 Algebra is the generalization of arithmetic, so anything we do in arithmetic has an analog to the things that we do in algebra. The division is no exception. So, for example, let's take a look at the division of two polynomials. If I'm dividing, what I'm really doing is I'm looking for some sort of missing fact. So, for example, 12 divided by 4, I have my dividend, that's 12, the thing I'm dividing, for my divisor, what I'm dividing by, and my quotient is going to be what do I multiply 4 by to get 12? 4 times what gives me 12? And this viewpoint means that I can reframe a question. So, for example, if I want to take a polynomial quotient, 3x cubed minus 4x, or plus 6x minus 20, divided by x minus 2, well, rather than thinking about this as a division, what I can do is I can ask myself the question, what do I multiply x minus 2 by to get x, 3x cubed minus 4x squared plus 6x minus 20? What can I multiply this by to give me this? And, well, I can find products using the area of a rectangle. I can find quotients in the same way. So, let's take a look at our problem. And, again, we can reframe this question as what can I multiply x minus 2 by to get this product? Now, when we multiplied two polynomials, we treated each factor as one side of a rectangle. And the product was the area, was the sum of the individual rectangles that we produced. So, here I want what and x minus 2 to be my two sides of the rectangle. So, I'll make x minus 2 one side, and I'll leave my what as so far undetermined. And, when we set this up, what we did is we divided our rectangle into a number of pieces where our, or each term of our factors was one side of a rectangle. So, here I have the factor x minus 2, so that means I'm going to split this rectangle horizontally. And, that gives me two rectangles. One rectangle has width x, the other rectangle has width minus 2. Now, the other side, I don't know yet, so I don't know how to do those splits. So, I'll leave that unsplit at this point. Next thing we did when we multiplied is we found the areas of a whole bunch of rectangles, we added them up. And, because the terms along what we loosely called a diagonal had the same degree, we were able to add along the diagonals and get our sum along this outer edge. So, that was our product, that was the area of the rectangle, and again, x minus 2 times, I don't know what, should give me a product, 3x cubed minus 4x squared plus 6x minus 20. So, I'll go ahead and write that product along the outer edge of our rectangle. Now, each of the term in our product, each of these terms, was the sum of the areas of a sub-rectangle, a set of sub-rectangles, and they formed along what we loosely called the diagonals. So, this 3x cubed here came from the first diagonal here, which we might represent this way, and this is the only thing that's making up this first diagonal, and that has to be a 3x cube. There's no other choice there, so I'll drop that 3x cubed into there. And remember, this is the area of that rectangle, so that this area had to come from whatever the width is by whatever the height is. Well, the height of that rectangle is x. So, if I know the height is x, then the width, in order to get area 3x cubed, the width of that rectangle has to be 3x squared. So, I know the width of this rectangle. We'll check 3x squared width by x height gives me area 3x cubed. Now, I know the width of this rectangle, which also means I know the width of this rectangle, which means I know the area of this rectangle right below it. This is 3x squared wide by minus 2 high, so this area here minus 6x squared. Now, our next term of the product minus 4x squared had to be the sum of the terms along, again, along this diagonal here, which means that minus 6x squared plus whatever is in here had to give us minus 4x squared. Well, I need to figure out what that is, and so what do I have to have in here? Well, what's in here has to be 2x squared. And we'll check 2x squared minus 6x squared does, in fact, give us minus 4x squared, so this is definitely the term that has to go here. And again, this is the area of that rectangle. We know the height is x, so we know the width has to be 2x. So, again, quick check. Width of 2x, height of x, 2x by x does give me an area 2x squared. I know the width of this rectangle, so I know the width of this rectangle here, and so the area 2x by minus 2, the area here, has to be minus 4x. And again, the sums along the diagonal have to give me by sum here. So whatever this is, plus negative 4x has to give me 6x, and that means that this term in this last box here has to be a 10x. And so I know this is 10x. Again, quick check. 10x minus 4x does, in fact, give you 6x. So if this area is 10x, that's x by 10. And again, I know the area with 10, height negative 2. I know the area this last rectangle here has to be minus 20. Now, let's take stock of what we have. Ignore the question for a moment and look at the picture. According to this picture, 3x squared plus 2x plus 10 times x minus 2 gives us 3x cubed minus 4x squared plus 6x minus 10. Now, go back to the question, well, go back to the reformulated question. We wanted to know what times x minus 2 gives us our dividend. Well, what is apparently 3x squared plus 2x plus 10? And so that allows us to answer this product, quotient. Well, it can do much more complicated quotients this way. Divide x to the fourth minus 3x cubed minus 13x squared plus 19x minus 10 divided by x squared minus 5x plus 2. And again, we want to reframe the question, what times this gives us this? And again, we can do products by finding the areas of rectangles. And in this particular case, we want an area, we want a rectangle where one side is what, and the other side is x squared minus 5x plus 2. So we'll draw that. And again, when we set up our products this way, we partition the rectangle so that every term of every factor corresponded to one side of a rectangle. So I'll split the rectangle into one, two, three parts because there's one, two, three factors. One of these rectangles has height x squared, height minus 5x, height 2, and there's my three rectangles. I did a whole bunch of products, found a whole bunch of areas, and added along, again loosely speaking, the diagonals. And I wrote the sum of those terms along the outside, and that was my area, that was my product. So here's my product, and so I'll write that along the outside. There. So again, going through our analysis, this first term came from the first diagonal, which just consisted of the single rectangle here. This area must have been x to the fourth. Well, I know the width x squared, so I know the width of the rectangle also has to be x squared. The next term, well, I know the width x squared, so I know the areas of these two rectangles as well. This is x squared by minus 5x, so that's minus 5x cubed. This last rectangle here, x squared by 2, that gives me area 2x squared. My next term, 3x cubed negative, well that came from the sum of the terms along this diagonal here. So I'll go ahead and draw in that diagonal. I'll look for what I need to add here. In order to get minus 3x cubed, these two have to add to minus 3x cubed. So that means this term here has to be 2x cubed. And again, I know the height of this rectangle. So I know the area, which means I know the width of the rectangle as well. That width has to be 2x. So I know the width of the rectangle. I know the width 2x. I know the height minus 5x. So I know the area minus 10x squared. I know the width and height. So I know the area. And now my next term came from this diagonal. Again, the one true diagonal we have in this problem. So here I know that this plus this plus whatever this is had to give me minus 13x squared. So that means that this has to be minus 5x squared. We'll check. This is minus 15 plus 2x squared. That does give us minus 13x squared. I know the area. I know the height. So I know the width has to be minus 5. I know the width. I know the height. So I know this area. I know the width minus 5. I know the height too. So I know this area. And let's see. This term had to come from the sum of these two along that diagonal. And it does. This term had to come from the last diagonal. And it does. And again, taking stock of what we have. Ignore the question again. What we have here tells us that x squared plus 2x minus 5 times x squared minus 5x plus 2 gives us x to the fourth minus 3x cubed minus 13x squared plus 29x minus 10. It says that this times x squared minus 5x plus 2 gives me my dividend. And so that answers my question. What times x squared minus 5x plus 2 gives us this? Well, here's our what? x squared plus 2x minus 5. And that is our quotients.