 Welcome to episode 10 of Math 1050, College Algebra. I'm Dennis Allison, and I teach in the math department here at UVSC. I have a bit of a cold today, so I was a tenor. I may be a baripuntone today, and tomorrow, if I'm a base, we're in big trouble. Let's look at the objectives for the course today. We want to look at, first of all, what's called the division algorithm. Then we'll talk about long division and a shortcut called synthetic division. Actually, those, I think, were covered in intermediate algebra, so that's probably just a review for you. Then we'll look at the remainder theorem and the factor theorem, and finally, we'll talk about constructing a polynomial from its roots. Now, you might say, what's the purpose of all these topics? Where are we going with this? Well, if you remember in the last episode, just at the end of the episode, we were discussing how you graph a polynomial if it's factored using its x-intercepts. Let me just give an example. Suppose I have the polynomial function, I'll call it p of x, equals, let's say, x plus 2 squared times x minus 1 times x minus 4 cubed. Now, what I'm doing here in this example is I'm purposely putting in a factor with multiplicity 1, a factor with multiplicity 2, because there are actually two factors of x plus 2, and a factor of multiplicity 3. And I want to demonstrate what is the effect of those powers, those exponents, on the graph. Now, the first thing I would do would be to locate the x intercepts. And if you remember, the x intercepts are the places where the function value, p of x is 0. Well, now, p of x is 0 whenever one of these factors is 0. So that could be when x is negative 2, when x is 1, or when x is 4. So the x intercepts would be, and I'll put them over here, x equals negative 2, x equals 1, and x equals 4. Now, this helps me draw the graph, because if I locate those points on the x-axis, I'll just put a heavy dot on each one of those points. So I'll put a heavy dot here at 4, I'll put a heavy dot here at 1, and a heavy dot here at negative 2. Now, because when I multiply this out, I think I'm going to get x to the sixth power, it looks like. Because you see, I'm going to get an x squared there. I have an x here. And when I cube this factor, I'm going to get an x cube. So when I multiply those lead terms, x squared, x, and x cubed, this is going to end up being a positive x to the sixth plus, and then a whole bunch of other terms. Now, what's significant about this is if I have a polynomial of even degree, then you remember both sides of the graph, the in behavior, has it either both sides going up or both sides going down. In this case, do you think both sides will go up, or will both sides go down, based on that lead term? Well, if the coefficient's positive, both sides go up. If the coefficient's negative, the graph has been inverted and both sides go down. So I'm expecting to see a graph that goes up over here, and when I finish, I should have the graph going up over there too. OK, well, let's see. If it's going up on that side, then if I go backwards, if I come down from up above, I guess the question is, how does it approach this x-intercept? And if you remember, in the last episode, we said that what the way we determined the behavior at the x-intercept is we look at the multiplicity of that factor, and the multiplicity there is 3. And therefore, it behaves something like a cubic function. That is, it comes in, it turns, levels off, and then it goes down again. So I get this little bit of a ripple in the graph. Now, we don't know how far it goes down, because we're not making a graph that's that accurate. I just am looking for the general shape, but I know I have an appointment to come back and touch the x-axis when I get here to 1. Now, what happens at 1? Will it have the same ripple effect there when I get to 1? Well, what I do is I go to the factor that produced the 1, and that factor has multiplicity 1. And therefore, it passes through just about like a straight line. So when my curve's going to come up and it's just going to pass right on through, there's nothing special that happens in the behavior there. And I'm going to go up. But eventually, I have to come back to the negative 2. So I'm going to turn right about here and come back to the negative 2. Now, what happens at negative 2? Well, let's see. If I look at the factor where the negative 2 came from over here, what is its multiplicity? Multiplicity 2. And Steven, do you remember what happens when the multiplicity is 2? What happens at that x-intercept? It touches the x-axis and turns back. Right. It just turns around sort of like a parabola, sort of a little mini parabola right there. But actually, in this case, it turns very quickly. It turns and it goes back up. And you notice we have both sides of the graph going up just like we predicted in the beginning. One thing to point out is when you draw these graphs, you don't want to make them bounce in like a bouncing ball. But it actually turns horizontally and it goes back up, so it makes a smooth curve as opposed to sort of a jagged curve like an absolute value, like the absolute value function. OK, well we've talked about this idea in the last episode. But this raises the question, what do you do if the polynomial hasn't been factored? Now this is a sixth degree polynomial. What if this had been multiplied out and we had not known what the factors were? It seems like we'd really be stuck in that case. For example, what if we were wanting to graph this polynomial function, and in fact in just a little bit we will, what if I graph p of x equals 2x cube plus 7x squared plus 4x minus 4? Now you see this time it's a whole new ball game because we don't know what the x-intercepts are because this hasn't been factored. And generally we don't have experience factoring cubic polynomials with the one exception of factoring by grouping. And occasionally factoring by grouping will help. But factoring by grouping won't allow us to factor this one. If you remember in factoring by grouping what you do is you pair up the terms two and two. Sometimes you group them in threes and ones, and you try to factor this, you try to factor this, and then overall you look for a common factor between the two. That will sometimes allow us to factor cubic polynomials, but I don't think it would in this case. So we're stuck in the very beginning because we don't know how to factor that. Well the lesson today, and the lesson that follows this, episode 11, will be about information for factoring higher degree polynomials. I remember when I was a student back in high school and so forth, I thought you could only factor quadratics. And I thought that it was just sort of serendipity whether you could factor higher degree polynomials. But in fact there's a whole theory for this. It's not very complicated. And that's what we'll be talking about today in an episode 11. OK, so to begin this, let me remind you of something that you've seen maybe back in middle school. It's called the division algorithm. And this is the division algorithm for arithmetic. And then we'll see what is the division algorithm for polynomials. Now the word algorithm means it's a procedure or a method. So if you solve, for example, a quadratic equation using the quadratic formula, that would be an algorithm for solving a quadratic equation. If you solve a quadratic equation by factoring, that would be a different algorithm for factoring quadratics. Well, there is an algorithm for division that goes like this. It says if you have two numbers, a and b. And let's say a is bigger than b. So let me put an inequality sign in there. Suppose a is bigger than b. Then I can write a as a multiple of b plus a remainder. I'll just put spaces in there to represent those numbers. So I can take a times a multiple of b and then there's some remainder left over. For example, what if I said 11 is equal to some multiple of 4 plus a remainder? This is multiplication right here. Well, I could put a 1 there, 1, 4. And then what would be the remainder in that case? Seven. Seven. Yeah, exactly. Another possibility is I could say 11 is equal to 2 times 4. But in that case, what would be the remainder? Three. Three. Now, this is the preferable way to express this relationship because I have a remainder that's smaller than the number 4. So rather than using the equality there, I'll use the equality here. This is referred to as the algorithm for dividing 4 into 11. Basically, 4 goes into 11 two times with a remainder of 3. We don't usually think of division as being, whoops, that should be a plus right there. We don't usually think of division in terms of writing products and sums. But the idea is I would like to find a number that is a coefficient of 4 so that my remainder is smaller than 4 but non-negative. Let's take one more example of that and then we'll look at this for polynomials. What if we take a number like 37 equals some multiple of 5 plus a remainder that goes right here? Well, I could put in a 1, a 2, a 3, a 4. But what's the biggest integer I could put in there so that I get a remainder that's either 0 or larger but smaller than 5? 7. Yeah, I think I could put a 7 here. 7 times 5 and the remainder would be what? 2. Would be 2. OK. The number that goes here is referred to as the quotient. And the number that goes here, just as I've been referring to it, is referred to as the remainder. And the way you normally write this is you say 5 is divided into 37 and it goes 7 times and then I subtract off 35 and I get a remainder of 2. And that remainder is normally expressed as a fraction 2 over 5. So what we do is just write that as 7 and 2 as 2 fifth. So this remainder comes up as the numerator of a fraction. OK, now back in middle school, you may have seen some expressions like this, but I want to now write this division algorithm for polynomials. And let's go to the graphic that's on the screen to express this. It says for two polynomials, p of x and d of x, I'm calling it d of x because that would be the one that I would want to divide by, for p of x and d of x, where d of x is not 0, p of x can be expressed uniquely in the form, p of x equals q of x times d of x plus r of x, where q and r are polynomials and either r is 0 or its degree is less than the degree of d of x. Now that looks pretty complicated. Let me just give you an example of what that refers to. Suppose I were to have the polynomial x squared plus 2. And I want to put a multiple of a divisor, let's say the divisor this time is x minus 1, plus a remainder. Now the polynomial that goes here in that graphic I call q of x, what do you think I called it q of x? Quotient. Because it's the quotient. Yeah, it's called the quotient. And the one that goes over here, the polynomial that goes over here I call r of x because it's the remainder. So the question is, can I find a polynomial that goes here so that my remainder will be a polynomial that's either 0 or it will have a degree smaller than my divisor? This is my divisor, by the way I call that one d of x in that relationship. And so let's see, can I find a polynomial that goes here so that I get a remainder with degree smaller than x minus 1? Well I think what I should put in here is x plus 1. Now what's the product of x plus 1 and x minus 1? x squared minus 1. x squared minus 1, right, because that's the difference in two squares. Let's just see what the remainder would be in that case. x squared plus 2 equals this product, well that product is x squared minus 1 plus my remainder. Now can you see what I should add on to this to get x squared plus 2? 3. 3, exactly. Now I don't get a remainder that's 0, but I do get a polynomial you might say whose degree is smaller than the divisor. Because this divisor has degree 1 and constant polynomials other than 0 are said to have degree 0. So this has degree 0, its degree is smaller than that divisor. OK, as a general rule in the division algorithm, if you have a polynomial p of x and if you choose a divisor polynomial d of x, like in this example I chose d of x to be x minus 1, then according to this algorithm there should be a unique polynomial that I can put here so that I get a unique remainder over here. And this remainder will be either 0 or have degrees smaller than d of x. So this polynomial I call q of x. This polynomial I call r of x. OK, now that's fairly abstract, but this leads us into what we call long division. So let me do an example of long division and see how we would actually divide polynomials and express them in the form of this division algorithm. Suppose I wanted to carry out this division, 4x cubed minus 2x plus 3 divided by x plus 1. Now I'm not sure that x plus 1 will divide that exactly. I may have some remainder, that is some rational expression I'll have to add on at the end, but I'd like to carry out that division. So the way I do it is I put x plus 1 outside the division sign, and inside I put my polynomial 4x cubed. I better leave a space for the second degree term. I'll move the minus 2x and the plus 3 over here. At the moment I don't have any x squared terms, but I might have some in the course of the division, so I have to allow for that. Now when I divide these two polynomials, what I do is I divide the lead term here into the lead term there. I divide x into 4x cubed, and x goes into 4x cubed, 4x squared times. I'll put that in the x squared column. Then I multiply 4x squared times x and 4x squared times 1, and I get 4x cubed plus 4x squared. Now let me ask you, do you think I should add these or should I subtract these? Subtract them. Yeah, I think we should subtract them, so I'm going to put parentheses on there and put a minus in front of that. When I subtract 4x cubed minus 4x cubed, of course I get a zero here, so I won't write down anything. And over here I'm subtracting zero minus 4x squared is minus 4x squared, since I'm subtracting that from a zero that we don't see. Then I bring down the next term, the minus 2x, and I'll divide x plus 1 into minus 4x squared minus 2x. Now the way I do that is I actually divide the two lead terms. I divide x into negative 4x squared. How many times will that go? David, what would you say? I'll divide x into negative 4x squared. I want to say four of it. Let's see now. Well, I want to divide x into negative 4x squared. I'll just come over here on the side. Sorry, just once. Sorry about that. Yeah, so what would that ratio be? Two to one or one to two? No, let's see. If I divide by x, it's going to be a negative 4x. That reduces to be negative 4x if I cancel that off. So right here I'll put a negative 4x. So when I divide this guy into that guy, what I'm thinking is negative 4x squared over an x. OK, now I multiply negative 4x times x plus 1. And I get negative 4x squared and I get negative 4x. And I'll subtract those off. You see my purpose in this is to always get these first terms to cancel off here. I got them to cancel off there. And as that cancels, then I can bring down the next term over here, and that forces me to move across the problem from left to right. Well, when I subtract negative 4x squared from a negative 4x squared, I get a zero. And if I take negative 2x and I subtract a negative 4x, let's see, negative 2x, I'm going to subtract a negative 4x. That's going to give me, who can say how much will that be? 2x. 2x. That'll be a 2x. And then I'll bring down the 3. OK, now here we go again. I want to divide x plus 1 into 2x plus 3. How many times will my divisor go into this expression? Twice. Two times. Right, very good. So I'll put a 2 here. When I multiply, I get a 2x plus 2. And you see, once again, I'm wanting those two lead terms to cancel off. And when I subtract here, I get a 1. That's my remainder. I have this quotient and I have this remainder. So the answer to this problem up here of what is the cubic polynomial divided by the linear polynomial, the answer is 4x squared minus 4x plus 2 plus. Now, the way I express the remainder is in the form of that rational expression, 1 over x plus 1, sort of like I did when I divided 5 into 37 a little while ago. If the remainder is 0, then I don't have to put a fraction there. I don't have to put a rational expression there. And I get a nice polynomial answer. And what that means is x plus 1 is a factor of 4x cubed minus 2x plus 3. If I get a remainder 0, it means this guy is a factor. And this is one way of finding factors is to see if they will divide evenly. OK, I'm going to do one more division problem, long division problem. And then I want to show you a shortcut for this process. In the next case, suppose I want to divide x to the third power plus 3x squared minus 7x plus 1. And let's say I'm going to divide by x squared plus x minus 3. OK? What makes this problem a little different is I'm not dividing by such a simple divisor like I did before. What was it x plus 1? Now I'm dividing by a quadratic expression. But the process is still the same. So if I come down here to carry out my division, outside I put my divisor. And inside I put my numerator. And I don't need to leave any spaces because every term is represented. The cubic, the quadratic term, the linear term, and the constant term. OK, so when I do this division, I compare lead terms. x squared goes into x cubed. How many times? x. x times. I'm going to put an x over here in the x column. Now you might say, Dennis, is it OK if we put the x right up there in front? Well, yes it is. Where you place the x really doesn't matter. But to me, it seems reasonable to put the x's in the same column with all the other x's. So when I multiply, I get x cubed plus x squared minus 3x. And I have to subtract that off. And I've done this in such a way that my x cubed terms will cancel. We always want those to cancel, which allows me to move over and bring down the plus 1. When I subtract in the quadratic column, what will the difference be? 2x squared. 2x squared. And be careful. When you subtract here, what do you get? Negative 6x. Not negative 6x. Negative 4x. Negative 4x. Yeah, negative 4x. Because you see, we're subtracting a negative 3, so that really makes that a plus 3. So I have negative 7 plus 3 is negative 4. So I'll bring down the plus 1 now. And starting over again, I divide my divisor into 2x squared minus 4x plus 1. I think that goes into it two times. So I'll put a plus 2 over here. And I take two times my divisor. I get 2x squared plus 2x minus 6. And when I subtract, the quadratic terms cancel off. When I subtract here, what will I get? That is the negative 4x. And I'm subtracting off the 2x. Negative 6x. Negative 6x. You know, the reason I'm asking you these questions is because when people make mistakes in division, it's generally with the subtraction. And when I'm balancing my checkbook, if I make a mistake, it's usually with when I'm trying to subtract off a check balance, I make a mistake in subtraction. So I'm purposely asking you about those subtractions. And if you get them wrong, that's fine. Because I think some of the people at home are probably wondering the same thing, how we're getting those. So we want to sort of go over that slowly. When I subtract over here, what will I get? 7 plus 7. Yeah, plus 7, exactly, plus 7. And you see, I'm no longer able to divide my quadratic into this expression because this has a degree too small. So that's my remainder. And that tells me that the answer to this problem is x plus 2 plus a negative 6x plus 7 over the divisor x squared plus x minus 3. OK, now what does this have to do with that division algorithm that I wrote on the screen here a little while ago? Well, you see, what this tells me is that my p of x, which is the numerator, and my d of x, which is my divisor, can be expressed in this form. We said p of x was equal to q of x times the divisor d of x multiplied together plus the remainder, r of x. And can anyone refresh my memory? What was the restriction on the remainder? How was it related to the divisor? It has to have a lesser degree. It has to have a smaller degree than the divisor, or it's 0. You might say, well, if it's 0, it doesn't have a smaller degree. Well, if this is equal to 0, it doesn't have a degree officially. A constant other than 0 is said to have degree 0, but the polynomial r of x equals 0 technically has no degree to it. So if I put x cubed plus 3x squared minus 7x plus 1 here, I'll need to put a quotient times the divisor. A quotient times the divisor. Well, there's the divisor. It's x squared plus x minus 3 plus a remainder. Well, the quotient is? x plus 2. Exactly, x plus 2. That's what goes there. That's the quotient. That's what was up above my division when I carried out the division. And the remainder is what? It's that, remember, the negative 6x plus 7. That's what came out at the end of the division. So I'll put a negative 6x plus 7. And by golly, it does have degree smaller than the divisor. The divisor has degree 2. The remainder has degree 1. Now, if you multiply this out, add on the remainder. Collect all those terms. You will get the original polynomial. Now, if it had turned out that the remainder had been 0, then you see I could have just erased that portion of the problem. And that would tell me that this quadratic is a factor of the original cubic. It would also say that the quotient is a divisor of the cubic. But we didn't get a 0 there, so we don't have any factors here. OK, now, we would certainly like to shorten this painful process of long division. And let me now show you a shortcut that applies in many cases. It doesn't apply in every case, but we'll explain the idiosyncrasies of that. Suppose I wanted to multiply, let's take x to the fourth power minus 2x cubed plus x squared minus 3x plus 2. And I want to divide this by x minus 2. Now, the process I'm going to show you here is referred to as synthetic division. What I was just using is referred to as long division. Long division works all the time, regardless of whatever you're dividing by. Synthetic division only works if you're dividing by a linear polynomial. That is a first degree polynomial. And the coefficient of x has to be a 1. So this could be x minus 2. It could be x plus 3. It could be x minus 7. But it has to just be a single x and then plus or minus a constant. OK, I'm going to do the division on the left. And then we're going to see how we could shorten this. x minus 2. And here I'll put x to the fourth minus 2x cubed plus x squared minus 3x plus 2. OK, I'll have to write sort of small so we can squeeze this in. x goes into x to the fourth x cubed times. So I get x cubed minus x to the fourth minus 2x cubed. And when I subtract, what do I get? We get 0 because these two terms cancel. And these two terms cancel. We get nothing other than 0. So rather than bringing down one term, I'll take the opportunity to bring down two terms because I want to divide my divisor into a pair of terms over here. x goes into x squared x times. I'll put that over here. x times. When I multiply, I get x squared minus 2x. And when I subtract those off, who can tell me what will I get when I subtract those off? Negative x. I'll get a negative x right there. Exactly, negative x. Because that's negative 3x plus 2x is negative x plus 2. And x minus 2 goes into negative x plus 2. How many times? Negative 1. Negative 1 time. I'll put a negative 1 right there. And I get a negative x plus 2. And this time, everything cancels off. I get a remainder of 0. First thing that tells me is x minus 2 is a factor of this fourth degree polynomial. It's a factor of it. Now, that's going to be very useful information to be able to identify factors. But more importantly, I want to find a way of shortening all this expression. So what I'm going to do is I'm going to rewrite this by putting coefficients only and leaving out the x's. So right here, I'm going to put a 1. Right here, I'm going to put negative 2. Right here, I'm going to put a 1. Right here, I'll put negative 3. And right there, I'll put a plus 2. I think I'll leave out the plus signs because if I see a 1 or a 2, I know it's a plus 1 or a plus 2. Let me move that over to kind of center it there. Because I'm thinking, you know, when I write these down, it's the coefficients that are important. And I can remember that this is the constant term, the linear term, the square, this is the cubic, this is the fourth power term. I'll also put a 1 over here. And I'll also put a 1 right here. And I'll put negative 2 there. And I'll put a 1 here and a negative 3. Here I'll put a 1. Here I'll put negative 2. You see what I'm doing is trying to leave out any extra symbols that I don't have to write down every time. Here I'll put a 1. I'll leave out the plus. Here I'll put a 1, negative 1. And here I'll leave out the plus. And up above, I'll put a 1 for the x cubed. Let's see to get it in the right column, I'll put it there. I'll put a 1 in place of the 1x. And you see, I know that's 1x because it's in the x's column and then negative 1 over here. So this is sort of a shorter version of what I just did. You know, furthermore, it seems to me that this 1, 1 times 1, that's a duplicate right there, so let's just leave that out. And over here, I knew that I was going to get a 1 below that 1 because I always get these first terms to cancel, so I'll leave that out. And right here, I always get those two to cancel, so I'll leave off the negative 1 because I know that's supposed to cancel off. How can I reduce it further? Well, you know, there's a 1 here and then I brought it down. What if I don't even bring down the 1? What if I just leave it off and I just look up there to find the number? Same thing here, the negative 3. If I look up there, there's the negative 3. I'll leave that out. And so it looks to me like what I can do is just press this all together. Oh, one more thing. I'm going to leave off the plus 1 right here because you remember we said this only works whenever you're dividing by x plus 2, x minus 3, whatever. There's always a 1 there. So now I'm going to push this all together and make it look very thin and it's going to look like this. I'm going to erase this bottom portion so I can squeeze this in. In synthetic division, I draw my division symbol that way and I put the coefficients in directly across the top. And outside, I put a negative 2. Now I begin by bringing the 1 straight down and I multiply here, negative 2 times 1 is negative 2. And I subtract and when I subtract those, I get a 0. If you remember, when we subtracted right here a moment ago, we got a 0 and I brought down the next two terms, that's the 0. The myth there was a 0 remainder. Now I multiply negative 2 times 0 and I get a 0. And when I subtract, I get a 1. Then I multiply negative 2 times 1 and I get a negative 2. When I subtract, I get negative 1. And then finally, negative 2 times negative 2 is 2. And when I subtract, I get a 0. This is my remainder right here. And this is my quotient. You might say, how's that the quotient? Well, look up above, look up here. I was getting a 1, a 1, and a negative 1. And I could put a 0 to hold that space open. This is my quotient, but it's represented only with coefficients. So what this tells me is the answer to my division problem is, let's see, constants, linear, quadratic, cubic. That's a 1x cube. There aren't any quadratics. There's a linear term plus x. And there's a constant term, negative 1. And the remainder is 0, so I don't have to add on any term over there on the n. And I have found the quotient of these two polynomials using what's called synthetic division. Now, you don't have to remember why synthetic division works. I just merely want to give you an indication of how it works. Let's go to the next graphic, and we'll do a synthetic division problem that's on the screen there. This says that when you are using synthetic division, when dividing by x minus c, c is a constant, we can use synthetic division as in this next example. Now, let me just copy that down before we take that off the screen. We're dividing 2x cube plus 7x squared minus 4x minus 11. And that's divided by x minus 3. OK, let's try solving that example here on the green board. You notice, first of all, we don't make the regular division sign. I make the squared off arrangement. This is just sort of standard. You'll see this in your textbook. And then across the top row inside the box, I put the coefficients 2, 7, negative 4, negative 11. And my divisor goes out here. And my divisor is x minus 3, but I'm throwing the x away. In fact, I'm throwing the coefficient away. I put a negative 3. And I'm just going to make a note out here that I'm going to be subtracting when I get to these columns, just like I do in regular division. So the process begins this way. I'll always bring the first number straight down, whatever it is. That's a 2 there. And then negative 3 times 2 is negative 6. And I subtract here and get 13. And then negative 3 times 13 is negative 39. And I subtract here. Got to be kind of careful. I subtract here. So that's actually negative 4 plus 39 is 35. And then negative 3 times 35 is negative 105. And I subtract here. And that's where I put my remainder. So I'll put a little box around it to separate that number from the others. When I subtract, let's see, that's negative 11 plus 105 I think is 94. So what's the answer to the problem? Well, this is the remainder. This is the quotient. That says 2x squared plus 13x. Let me make that 2 a little better there. Plus 35 with a remainder of, I'll have to put this right below it. I'll say with a remainder of 94 over x minus 3. So this is much quicker, I think, than the long division process. And we'll do another example or two here in a minute. One thing to point out. You remember this 2 represented 2 cubed, and this 7 represented 7 squared. But when you look at the answer, I'm saying this 2 represents x squares. So you say, wait a minute. You had x cubes. Here you have x squares. How did that happen? Well, if this is the remainder and this is my quotient, I have to count from here constants, linear, quadratic. So those two don't represent the same degree polynomial. This is always going to be one less. Now, how could I make this even faster? I mean, this is fairly fast. Well, most people, if they make a mistake in synthetic division, it's with this subtraction here. Because you always make more mistakes with subtractions than with addition. So I would like to change this to addition. And I can do that if I merely change the sign of that number and make that a plus 3. So when I look in this factor, x minus 3, I throw away the x. I change the sign to a plus 3, and then I can add. Now, I'll just try this right below, 2, 7, negative 4, negative 11. And over here, I'm going to add only for convenience, because I figure I make fewer mistakes when I'm adding than when I'm subtracting. We start off the same way. I bring a 2 straight down. 3 times 2 is 6, and I add. I get 13. Yeah, that's right. 3 times 13 is 39, and I add, and I get 35. 3 times 35 is 105, and I add, and I get 94. Well, by golly, it's the very same result. What I got here and what I got here are the same. So the actual synthetic division algorithm has you change this sign so that you can add. Another way to look at it is if you look at the divisor, x minus 3, this polynomial, this linear polynomial, has a 0, and the 0 of that polynomial is 3. See, if you're plugging a 3, you'd make that 0. So it's the 0 that you write out here, and then you add. If you don't change the sign, you have to remember to subtract. And if you want to subtract, that's fine with me, but I think you'll make fewer careless errors. I know I will if I do the addition route. OK, let's take another synthetic division problem. Let's see. Let's suppose this time we have 3x to the fourth minus 5x cube plus x minus 11, and I'm going to divide by x plus 2. x plus 2. I have no idea what this answer is. I'm just making it up, but we can carry out the division nonetheless. So if I'm going to divide, I'm going to use synthetic division. You remember, I can only divide if that's a linear factor with a x coefficient of 1. So I'm going to put the 0 of that factor in front. What's the 0 of x plus 2? Negative 2. Negative 2. Or another way of looking at it, I'm writing down a 2 and I'm changing the sign. And then across the top row, I put 3, negative 5, 0, 1, and negative 11. The 0 is holding open the fact that there are no x squared terms. If you leave that out, these numbers move over, and that 3 ends up looking like it's the coefficient of x cube, not x to the fourth. OK, here we go. I bring down the 3. I multiply. I add. And I multiply. And I add. And I multiply. And I add. And I multiply 86. And I add. What am I going to get there? 75. 75. The remainder is 75. What that tells me is x plus 2 is not a factor of the polynomial on top. But furthermore, it tells me that this quotient is 3x cube, let's see, constants x, x squared, x cube, 3x cube, minus 11x squared, plus 22x, minus 43, plus, and then I'll squeeze in here, 75 over x plus 2. OK, that's synthetic division. Now, if I were to write this in the form of the division algorithm, I would write it this way. I would say that 3x to the fourth minus 5x cube plus x minus 11 is equal to the quotient, OK, here's the quotient without the remainder term, times the divisor plus the remainder. And I'll just put the remainder right below there, plus 75. I'll leave this to you to verify. If you multiply this four-term polynomial times this binomial, if you multiply that out, add on 75, you will get this fourth-degree polynomial that I've given over here. OK, now, what if I wanted to carry out this division? Let's say we have 2x cube plus 3x squared minus 17x plus 12, and I want to divide this by something that doesn't have coefficient 1. It's linear, but other than coefficient 1. Let's say it's 2x minus 3. Now, when I see a 2 there, I think, wait a minute, I can't use synthetic division here because I'm only supposed to do that when I have a 1. So what I'm going to do is rewrite this as 2x cube plus 3x squared minus 17x plus 12 over 2 times x minus 3 halves. I'm factoring the 2 out of that. So I'm going to use synthetic division to divide by x minus 3 halves, and afterwards I'll divide by 2. So I'll go back and divide by the 2 separately. So here's how I'll do this. I'm going to put 3 halves out in front. I'll put the coefficients inside 2, 3, negative 17, and 12. There's a 12 there that's kind of hard to read. And when I finish, I'm going to divide my quotient by 2 because I have an extra 2 in the denominator. So I bring the 2 down, 2. I multiply 3 halves times 2 as 3. And I add, and I get a 6. And 3 halves times 6, let's see, 3 halves times 6 is 9. So when I add, I get a negative 8. And 3 halves times negative 8, let's see, can someone help me out there? How much is 3 halves times negative 8? Negative 12. And when I add, I get a 0. Whoa, hey, we're all excited now. Because what does the 0 mean? No remainder. It means there's no remainder. But what does it tell me about this factor? It's the visor of the top one. Yeah, actually, what does it tell me about this divisor? It's a factor of the numerator. Yeah, I got the words backwards. But we're not quite finished because I've just divided by x minus 3 halves. I also have to divide by 2. So when I divide by 2, that means I'll divide this by 2. 1, 3, negative 4. The remainder's still 0. And that tells me that the answer to this problem is x squared plus 3x minus 4. That's the, if I erase what's in the middle here, I've now divided by the 2. And this cubic polynomial divided by this linear polynomial is, gives me that answer. So I had to bring the 2 out and treat it separately from this linear factor with the lead term coefficient 1. OK, that's about everything I can tell you about synthetic division. Now let me show you where it leads us. Let's go to the next theorem that's on the graphic. This theorem has a name. It's called the remainder theorem. And this is one of the things I need to point out to you somewhere along the way. When you come across a theorem that has a name, what that means is it has a name probably because you're going to be referring to it a lot. You have a name because people refer to you a lot. Your parents gave you a name. They didn't just say, hey person or whatever, but they gave you a name like Jeff or Stephen. So when a theorem has a name, it must be because it's something you're going to refer to more frequently. So the name of this theorem is the remainder theorem. It says when a polynomial p of x is divided by x minus c, the remainder is p at c. Let me explain what this means before we go to that example. Suppose I have a polynomial. And let's say the polynomial is x squared. Let's say x cubed plus 2x squared minus x minus 4. And suppose I'd like to know what is p at 2, p at 2. Well one way to find p of 2 is just to plug the number in. Let's do that. Right here I'd get 2 cubed or 8. Right here I'd get 2 times 4 is 8 minus 2 minus 4. How much will that be? 10. That'll be 10. So p of 2 is 10. Here's another way you can get that answer. Use synthetic division and divide x minus 2 into this polynomial, x minus that number. So if I divide x minus 2 into that polynomial, I'll put a 1, a 2, a negative 1, and a negative 4. And if I'm dividing by x minus 2, I'll put its 0 or 2 outside. You notice that's the same number that I would like to plug in. And you know what? The remainder here should be 10. The remainder will be the function value. Let's try this. I bring the 1 straight down and I multiply. And then I add these, and I multiply. And then I add these, and I multiply. And I add these, and I get a 10. Now the remainder you get will always be p evaluated at this number right here. Now that seems pretty incredible that that could be true. Why is that the case? Well let me explain, and then I'll work another example like this. This goes back to that division algorithm that we wrote on the screen at the very beginning. You remember that we had p of x is equal to a quotient times the divisor. And in that example, I was just dividing by x minus 2 plus a remainder. Now this remainder would have to be constant because its degree is smaller than the divisor. The divisor has first degree, so this has to be 0 degree. Look what happens if I substitute in a 2 here. If I plug in a 2, I get q evaluated at 2 times 0 plus the remainder. Well it doesn't really matter what q at 2 is. When I multiply, this says p of 2 is the remainder. That's what the remainder theorem says. The remainder upon division by x minus 2 is the same thing as the polynomial value at 2. Let's take an example. I'm going to ask you to give me a polynomial this time. I'd like for someone to make up a polynomial, not terribly complicated, but not exactly trivial either, a polynomial that's, let's say, a fourth degree polynomial. Susan, can you give us a fourth degree polynomial, just whatever? 2x to the fourth. 2x to the fourth. Minus 3x squared. 3x squared. Plus x minus 1. Plus x minus 1. And suppose we'd like to know, what is the value of this polynomial at, let's say, negative 3? Well, here's the most primitive way to do it. You could actually plug in the negative 3 in three different places and multiply that all out and expand it. But you see, taking negative 3 to the fourth power, taking negative 3 squared, and doing all that arithmetic is sort of tedious. We can surely do it. Here's a faster way to do it. I'm going to use synthetic division. And I'm going to put negative 3 outside. And I'll put 2, 0, negative 3, 1, negative 1. And I'm looking to see what the remainder is. Why is it 2, 0? Well, you see, you didn't have a cubic term, so I have to hold that place over. That's a good question. In fact, I'm actually glad you gave us an example that was missing a term, so we could emphasize that, because it's easily overlooked. OK, I bring the 2 down. I get negative 6. And let's see, I'm adding. Because I'm not dividing by the plus 3, but the negative 3. So when I add, I get a negative 6. Multiplying, I get 18. Adding, I get 15. Multiplying, I get negative 45. Adding, I get negative 44. Multiplying, let's see, I'm going to get a positive number. 3 times 4 is 12. 3 times 4 is 12. And one more is 132. And I add, and I get 131. So how much is P at negative 3? 131. 131, that's the answer. Now, you might say, well, Dennis, you know, if it's all the same to you, it seems to me it's just simpler just to plug in the negative 3 and get it over with, even if the arithmetic is a bit more complicated. OK, well, imagine this isn't a fourth-degree polynomial. What if it's a 14th-degree polynomial? So it's really long. And I want to plug a negative 3 into the 14th power, the 13th power, the 12th power. You get the picture. This gets no more complicated, except the numbers can get bigger right here, like I was up to 132. But that's nothing like taking negative 3 to the 14th power. I mean, come on, give us a break. So this is much quicker for big polynomials. When the polynomial has a relatively small degree, like this one does, I don't think there's any clear advantage of one over the other. OK, let's go to the next graphic. And look at a theorem that's called the factor theorem. We make heavy use of this in the next episode. Now, the factor theorem says that x minus c is a factor of the polynomial p of x, if and only if p of c is 0. Now, what lies behind this theorem, as you see, the remainder theorem just told us p of c would be the remainder. But if the remainder is 0, if p of c equals 0, the remainder is 0, and when the remainder is 0, that means that x minus c is a factor. That kind of sums up several ideas that we've mentioned here today. OK, let's take this problem. I'm wondering the answer to this question. Is x minus 1 a factor of this polynomial expression? x cubed minus 3x squared plus 3x minus 1. Is this a factor of that? Well, one way to find out is to divide x minus 1 into the cubic polynomial, see if it divides it evenly. But another way to do it is to say, what's the remainder if you divide x minus 1 into this cubic? If the remainder is 0, this is a factor. So to find out what the remainder is, I'm going to evaluate the polynomial at 0 or the root of x minus 1. If I plug in a 1, I get 1 minus 3 plus 3 minus 1. The 3's cancel, the 1's cancel, I get a 0. So the function value at 1 is 0. And therefore, we can say x minus 1 is a factor of x cubed minus 3x squared plus 3x minus 1. Yeah, it is a factor. So this time, rather than doing the division, I'm calculating a function value. So I can sort of play these off against each other. OK, the last item of business here today is to construct a polynomial when I'm given its roots. And all of the information that we've gone over in this episode comes to life and has purpose in the next episode. Let's go to the next graphic. This is the last graphic of this episode. And it says to find a polynomial of degree 3 that has roots 2, 5, and negative 4, we must first recognize that the roots of the form x equals c leads us to factors x minus c. Now, let me just work that example right here on the green screen. We have three roots, 2, 5, and negative 4. I'm looking for a polynomial that has these three roots. Well, if 2 is a root, then x minus 2 has to be a factor. And if 5 is a root, then x minus 5 has to be a factor. And if minus 4 is a root, what must be a factor? x plus 4. x plus 4. Now, if I multiply those out, I will have a polynomial that has these three roots. Let's see, if I multiply it out, that'll be x plus 2 times, if I multiply these two, that'll be x squared minus x minus 20. And then if I multiply these two out, I'll get x, I'll just multiply here, here, and here. I'll get x cubed minus x squared minus 20x. Then I'll multiply here, here, and here. Minus 2x squared plus 2x plus 40. Going back and collecting like terms, I have x cubed minus 3x squared minus 18x plus 40. This is a polynomial that has three roots, 2, 5, and negative 4, like we asked. However, this isn't the only polynomial that has those roots. For example, I could have put in x plus 4 two times. I could have used that factor twice, and I'd still have the same roots, but this would have ended up being a fourth degree polynomial. I think the problem on the graphic said, find a cubic polynomial, a third degree polynomial. This is the one we'd want. Another way I could come up with a different answer is I could put a coefficient in front here, like a 3. And in that case, what I would do would be to triple this answer, and I would come up with another polynomial, I'll call it p1, 3x cubed minus 9x squared minus 54x plus 120. So when I say find a polynomial that has these three zeroes, the answer is not unique, but this is probably the simplest answer, and any multiple of it, any non-zero multiple of it will be an answer. And I could even, if I was willing to go to higher degrees, I could put exponents on these factors, and I would still have these three roots with yet a higher degree. OK, let me summarize what we've done today. We started off talking about something that we call the division algorithm, and I think we used an example from like middle school. We took 37, and we divided it by 5. We got the quotient, we got the remainder. But we quickly moved on to polynomials, and we looked at the division algorithm for polynomials. From there, we did long division, we did synthetic division, which only works in certain cases. Then we looked at the remainder theorem, and finally, the factor theorem, and we concluded by reconstructing polynomials from the roots, from its roots. Now, if you stay tuned for episode 11, you'll see how we will use all this information to factor big and rather complex looking polynomials. I'll see you next time.