College Algebra: The Remainder Theorem

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Okay, so now I want to take a look at how to find the remainder of something when you're dividing by something of the form x plus a number or x minus a number. So let's just think about exactly what is going on when you do a long division. So if I were to take p(x) and divide it by q(x), what would that look like? Well, at the end of the day I'd have something like this: p(x) = q(x)(x - c) and then I'd add the remainder. So I'd add plus a remainder. So I'd take p(x), I'm going to divide it by x - c, that's the kind of thing I'm talking about, and when I divide that, what I would see is q(x) plus the little remainder term--with remainder r(x).
Well, let's think about this for a second. The remainder, in some sense, has to be smaller than the thing I'm dividing. Well, what does smaller mean here? We're talking about the degree. So this, for example, has degree 1, because the highest degree of the x's is one. So something smaller than that would be degree 0, which means there would be no x's at all. So, in fact, this thing right here, which looks like a little function, is really just a constant number. So when I'm dividing by x - c, the remainder is never going to have x-s in it, because if it had an x in it, I could keep dividing further. I have to keep dividing until I get something that's too small to be divided by x - c. Well, that too small thing will be just a number. So, in fact, this is a number.
Let me say that again. When you do long division you keep long dividing until the thing you have is actually smaller than the thing you're dividing by. And so since this is an x-stuff, I keep dividing through until I have no more x's left, and what remains is the remainder. So since I just have an x thing here, this remainder must be just a number. So, in fact, this is a number, and I don't know what the number is, but it's some number. Let's see if we can figure out what the number is. If we go back here and just plug in the value c for x everywhere, let's see what happens. What I would see is p(c) = q(c)(c - c) + r(c). But c - c is zero, so, in fact, this whole term is zero, and so I see that the r number, that thing that I know is a number, must equal p(c). So in particular what I see is the remainder will equal the polynomial p evaluated at the number c.
To sum up, if I have a polynomial, p(x), and I'm dividing it by x - c, the remainder will just be the value of the polynomial when evaluated at x - c. And this actually has a name. It's such an important result. It's called the remainder theorem. So the remainder theorem states that if you have a polynomial p(x), when it's divided by x - c, its remainder will always be p evaluated at c. Let me try to illustrate this now with an example.
So suppose that I have the following: p(x) = 2x³ - x + 1. And I want to know what is the remainder when that polynomial is divided by x - 2? Well, by the remainder theorem, the remainder should equal p evaluated at 2. Well, let's compute that and see what that number is. P evaluated at 2 is what? Well, it's 2³, which is 8. 8 times 2 is 16. 16 minus 2 is 14, plus 1 is 15. So the question is, is that really the remainder? That's what the remainder theorem tells us the answer should be. Let's check it using synthetic division. How do I do synthetic division? I want to take this polynomial divided by x - 2. So remember what you do. You first flip the sign of this. So that is a plus 2, draw the little half house. The coefficient's 2. Notice there's no x² term, so I put a zero. There's an x term - 1 and there's a constant term 1. I bring down the 2. 2 times 2 is 4. 4 + 0 = 4. I multiply this and get 8. I add these and get 7. 7 times 2 is 14. 14 and 1 is 15, and remember, that represents the remainder, because this is the answer: 2x² + 4x + 7, remainder 15. Hey, remainder 15. That's what we said. So you see, an easy way to find the remainder is just to either do this out or to plug in this value, and we get the same answer. Notice, by the way, that this was actually pretty easy, whereas when I plugged this in I had to take 2³ and then multiply it by 2, subtract 2, add... Here, there was no cubing at all, and I was able to get the answer.
Let's try this again and see, that, in fact, really to find the remainder I just take the polynomial and evaluate it at this point here. One last example. Suppose that I have p(x) = -x² + 10x - 3 and I want to find out what's the remainder when I divide that polynomial by the polynomial x + 4. Well, what do I do? Well, the remainder theorem says it's just going to be p evaluated at--now, I've got to be a little bit careful here, because the remainder theorem says if you divide by x - c, then you look at p(c). But I have to write this therefore, in the form x minus a number. So, in fact, what I have to evaluate at is -4. So I put in -4 and what do I get? Well, I get -4², which is 16, and then there's a negative sign out in front, so that's -16, and then I have -4 times 10, which is -40, and then I have a -3. So what does that equal? Well, that equals -56 and 3 is -59. So that should equal the remainder. Let's see if that's really right or not by doing synthetic division.
So I take a look at this, I switch the sign, so I write a -4. I write down the coefficient, -1. Everyone's represented. 10 - 3, so there's no problem there. I bring down the -1, multiply through, -1 times -4 is 4. I add, I get 14. -4 times 14 is -56, and -56 and -3 is -59. Look, that's the remainder and that's what we got just by plugging this in. So we see the remainder theorem really works. Namely, if you have a polynomial and you want to divide it by something of the form x minus a number, then to find the remainder, it's the exact same thing as just plugging in that number into the polynomial and seeing what the answer is. What's the point of all this? Well, the point of all this is that sometimes actually evaluating a polynomial could be pretty difficult if the exponents were really high, for example, it might be hard to compute -4 to a really big power. But since that is the exact same thing as the remainder, all we have to do is do a synthetic division, and we can actually start to evaluate polynomials by just doing synthetic division. That is to say if I want to find p(-4), it might be too hard to plug in, but it might be easier just to do a synthetic division and then see what the remainder is, because the remainder will actually equal the polynomial evaluated at -4.
So there's a connection between evaluating polynomials and looking at a remainder, and we'll exploit that connection up in the next lecture, and I'll meet you there.
Polynomial and Rational Functions
The Remainder Theorem
The Remainder Theorem Page [1 of 2]