 So, let's take a look at how we divide polynomials using something called the division algorithm. So, an important idea to understand is arithmetic is bookkeeping. And what that means is that every algorithm that you learned in arithmetic and in polynomial arithmetic is a bookkeeping trick to keep track of how much of which quantity. And this is an important idea because success in mathematics does not depend on how many theorems you remember. It does not rely on how many formulas you know, and it's not based on how many algorithms you can apply. It is centered around understanding fundamental concepts. And with polynomial arithmetic, the fundamental concept is arithmetic is bookkeeping, and everything you do is centered around that. And in fact, if you really want to do yourself a favor and really understand division, skip this video. Now, keep in mind you'll still need to be able to divide polynomials, but in the long run, you're going to be better off if you ask yourself the question, how do I organize the repeated subtractions that I have to do with division? So if you're still watching, let's look at how this organization works. When we divide whole numbers, we organize our subtractions. So when we divided 137 by 7, we began by subtracting 10 divisors of 7, and then we subtracted 9 divisors of 7. And altogether, we subtracted 19 divisors of 7, and had a remainder of 4. Now you might be familiar with something called the long division algorithm, and the thing to recognize is that all that the long division algorithm is, is it's a rearrangement of these subtractions and these record of the divisors. So we'll write our dividend 137 here, we'll write our divisor 7 here. So my first step was subtracting 10 multiples of 7, so we'll keep track of that here. And then we'll do the subtraction below. Our next step was to subtract 9 divisors of 7. So we'll record that, and now we have subtracted 19 divisors of 7, doing the subtraction. And so 137 divided by 7 is 19 with remainder 4. So the important thing to recognize here is that the familiar long division algorithm really is just a reorganization of this process of repeated subtraction. We can do the same thing with polynomial divisions. So when we divided 8x plus 6 by x minus 3, our subtraction looked like this. But if we rearrange things so they look like the long division algorithm, the divisor is going to come out front here, and we'll keep track of the quotient above and subtract below. Or when we did this division, we subtracted 2x divisors of x plus 4, and then we followed that up by subtracting an additional negative 13 divisors of x plus 4. And we got our remainder and our quotient. So again, we'll rearrange the pieces. We'll move the divisor to the front, and the tracking for the number of divisors will move to the top of the division bar. And so we have our quotient and our remainder. Again, the thing you want to remember is that arithmetic is bookkeeping. And when we divide, we are subtracting multiples of the divisor. So we can think about this as subtracting multiples of x minus 2 from this 5x squared plus 3x minus 7. Since our divisor is a first degree polynomial, we need to subtract until we end up with a zero degree polynomial, a constant. So that means we definitely need to get rid of the 5x squared and the 3x terms. So I want to multiply x minus 2 by something. That'll get me 5x squared, so I can then subtract it. So I can multiply x minus 2 by 5x. So I'll multiply 5x by x minus 2 to get 5x squared minus 10x. But now I want to subtract, so I'll change the signs and add, which gives me 13x. Well no, it actually doesn't give us 13x. Remember arithmetic is bookkeeping. We had a minus 7. It can't just disappear. So we actually get 13x minus 7. So we've gotten rid of the x squared term. Now we have to get rid of this 13x. So what could we multiply x minus 2 by to get us a 13x? And the answer is 13. So we'll take 13 copies of x minus 2. That's 13x minus 26. We'll subtract by changing the signs and adding. And now we have a constant, a zero degree polynomial, which means that we're done. Well, almost. We still have to write down our quotient. So our quotient is going to be 5x plus 13. Our remainder 19 can be expressed as the fraction 19 over x minus 2. Again, the long division setup is just a way of keeping our subtractions organized. So here I want to subtract multiples of x squared minus 30x plus 7 until I get a low degree polynomial. And because our divisor is of degree 2, then our remainder should be of degree 1. So we'll set up the division. And first we want to get rid of the x cubed term. And since my divisor is an x squared, if I multiply it by x, I'll get an x cubed term. Multiplying our divisor by x gives us x cubed minus 3x squared plus 7x. We want to subtract this, so we'll change the signs and add. And so now I don't have an x cubed term, but I do have 11x squared. So to get an 11x squared from our divisor, we need to multiply by 11. And so that'll give us 11x squared minus 33x plus 77. We'll subtract by changing the signs and adding, not forgetting that we have a 5 here. A arithmetic is bookkeeping. And now what we have is a first degree polynomial because our divisor is a second degree polynomial. We have a lower degree and we're done with the division except for the statement of the final answer.