 So again, algebra is the generalization of arithmetic, and so sometimes we run into some problems with division. So while I can try to answer division question by looking for a missing factor, sometimes I need a little bit more. So for example, let's take 17 divided by 5. And so I can reframe this as the question, what times 5 gives us 17? And so since we know our arithmetic facts, we know that 3 times 5 gives us 15. And 4 times 5 gives us 20. And so there's nothing we can multiply 5 by that'll give us exactly 17. So, well, we might look at it this way. I might note 3 times 5 gives us 15, which is close to what we want, but it's too short of what we want. So what I need is I need 3 5s and 2 more. And so I can interpret that as by saying that 17 divided by 5 is going to be 3. That's how many 5s I need. But then there's a remainder. There's this additional bit of 2 that I need to make 17. And so I can think about the remainder as the additional amount that I need to make up the dividend. And we can apply this problem to polynomial quotients. Now, fair warning, we'll have to draw some pictures, but that's not too difficult a task because all of the pictures we're drawing are going to be rectangles. So let's take a look at the quotient x cubed plus 4 x squared minus 8 x plus 12, the whole thing divided by x squared minus 3x plus 3. And again, the thing that I really want to do is I can reframe this as a question. What do I multiply this by to get, well, if I can't get this exactly, I want to get as close as I can to it. So possibly I might need some extra pieces to make this exactly. So let's go ahead and set up our rectangle. And again, this question is really an area problem. The area of a rectangle, that's what? By x squared minus 3x plus 3. And then maybe I'll add in a little bit more pieces of area. So I'll set up a rectangle where one side is x squared minus 3x plus 3. The other side is what? And this area is this sum. So there's our area. There's our rectangle. So again, I've set up my rectangle. I've partitioned the factor so that now I have 3, 1, 2, 3 rectangles. Each will side x squared minus 3x 3. My area of the whole thing I found by adding along the diagonals. I'm going to write along the outer edge here. And we'll go through our process. So here our first term came from the first diagonal, which means that this first diagonal, which only has the one term, had to have an x cubed in it. I know the area. I know the height. So I know the width has to be x. Now I know the width of all these rectangles. So I know the width. I know the individual heights. And so I can find the area of this rectangle. That's x by minus 3x and x by 3. That's going to be 3x. So now I know these areas. This next term came from the sum of the terms along this diagonal. So I know this term, this term has to have been 7x squared in order for these two terms to add up to 4x squared. Well, I know the area of this rectangle. I know the height of the rectangle is x squared. So that means I know the width has to be 7. And so once I know the width, I also know the areas of these two rectangles. So this is 7 by minus 3x. This is 7 by 3. And so I can fill in those areas. This term here has to come from the sum of the terms along this diagonal. And so that means this term here, minus 8x, has to be all of these three terms added together. And so that means this one has to be 10x. And again, quick check, 10x minus 21x plus 3x does in fact give us negative 8x. However, at this point, we have a problem because if I want to find the width of this rectangle, well, the height is x squared, the area is 10x, the width of this rectangle is not going to be a term of a polynomial. And so I should stop. Now the analog to this, by the way, if you're doing long division, is that if you do long division, you have a whole number portion, but then you get to the last whole number part you could make, and then the only way to proceed from that point is to introduce decimals. Well, for a variety of reasons, we don't want to introduce decimals to our polynomial expansion. So we can stop here, but we're not quite done because we don't quite get the area that we want. This area here is minus 8x, but we have to get this plus 12 as well. So this 12 here had to have come from the terms along these last two diagonals, this last diagonal here, and so this is 21, and this has to be minus 9. So again, ignore the question for a moment. Taking stock of our figure here, we can read this as follows. The rectangle here, which is x by x plus 7 by x squared minus 3x plus 3, this rectangle here, x plus 7, times x squared minus 3x plus 3 plus this bit here, 10x minus 9. If I put all of these things together, I get my x cubed 4x squared minus 8x plus 12. So what times x squared minus 3x plus 3 is this? Well, what? x plus 7, and the additional pieces that we need, 10x minus 9. And so that allows me to answer my reframed question, and I can also write this in a slightly different form. I may want to write it as the quotient gives me x plus 7, and the additional bit is my remainder. So the quotient is x plus 7 with the remainder 10x minus 9, and there's my polynomial quotient.