 So far we've learned how to add, subtract, and multiply polynomials. Let's try to divide them. So what is division anyway? When we divide whole numbers, we subtract the divisor until we can't. So for example, if I want to divide 20 by 7, what I'll do is I'll start with 20 and I'll start subtracting 7. So I'll subtract 7, which gives me 13. I can subtract 7 again, which gives me 6. I can subtract 7 again. Oh wait, I can't. I don't have enough. And so I have to stop here. Since we are able to subtract 7 twice and got 6 as our leftovers, we can say that 20 divided by 7 is 2 with remainder 6. Better yet, remember that if a divided by b is q with remainder r, then we can also say a divided by b is q plus the fraction r over b, remainder over divisor. And so we can also say that 20 divided by 7 is 2 plus 6 7s. The thing to recognize is that when we divide polynomials, we do exactly the same thing. So let's say I want to divide 4x plus 7 by x. So we'll see how many times we can subtract x from 4x plus 7. So we'll write down our 4x plus 7 and we'll see how many times we can subtract x. So we'll subtract x. We can subtract x again. We can subtract another x and one last x. And at this point, we've run out of x's, so we'll stop our subtraction. So we've subtracted x one, two, three, four times. And what's left over is 7. So we can say that 4x plus 7 divided by x is 4 with remainder 7. And again, we can also express this as a fraction. 4x plus 7 divided by x is 4 plus 7 over x. We can do every division this way. So if I wanted to do 137 divided by 7, I could begin by subtracting 7 and keep going, keep going, keep going. And the only problem with this is it would take us a long time to do all these subtractions. So we try to make our operation more efficient. So rather than spending a week or so subtracting 7's, we can subtract sets of 7's. This is sometimes known as chunking. In other words, we're going to take a set of 7's and remove them as one chunk. So suppose I subtract 10 divisors of 7. Why 10? Why not? More importantly, I know that 10 divisors of 7 is 70, and I know I can subtract 70 from 137. So I'll subtract 70, and I can subtract a bunch more 7's. This time, I'll subtract 9 divisors of 7. Again, why 9? Because I know that 9 divisors of 7 is 63, and I can definitely subtract 63. And if we do the subtraction, we have a leftover of 4. And all together, we've subtracted 10 plus 9, 19 divisors of 7, leaving 4. And so we can say 137 divided by 7 is 19 plus 4 7's. Now there is one important change we have to make when we're dealing with polynomials. With whole number division, we stop when the next subtraction will give us a negative number. With polynomial divisions, we must continue until the remainder has a degree less than the divisor. So let's divide 8x plus 6 by x minus 3. Rather than repeatedly subtracting x minus 3, we'll subtract multiples of x minus 3. What, multiple? Well, if you don't play, you can't win. So let's subtract, oh, I don't know, 5 divisors of x minus 3. Now let's think about that. 5 divisors of x minus 3 will be 5x minus 15. Now if we do this subtraction, we're going to subtract 8x minus 5x, and we'll still have 3x left over. We'll still have a first degree polynomial, so we won't be done. So let's see if we can subtract more. How about, oh, I don't know, 10 divisors of x minus 3? Well, that'll be 10x minus 30, but now we have 8x, and we're trying to take away 10x, which is more than what we have, so we don't want to do that either. So what if we subtract 8 divisors of x minus 3? 8 divisors of x minus 3 will be 8x minus 24, and now when we subtract, we'll have 8x minus 8x, so our x terms will drop out, and we'll have a polynomial of degree 0, a lower degree polynomial, which is exactly what we want. Well, let's not celebrate just yet. We still have to do the subtraction. To do the subtraction, it's convenient to remember that we can subtract by adding the additive inverse. So that means instead of 8x minus 24, this is really 8x plus negative 24, and again, because we're subtracting, we'll change this to adding the additive inverses, so we'll add negative 8x plus 24, and that gives us 30. And we can summarize our results. What we've done is we've taken 8x plus 6, we've subtracted 8 divisors of x minus 3, and left a remainder of 30, so that means our quotient will be 8 plus 30 over x minus 3. What about the division 2x squared minus 5x minus 7 divided by x plus 4? So remember, we can't stop until we end up with a remainder that has a degree lower than the divisor. In this case, our divisor is x plus 4 is a first degree polynomial, so we need to keep subtracting until we have a 0 degree polynomial a constant. But there's a problem. We're starting with a second degree polynomial, and no matter how many times we subtract a first degree polynomial, we'll never end up with something that isn't a second degree polynomial because we have this x squared term that will never go away, unless we subtract x times our divisor x plus 4, because that'll give us an x squared term, and we'll have some hope of getting rid of this 2x squared term. So let's start by subtracting 2x divisors of x plus 4. So 2x divisors of x plus 4 is the same as 2x squared plus 8x. We'll subtract by adding the additive inverse. That gives us minus 13x, and we'll pull down this minus 7. It was always there to begin with, but we didn't bother writing it. Now we want to subtract some more multiples of x plus 4, so we want to subtract negative 13 divisors of x minus 4. We'll change our signs and add, and since our remainder is a 0 degree polynomial, we can stop. So altogether, we've subtracted 2x minus 13 divisors of x plus 4, and we have a remainder of 45, so we can say our quotient is 2x minus 13 with remainder 45, or we can write it in fraction form 2x minus 13 plus 45 over x plus 4. The important thing to recognize here is you can use the preceding approach to find any polynomial quotient. And conversely, every algorithm for dividing is just a reorganization of the preceding work. We'll take a closer look at that process in the next video.