questions:

1)How is an equation different from a function.

2) I can correctly operate exponents and logs but still don't see why logs are necessary? Why not just use exponents - is it because log tables exist? How can I better understand this concept?

1 answer

Last reply by: Professor Selhorst-Jones

Mon Oct 20, 2014 11:59 AM

Post by Saadman Elman on October 19, 2014

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