 Well, welcome back, class, to episode 11. This time, we will be discussing finding roots of polynomial functions. This is episode 11 of Math 1050, College Algebra at Utah Valley State College. And I'm Dennis Allison. I teach in the mathematics department here. What we'd like to do today is to see how we can take a polynomial of degree higher than 2. That is something bigger than a quadratic. And how we can factor the polynomial so that we can find the factors. From the factors, we can find the x-intercepts. And from the x-intercepts, we can draw the graph. So there's sort of a chain of ideas that we're going to go through. Now along the way, we'll be using synthetic division. We'll be using the factor theorem from the last episode. And we will ultimately be able to factor some rather large and complex looking polynomials. Let's begin with the polynomial. Oh, let's look at the objectives, first of all. We'll be discussing today Descartes' Rule of Science. This is the same Descartes who invented the Cartesian plane back in the 17th century. We'll also look at something called the rational root theorem. We'll look at the factor theorems. I think that's a misprint on the screen. We'll look at the factor theorem and how it applies. And then finally, we'll look at some applications of all of this to, shall we say, real-world applications. OK, here we have a cubic polynomial. And I think I can factor this by grouping. In the last episode, we talked about factoring by grouping. And I think I should show you an example of that just to refresh your memory how this goes. In factoring by grouping, the idea is that you group terms into, in this case, two terms and two terms. And I'm looking for a common factor in each pair. In the first pair, I think the common factor is 4x squared. And if I factor out a 4x squared there, that's going to be x minus 2. And in the second pair, the best I can do is factor out a negative 1. And that's going to give me x minus 2. OK, now when I look overall at this polynomial, I have a common factor of x minus 2 that I can bring out in front. And that leaves me with, let's see, who can tell me what would go in that space? 4x squared minus 1. 4x squared minus 1, exactly. And now I can factor this further because this is the difference of two squares. So I can factor that to be x minus 2 times 2x plus 1 times 2x minus 1. So in other words, I've been able to factor this polynomial by using factoring by grouping. But if you just make up a polynomial, say a cubic polynomial at random, then the chances of you coming up with an x minus 2 that's a common factor on each in each half would be rather slim. So factoring by grouping will sometimes factor a polynomial, but it's not going to solve all of our problems. By the way, if I were going to graph this polynomial right here, where would it cross the x-axis? Negative 2 and positive 2. Positive 2, yeah. Negative 1 half. Negative 1 half and positive 1 half. And positive 1 half, exactly. So those would be my 3x intercepts. And let me ask you another question about this while we have it up here. What is the multiplicity of each root? 1. Each one is multiplicity 1 because you see each factor appears only one time. So that tells me that when I draw the graph of this polynomial, it's going to go right through 2. It'll go right through negative 1 half, and it'll go right through plus 1 half. Let's just try graphing that before we leave it because we'll be graphing other polynomials later today. This is a good way to sort of introduce that. OK, we said that the x-intercepts were going to be at x equals 2 and x equals plus or minus 1 half. And just as importantly, we need to know that the multiplicity of each root is 1 for each one of those. So when I go to draw the graph, I'll locate those x-intercepts at 2, at 1 half, and at negative 1 half. And let's see, because this lead coefficient is positive, that says the in behavior on the right-hand side is going up. So I'm going to draw my graph from here going up. Because there's a 4 on there, you might say, isn't there a stretch of 4? Well, it's rather difficult to indicate that anything's being stretched for. It's just very steep. So I'm going to disregard the 4 in the interest of saving time. And let's see, by the way, what's going to be the in behavior on the left-hand side over here? Will it be going up or will it be going down? It'll be going down. And how do you know that, Stephen? You're absolutely right. Because the lead term has an odd power. Right. This is an odd-degree polynomial. And so the in behavior is opposite on left and right. So if it goes up on the right, it goes down on the left. And once again, I know it goes up on the right, because that's a positive x cubed. And you see, if you start picking rather large positive values like 5, 10, 20, whatever, that cubed term dominates all others. And if that cubed term wants to be positive, everything goes positive. So the graph goes up. This one dominates. In the middle, when you're close to 0, the other terms have a little bit of voice. And they're the ones that make it wiggle and go up and down. On the other hand, if I pick a negative number like negative 5, negative 10, negative 20, the cube of a negative number gives me a negative answer. And if that wants to be negative, the other terms can't stop it, because that power is so high. OK, back over here to my x-intercept. The multiplicity is 1. So my graph goes right on through that point. It comes down. I don't really know how far, probably more than that. But then I come back up. I go right through that point, because multiplicity is 1. I turn. And I go right through this point. And now we go down. This is the graph of p. Now, what I've lost to your member is some accuracy, because I don't really know where that lowest point is, where this highest point is. But I'm just getting the general shape. That's all we're after. OK, but now what do we do if we have a polynomial function that will not factor by grouping? And if I want to sketch a graph of it, I need to be able to factor it. And so we now want to discuss how you can factor higher degree polynomials, that is degree higher than 2, using a combination of tools that we have learned about and that we're going to learn about today. Let's take, for example, this polynomial. Suppose I have p of x equals, I'll just make something up here. Suppose this is 5x to the fourth plus 3x cube minus 2x squared minus 7x plus 1. Now, if I count the sign changes as I move from left to right, I notice, let's see, plus, plus, minus. I get a sign change right there, minus, minus, plus. I get a sign change right here. So I would say that this polynomial function has two sign changes. And if you count the number of sign changes, this gives you a clue as to how many positive x intercepts or positive roots this polynomial has. If I have two sign changes in this case, what I can conclude from that is there are either two or zero positive roots. You might say, wait a minute. Where'd the zero come from? Well, if you count the number of sign changes and if you reduce it by 2 to 0, then there are either two or zero positive roots. If I had had three sign changes, there would be either 3 or 1. I'd just keep reducing it by 2. If there had been four sign changes, then there would be 4 or 2 or 0. Again, I just keep reducing it by 2. Now, I don't know which one of those is the case, but it gives me an idea, sort of a premonition, of how many roots to be looking for. There's a name for this rule. This is referred to as Descartes' Rule of Signs. Descartes' Rule of Signs. You will see me using this quite a bit in some of the examples that come up, so we'll have ample opportunity to see how this goes. By the way, this only told me how many positive roots there are, how many negative roots are there. Well, if I substitute in a negative x in place of an x, then I get 5 times negative x to the fourth plus 3 times negative x to the third power minus 2 times negative x squared minus 7 times negative x plus 1. And if I simplify that, let's see, to the fourth power, that'll still be 5x to the fourth. But if I cube a negative, that's going to be a negative 3x cube. What will this term be? Positive is negative 2x. It's going to be a negative 2x squared, yeah. Because when you square the negative, you do get a positive, so there's no change in what I had before. This will be a plus 7x and a plus 1. Now, if you count the sign changes after you've replaced x with negative x, then that gives you an indication of how many negative roots there can be. Sign change here, sign change here, that's it. I have two sign changes once again. So because I have two sign changes this time, that tells me there are either two or zero negative roots in this polynomial function. OK. But let's see, why don't we discuss why if you put in a negative x, you would get the negative roots. Do you remember back in, I think it was like episode three, we talked about transformations of functions where you can translate a function up or down. You can shift a function left or right. You can stretch it, compress it. Do you remember what happens if you replace x with negative x? Do you remember what that does? Maybe that's something we didn't talk about. What it does is it flips the graph around the y-axis. So when I put in a negative x in place of an x, it turns the graph completely around. And so what used to be negative roots become positive roots. And so by turning the graph around, I'm now counting positive roots, but the graph has been reversed. So for the original function, those are the negative roots. So graphically, this flips the graph around the y-axis, I count the number of positive roots, and they become the negative roots of the original function. OK, let's take a problem to solve now. And I'm going to be using Descartes' Rule of Signs. And I'm also going to be using the rational root theorem. And I'm going to go over here to the marker board so that we can see how to solve a problem. Let's put the next graphic up on the screen. And I'm going to be using this theorem along with Descartes' Rule of Signs. And the theorem says this. It's called the rational root theorem. It says, let p of x be a polynomial of the form a sub n times x to the n plus a sub n minus 1 x to the n minus 1 all the way down to a sub 0. Suppose this is a polynomial with real coefficients. And the coefficients are a sub n, a sub n minus 1. And then the constant polynomial term is a sub 0. Now, a rational root of p must be of the form p over q, where p is an integer divisor of a 0, and q is an integer divisor of a sub n. OK, now you know that's kind of a mouthful when you see that on the screen. Let me just explain to you what that's all about. First of all, I talk about a rational root p over q. What I mean by a rational root is it's a number that is a rational number that is a 0 of the polynomial. For example, 1 half is a rational number, negative 7 thirds is a rational number, and 6 fifths is a rational number. And does anybody know why those are called rational numbers? Because they can be written as a ratio. Yeah, because you see in here we have the word ratio. These are ratios of integers, 1 divided by 2, negative 7 divided by 3, 6 divided by 5. Those are called rational numbers. And even a number like 4 is a rational number, because it is a ratio of two integers. It's 4 over 1, or you could also write it as 8 over 2, or you could write it as negative 12 over negative 3. There are lots of ways of expressing 4 as a rational number. So when I say a rational root, I mean basically it's any fraction or any integer, positive or negative. So the rational root theorem gives me a way of finding all the rational numbers that are roots of a given polynomial. And here's how we do it. Let's go to the next graphic. And here's an example of a problem that I'd like to solve. Find the rational roots of the polynomial function p of x equals x to the fourth minus 6x cubed plus 8x squared plus 6x minus 9. Then factor the polynomial and we'll sketch its graph. Now while that graphic's up there, let me just copy that down up here so we can solve it. Let's say we have x to the fourth minus 6x cube plus 8x squared plus 6x minus 9. OK, let's come back to the board. And when we first look at this, we think, oh my gosh, there's no way we could factor that. It's way too complicated. But I think if we follow a certain procedure or algorithm, we'll find that it factors fairly nicely. In this case, my lead coefficient here, the coefficient there is a 1. So I would say that a sub 4 is 1. That's the leading coefficient. And the constant coefficient over here, that's what I call a sub 0 in the theorem that was on the screen a moment ago as negative 9. Now if I'm looking for a rational root, say p over q, then p must be a divisor of the constant term, negative 9. And q must be an integer divisor of the lead term, 1. So let's just list what are the possible choices for p and q. Well, let's see, what integers will divide negative 9? Well, there's plus or minus 1. There's plus or minus 3. And there's plus or minus 9. All of those integers, six of them, all of those integers will divide negative 9. You notice I don't include their 4, because 4 won't divide negative 9 evenly. I don't include 7, because 7 won't divide it. So only those six numbers. Now for q, these must be divisors of the number 1. So that would only be plus or minus 1. Those are the only integers that will divide 1. So if I list my choices for p over q, I think I'll come over here and list those to the side. So I have some room to work in a moment. p over q, well, if I take every choice of p and put it over every choice of q, that'd be either plus or minus 1 over 1, plus or minus 1 over negative 1, I think those ratios end up just being plus or minus 1. Then there's plus or minus 3 over either plus or minus 1. And I think that's just plus or minus 3 in this case. And there's plus or minus 9 over plus or minus 1. So I end up with a list in this example that ends up just being the same thing as the p's, because my denominators are just either 1 or negative 1. So if I'm looking for a rational number that is a root of this polynomial, it has to be one of these rational numbers. And in fact, these are all integers, because my q's were all 1's. Now, which one of those numbers would be a 0 or a root of the polynomial? Well, for that, I'm going to go back and recall the factor theorem from the last episode. And if you remember, we said that the factor theorem says x minus c is a factor of p of x if and only if p at c is 0. If and only if p of c is 0. You say, well, Dennis, how will you know if p of c is 0? We're like, how would I know that if p of 1 is 0? One way to do it would be to plug it in. But another way to do it would be to use the remainder theorem. Now, the remainder theorem that we also saw in the last episode said that if you're given a polynomial function, p of x, and you want to calculate p at c, it's merely equal to the remainder when we divide by x minus c. So I'm going to put these two ideas together to find a shortcut for determining which one of those are roots of the original polynomial. Let me erase this, and we'll see how this goes. So the first question is, is p at 1 equal to 0? I'll put a question mark there, because I don't know. But the fastest way to find out what p of 1 is is to use synthetic division. So I'm going to use synthetic division and put 1, negative 6, 8, 6, and negative 9. And out here, I'll be dividing by x minus 1. And if I want to add, I should put in the 0 or the root for x minus 1, which is 1. So I'll just put that number right here. And my division says, first of all, bring down a 1. 1 times 1 is 1, and I add now, and I get negative 5. 1 times negative 5 is negative 5, and I add. 1 times 3 is 3, and I add. 1 times 9 is 9, and I add, and my goodness, right off the bat here, I come up with a 0. So that tells me that p of 1 is 0. The answer for this is yes. And of course, we're very excited about that. You say, well, how's that going to help you? Well, if 1 makes this polynomial 0, then x minus 1 is a factor. So I know right away that this polynomial factors into x minus 1 times something. Now, I shouldn't have erased so much there, but this is the quotient when you divide by x minus 1. So this will be x cubed minus 5x squared plus 3x plus 9. I've been able to factor my fourth degree polynomial into a linear and a cubic polynomial. Now, the next step is to turn my attention to the cubic and see if I can factor it. You notice there's still a 9 here. There's still a 1 here. So this is still giving me all possible rational roots. And so I go into this list, and I just pick a number and decide which one I'd like to see if it's a root. Now, just because x minus 1 is a factor doesn't mean it couldn't be a factor again. So I think I'll test that plus 1 one more time. So I'm now using this polynomial 1, negative 5, 3, and 9. And I'll put a 1 outside again, because I'm checking to see if I could maybe have multiplicity 2 on this factor. So I bring down a 1. And I proceed with synthetic division. It goes like this. Negative 4, negative 4, negative 1, negative 1, and 8. What's the most significant thing about 8 that you can say in this problem? It's not 0. 8 is not 0. Yeah, you thought about that. 8 is not 0. And so because I didn't get 0, that means x minus 1 is no longer a factor of the cubic. It was a factor of the fourth degree polynomial. OK, so I think we've gotten all we can out of the plus 1. Let's go to the negative 1. And my coefficients are 1, negative 5, 3, and 9. And now I'll use synthetic division to check this root or this 0. By the way, if this turns out to be a 0, what is the factor that corresponds to it? x plus 1. x plus 1 would be my next factor. But we have to see what the remainder is going to be. So I bring down a 1, I multiply, I add, I multiply, I add, I multiply, uh-oh, look what happens. We get a 0. We're really getting excited now. So that tells me that I have another factor. And the new factor is x plus 1. And you might say, now, there's a negative 1 there. Shouldn't you be writing x minus 1? Well, if you remember in the previous episode, we said that we would change the sign of that number from plus 1 to negative 1 so that I could do my addition. So when I go back, I have to change the sign back. And I write that as x plus 1. It was also, I could take the shortcut of using addition here, times something on the end. Well, these coefficients tell me what is the quotient when I factor out the x plus 1. So what polynomial should I put there? Can you guess? Let's see, Susan, what would you say? 1x cubed minus 6x squared plus, no, 1x squared. Let's see, now, that's the remainder. So I take this to be the constant term. The linear, this is a quadratic. 1x squared minus 6x plus 9. Right, there you go. x squared minus 6x plus 9. The 1 up here refers to a cubic. The 1 down here refers to a quadratic. Because this is after the division has taken place. So I have a linear factor, a linear factor. And now I turn to this one. But you know, now that this is a quadratic, I can factor this by sight, knowing what we know from elementary and intermediate algebra. So I don't have to use the synthetic division anymore. I just factor this straight away. And can anyone tell me how you factor x squared minus 6x plus 9? x minus 3 times x minus 3. Yeah, x minus 3, I'll just say squared. So there are three roots for this polynomial, for the original polynomial. The roots are 1, negative 1. And if I list those roots twice, I'll say 3 and 3. So there are actually four roots, but only three different ones. And you notice all of those did come from this list. And I just take out one root at a time. Every time I find a root, I get a factor that reduces the degree of the polynomial. And I keep reducing it until I get a quadratic. Then I factor that by sight, if I can. And I have the complete factorization. We have just factored a fourth degree polynomial. Now, I think I've mentioned this before. But when I was in high school, before I first saw this, I would have thought, hey, this is way too complicated for me. I can't do that. But it's really not that difficult if you know the rational root theorem and if you know synthetic division. And also, if you know the factor theorem and remainder theorems, you see, we're applying a lot of information to solve these. Let's take a new problem. OK, here's another example I'm going to take out of a college algebra textbook. We're given a polynomial function. I think I'll call this one f of x. So we don't call them all p. This says 2x cubed plus 7x squared plus 4x minus 4. Let me put a 4 on that. And let's say my goal here is to factor this polynomial. And then I'll draw the graph of this one as well. So I have to decide on what are the possibilities for p and what are the possibilities for q. Now for p, p has to be a divisor of the constant term, negative 4. So which integers will divide negative 4? 1 plus or minus 1, plus or minus 2, plus or minus 4. Right, plus or minus 1, plus or minus 2, plus or minus 4. And for q, I have divisors. I have to have divisors of 2. I think that would be plus r minus 1 and plus r minus 2. So we're obviously taking positive and negative divisors. So my choice is for p over q. If I put these numbers over these numbers, there's plus or minus 1, and there's plus or minus 1 half. There's plus or minus 2, and there's plus or minus 1. But we already have those listed, so I don't need to list them again. And there's plus or minus 4, and there's plus or minus 2. But we already have those listed, so there's no need to list those again. So it looks like I have only eight routes to test. So rather than testing all possible integers and rational numbers, I have to test only those eight. So already, we've reduced it to a very small pool. Now, which one should I test first? The ones or the halves or the 2s or the 4s? Well, I would say they're all equally likely at this point. So stick with the simplest numbers to divide by. I would go with the one. So I'm going to divide, actually, I'm dividing by x minus 1 into the polynomial 2, 7, 4, and negative 4. And the remainder is going to come up there. Well, let's see. I bring the 2 down, I multiply, and I add, and I multiply, and I add. I think we can see this is going nowhere. We're going to get a 9. And what's significant here is 9 isn't 0, so I can eliminate 1 from my list. We're down to seven more possibilities. OK, I'll erase this. And over here, I'll just put an x through the 1 to know that we're 2. So we remember that we've already checked that one. Let's try negative 1. And my coefficients are 2 and 7 and 4 and negative 4. So I bring down the 2 and I multiply, negative 2. Negative 5. And I multiply. And I add to get negative 1. And I multiply. And I get negative 3. Well, that means I can rule out the negative 1. So we're down to six more possibilities. Let's go to, I'd say let's go to the 2. Skip over the half for a moment. So let's try dividing by 2. Negative 2, 7, 4, negative 4. 2, 4, 11, 22, 26, 52. Doesn't look good, does it? Looks like it's going to be a 48. So I don't get a 0. I can eliminate the plus 2. OK, let's try going to the negative 2. If I put a negative 2 out here, we have 2, 7, 4 and negative 4. So 2, negative 4, 3, negative 6, negative 2. Negative 2 times negative 2 is 4. The remainder is 0. OK, that's what we've been looking for. I found a number that makes the remainder 0. So what that means is if I take f and evaluate it at negative 2, this function value will be 0. Now by the factor theorem, if negative 2 makes this polynomial 0, then x minus negative 2 is a factor. x minus negative 2, we're talking about x plus 2. So my factor is x plus 2 times, let's see, that was a cubic. So this would be a quadratic polynomial. And I can pick up the quadratic polynomial by just looking at this expression down here. That'll be 2x squared plus 3x minus 2. And even better, I've reduced the cubic to a quadratic. So if I can now factor that by sight, I have the complete factorization of my polynomial. And let's just see, x plus 2 times, let's see, there's a 2x squared, so I know that if I'm going to try something, I'll put a 2x and an x. There's a negative over here. What does that negative tell me about the signs in here? One's positive and one's negative. One's positive, one's negative. But I don't know if the positive should go with the 2x or if the positive should go with the 1x. Does anybody see a way to factor that to make this work out? I think if we put a plus 2 here, and if we put a minus 1 there, I think we have it. Now you notice I have the same factor twice. So this is x plus 2 squared times 2x minus 1. This is the complete factorization of the original polynomial. What would you say would be the roots of this polynomial? Negative 2 and 1 half? Yeah, negative 2. I'm going to write negative 2 twice because it comes from two different factors. And for this one, it's a plus 1 half. Now we're all of these numbers in my list of choices for p over q. Negative 2, yeah, there's negative 2 right there. And plus 1 half, yeah, there's plus 1 half right there. So we didn't get all of these ratios to be roots, obviously, but among those, we find all of the possible roots. Now the thing I wanted to do that goes beyond this is to draw the graph of the polynomial function. So let's just make a space here and see what the graph would look like. Now that we have factored the polynomial, I know that the roots are negative 2 and plus 1 half. So here's negative 2. And here is plus 1 half, right about there. This negative 2 has multiplicity 2. So that tells me that the graph is going to be turning when it gets to the x-axis there. And at plus 1 half, it's going to pass through the x-axis. So my graph, let's see, going back to the beginning, that's a positive 2 x cubed. So my graph goes up right here. And my graph passes right on through that point. But it turns, and it has to come back to negative 2. And it turns there, sort of like a parabola, and it goes down. And that is a rough sketch of the function f of x equals 2x cubed plus 7x squared plus 4x minus 4. Now, of course, it isn't totally accurate, because I don't know where that lowest point should be. There's no indication here where the lowest point should be. But what is important is I have very quickly now sketch the graph based on the factorization. So given a polynomial, I first look for possible candidates for rational roots. Then I test them by synthetic division to see what the rational roots are until I get a complete factorization. From that, I'm able to draw my graph. Let's take one more example. OK, for the next polynomial, let's choose one that's an even higher degree. This will be polynomial p of x equals 2x to the fifth plus 5x to the fourth minus 8x cubed minus 14x squared plus 6x plus 9. Now, here's my goal. I want to factor the polynomial, and then I want to sketch its graph. I'll be using the rational root theorem. I'll be using Descartes' Rule of Signs. I'll use Descartes' Rule of Signs for the first time here. And then we'll sketch the graph on that basis. Well, let's see. First of all, if I use Descartes' Rule of Signs, you remember this gives me an indication of how many positive and negative roots there are. So what I do is count the number of sign changes. So define the possible number of positive roots, or positive zeros. How many sign changes do you see there as you go from left to right, class? Two. Looks like two. There's a sign change here, and there's a sign change here. Otherwise, as I go across, the signs stay the same except for those two places. So two sign changes, that means there are either two positive roots or there are no positive roots. So either two or none. Two or none. Now, you might say, well, Dennis, how is this going to help you in speeding up the process of solving this problem? Well, if it turns out that I find a positive root, then I know that there's another one for me to look for. So I know that there's one more to look for. And once I found two positive roots, I don't need to check any more positive ratios, p over q. On the other hand, what if there had been no sign changes here? If there were no sign changes, there would be no positive roots. And therefore, when I go to my list of possible roots, p over q, I wouldn't check any of those positive ones. And that immediately cuts the list in half because I would know there are no positive roots. But in this case, we could have two. Now, how many negative roots? Well, do you remember what we have to do to find the negative roots? Change x to negative x. Right. We're going to look at p of negative x. So let me just write down p of negative x first. That's 2 times negative x to the fifth plus 5 times negative x to the fourth minus 8 times negative x cubed minus 14 times negative x squared plus 6 times negative x plus 9. Well, if I simplify that, this first term will be a negative 2x to the fifth. The next term is plus 5x to the fourth. The next term will be a plus 8x cubed. And then I have a negative 14x squared minus 6x plus 9. And if I count the sign changes this time, I'll have an idea or have some information about negative roots. Sign change, sign change, sign change. I get three sign changes. So the number of negative roots will be either 3 or 1. I'm not saying 3 is a root. I'm saying that there could be three negative roots or one negative root. But this tells me there is at least one negative root in this problem. OK, I think I'll just put this information over here on the side so that I can use that space to write. We said this was either 2 or 9 for the positive roots. And for the negative roots, it's either 3 or 1. Either 3 or 1. OK, so now we've gotten what we can from Descartes' rule of signs. So now I go to the rational root theorem. By the way, in some books, this theorem is called the rational zeros theorem. So it depends on what book you're looking at. For example, you may go to the library and check out some other college algebra books. You'd find all this information in there with more examples and more problems with solutions. But they may have the rational zero theorem listed. Let's see, what are the choices for p and for q? Well, p has to be a divisor of the constant term. So that would be plus or minus 1, plus or minus 3, plus or minus 9. q has to be a divisor of the lead coefficient. That's plus or minus 1 or plus or minus 2. So what are the possibilities for p over q? Well, actually, this time there are quite a few. I can put the ones over the ones, plus or minus 1. I can put the ones over the twos, plus or minus 1 half. 3 over 1, that's plus or minus 3. 3 over 2, that's plus or minus 3 halves. 9 over 1 is plus or minus 9. And 9 over 2, so plus or minus 9 halves. How many possible rational roots do we have listed there? 12. 12 of them. You might say, Dennis, 12. That's an awful lot of roots to have to test. Well, you know, without this, I'd have to test everything, 3 fourths plus 5, negative 7. I mean, all those rational numbers have been reduced to only 12. And if we're lucky, we'll come up with a root for the original polynomial fairly quickly. In the last example, it took me three or four stages before I found a root. This time, hopefully it'll go faster. Let's just see. Which of those numbers would you suggest that we test first? 1. 1, yeah. Why not go with the easy one? I don't know why anybody would want to check negative 9 halves first, unless you're just a sadist. So let's just go with the plus 1. So 1 goes here. The coefficients are 2, 5, negative 8, negative 14, 6, and 9. So I divide 2. 2 makes 7. Negative 8 and 7 makes negative 1. Negative 1 and negative 14 make negative 15. Negative 15 added to 6 is negative 9. Negative 9 and 9 makes 0. Well, we got lucky. Right off the bat, we have found a root. So I get a lot of information out of that. Not only is 1 a root, I get a factor. What is a factor of this polynomial? X minus 1. X minus 1. Very good. And not only that, but when you factor out X minus 1, what's the remaining polynomial going to be? 2x to the fourth. Yeah, see, it's going to be 1 degree less, 2x to the fourth. That's the fourth degree. Then what, Stephen? Plus 7x cubed. Right. Minus x squared. Minus x squared, right? Minus 15x. Minus 15x. Minus 9. Minus 9, exactly. OK, so I've been able to factor the fifth degree polynomial down to a linear and a fourth degree polynomial. So now my attention turns to this guy. I want to factor that one. OK, well, I go back to my list. And just because 1 made what was a root once doesn't mean it couldn't be a root again. So I don't want to accidentally pass over that. So I'm going to try dividing by x minus 1 again. I'll put a 1 out here. And I have 2, 7, negative 1, negative 15, negative 9, just like before. And I'm wondering if the constant term could be 0 again. It could happen. Let's see, 1 times 2 is 2. Adding, I get a 9. 1 times 9 is 9. Adding, I get 8. 1 times 8 is 8. Adding, I get negative 7. 1 times negative 7 is negative 7. Adding, I get 0. No, just trying to see if you're paying attention. No, I get negative 16. Oh, that's too bad. So what that means is I don't have any more factors of x minus 1. Well, OK, so we have dispensed with the plus 1. What do you think we should try next? Let's see, Matt. Well, you could try plus 3. Try plus 3? OK, why don't you say try plus 3? Well, one thing, it might be easier than a fraction. Also, if that ends up being a root, there are no more positive roots. Oh, OK, yeah. So going back to Descartes' rule of signs, we said there would be either 2 or no positive roots. We've already found one positive root. C1 is the root. x minus 1 is the factor. So we know that now there's going to be another positive root. So Matt's thinking on sort of like divide and conquer. He's going to start zeroing out on the other positive root. So let's go to 3, see what happens. You don't have to follow the same strategy at home if you want to try one of the other numbers next. That's fine. But Matt suggests that we go with the 3. Now, if I get a remainder of 0, then x minus 3 is my next factor. So I have 2, 7, minus 1, minus 15, minus 9. So I bring down to 2. I get a 6. I add 13. 3 times 13 is 39. I add and I get 38. 3 times 38 is, oh my goodness, that's going to be 114. This is going to be a 99. I think we can see that we're not going to get a 0 there. We don't need to bother multiplying the rest of this out. That's not going to be a 0. So I can rule out the 3. Let's see, we've eliminated the plus 1. And we found it was a root. We've eliminated the plus 3. So Matt, if we follow your advice, what should we try next? Well, you could try plus 1 half or plus 9. Yeah, we go to plus 1 half or plus 9. Now, see, when you get into the 9s, that means the multiplication is going to get pretty big. Why don't we try the 1 half and see what happens? If the 1 half doesn't work, we'll go to the 9 or the 9 halves. So I'm going to put a plus 1 half out here. And the coefficients again are 2, 7, minus 1, minus 15, and minus 9. So let's see what happens. I bring down to 2. I multiply. A half times 2 is 1. I add and I get an 8. And then a half times 8 is 4. And I add and I get a 3. And a half times 3 is 3 halves. And I add, let's see, OK, what is minus 15 plus 3 halves? This is minus 30 halves plus 3 halves. Negative 27 halves. Minus 27 halves. Minus 27 halves. And when I multiply there, I get minus 27 over 4. That's not going to be a 0. So we can rule out the 1 half. Well, let's see. Let's go to, I think rather than going to the 9s and the 9 halves, let's try some negative numbers, OK? Because I think the arithmetic is going to be simpler than if I go to those bigger numbers. So let's drop back and start looking for some negative roots. I'm going to go to negative 1. We're still trying to find roots of this polynomial. 2, 7, negative 1, negative 15, negative 9. Bringing down to 2, I get a negative 2. Adding is 5. I get a negative 5. Adding is negative 6. Multiplying, I get 6. Negative 15 and 6 is negative 9. When I multiply, oh, this is good. We're going to get a 0. So when we started looking for a negative root, we found one right away. This tells me that my factorization is x minus 1 times, this gives me an x plus 1, times. Now I need a new factor to go in here. Let's see, what would be the next factor? Jeff, what would you say? It'd be 2x squared. I can't. I think that would be a cube, actually. OK, 2x cubed plus 5x squared minus 6x minus 9. Minus 6x minus 9. OK, so now I turn my attention to the cubic factor. And you know at this point, a person might say, gee, should we start all over again? And should I try a plus 1? Well, you see, when I found x minus 1 was a factor and it wasn't a factor twice, it won't be a factor coming out of this one because it wasn't a factor a second time in the earlier polynomials. But I could have x plus 1 as a factor again. So I'm going to try dividing by negative 1 one more time into 2, 5, negative 6, and negative 9. The reason I do that right away is because many times I think students overlook the possibility that you could have a multiple root. And as soon as they take out the x plus 1, they move on to new numbers. And perhaps all of the other roots are still negative 1's. So we don't want to overlook these. Multiplying negative 2, I get a 3. Multiplying negative 3, I get negative 9. Multiplying, I get a 9, I get a 0. It is a multiple root. So this tells me I have x minus 1 times x plus 1 squared because now we have two of those roots times a quadratic. You notice we had a 5th degree, 4th degree, 3rd degree. Now it's a quadratic. And this quadratic is 2x squared plus 3x minus 9. What's good about a quadratic? You can factor it by site. Yeah, if it's going to factor, we should be able to factor it by site. So let's see if we can dispense with a synthetic division. We used it while we needed it, but we don't need it anymore. And I'm thinking that this will factor into a 2x and an x. And I think this should be a plus 9 and a minus 1. No, that's not going to work. How about 3 and 3? Let's see, a plus 3 over there and a minus 3 here. That'll do it. Yeah, that's going to give me a negative 3x and a plus 6x. Makes a plus 3x. Well, if we kind of sum it all up, what turns out to be the roots for this problem? They are 1. There's negative 1 and negative 1 again, because it came up twice. What's the root here? Plus 3 halves. 3 halves, yeah, 3 halves. You know, 3 halves would have been my next choice. If we had followed math procedure, I bet we would have found that root next. And then finally, negative 3. Finally, negative 3. So these are the roots. We have four different roots, but we do have a total of five roots. By the way, if you have a polynomial of degree 5, you would expect that you could have as many as five roots, maybe a like, maybe some of them are different. But you could have that many roots. If I were going to graph this polynomial, the graph of the function would look like this. Let's see, here's the x-axis. And my roots are at 1 and at negative 1 and at 3 halves and at negative 3. And this is a positive 2 x to the 5th. So I know my graph goes up on the right. And because this is an odd degree, I know it has the other in behavior on the other side. So I expect to see the graph going up here. And I expect to see the graph going down over there. Now when I come to 3 halves, it passes right through and comes back because this root has multiplicity 1. And at 1, I expect the graph to pass through because that root had multiplicity 1. But when I come to negative 1, I expect to see it turn and go back up. And then it's going to turn and go back down through that last intercept. And this is the graph of the function capital P. If I were to draw that on a graphing calculator, that's exactly what I would see, except I would have more accuracy on where those peaks and valleys are. I would do that. But I think we need to move on to the next graphic. The next graphic has to do with irrational roots of polynomial functions. And the purpose of this next example is to show you that not all of the roots have to be rational numbers. Some of them can be irrational numbers. And when I say irrational, I mean not a ratio, numbers that are other than fractions and integers. Like the square root of 2, like 5 plus the square root of 3, those numbers are irrational. So let's take that example. And I will copy it up here. And we'll see how it works. So this polynomial function in this example is f of x equals x cubed minus 2x squared minus x plus 10. OK, so let's try solving this one up here on the, oh, that should be a minus 4x squared right there. So let's try solving this to find its roots. I'm going to begin by listing all positive, the number of positive roots that it could have. And how many sign changes do you count there? Two. Two sign changes, yeah. So there's either 2 or none for the positive roots. To get the negative roots, I need to calculate f at negative x. And that's a negative x cubed minus 4x squared plus x plus 10. I went ahead and simplified that. I see only one sign change. So for the negative roots, it looks like there's only going to be one negative root. There's going to be exactly one negative root. OK, so with this information, let's try factoring it. So my choices for p are plus or minus 1, plus or minus 2, plus or minus 5, and plus or minus 10. Those are the divisors of plus 10. My choices for q are plus or minus 1. So my choices for p over q, if I take these ratios, is plus or minus 1, plus or minus 2, plus or minus 5, and plus or minus 10. OK, well, let's try these one at a time. I think we should begin with the plus 1. So for 1, I get 1, negative 4, negative 1, and 10. So I get 1, 1, negative 3, negative 3, negative 4. It doesn't look like we're going to get a 0 there. That's not going to be a 0. So let's go to negative 1. For negative 1, we have 1 minus 1, minus 5, 5, 4, minus 4. It's not 0. So both of the ones are out, the plus and the minus 1. Let's go to the 2. For the 2, we have 1, negative 4, negative 1, and 10. 1, 2, negative 2, negative 4, negative 5, negative 10. Oh, OK, we found a 0. So that tells me that x minus 2 is a factor. Oh, let's see. I better leave those numbers up there, because I want to use those in just a second. That's going to be x minus 2 times x squared minus 2x plus 5. That was the bottom row that I almost erased. Now, I don't believe this will factor by sight. So to be able to solve that, to find the roots of that, I'm going to set that equal to 0 and use the quadratic formula. So I want to find out, when is that quadratic 0? Well, x must be, let's see, negative b plus or minus the square root of b squared minus 4ac, all over 2a. So that's going to be 2 plus or minus the square root of 4 minus 4 times 1 times 5. Over 2a. So that's going to be 2 plus or minus the square root of negative 16 over 2. Now, of course, the square root of negative 16 is an imaginary number. And so these are complex numbers. So I have no more real roots for this problem. The only real root is 2. And the other two roots are complex numbers. So when I go to draw the graph, there's only one x-intercept, and that's going to be it too. OK, I'd like to finish today with an application in which we want to find the roots of a cubic equation, but the roots aren't so easily found is what we just did on the board there. OK, the problem is this. What if I take a circular cylinder, turn it on its side, and it's 4 feet, 4 feet wide. And on the end, I put half of a sphere here. I put a hemisphere. And on the other end, I put half of a hemisphere, or half of a sphere there, so that I get these sort of rounded ends on the end of the cylinder. I would like for the volume of this tank to be 100 cubic feet. Now the question is, what should be the radius of the hemisphere on either end? And that would also be the radius of the circular cylinder in order to have this volume. Well, we have to know that the volume of a sphere is 4 thirds pi r cubed. And we have to know that the volume of a circular cylinder tank is pi r squared times h. Now if I put those two together, the total volume gives me a function of r. That would be 4 thirds pi r cubed plus, now the h is 4, that's going to give me 4 pi r squared. Now these coefficients make it difficult for me to use the rational root theorem because these are not integers. I've got pi in there. So instead, what I'll do is use a graphing calculator. Now I have the calculator already set up to solve this if we can zoom in. Now I've already entered both of the functions that we want to use in the calculator. The first one is the volume function. I have 4 thirds times pi x to the third power plus 4 pi x squared. I'm using an x because the calculator won't accept r as a variable. The other function is the horizontal or the constant function, y2 equals 100. I want to see where the volume equals 100. So if I graph these, and I've already set the window, if I graph these, I have the two graphs intersecting at this point. My vertical axis is scaled off in 50 units. That's 50, 100, 150. So this is the horizontal line at 100. My horizontal axis is scaled in one unit, 1, 2, 3, 4, 5. So if I put this on trace, we are right about at that point. It looks like it's between 2.1297 and 2.18103. Let's try zooming in and take a closer look. So here comes a portion of the volume function, and here comes a portion of the constant function. I'll put it on trace one more time, and I'll move down to the intersection point. And it looks to me like we're just above 100 when we choose x to be 2.15. So roughly we want to make the radius 2.15 feet. Now, this is only an approximation. I could keep zooming in and get more and more accuracy on that answer, but this is an alternative to using the rational root theorem where things work out rather neatly. I'll see you next time for episode 12.