 The area model provides us with an easy way to multiply two polynomials. We can also use it to divide two polynomials if we keep in mind two ideas. The area of our rectangle is the product of its width and height. The area of a figure is the sum of the areas of all its parts. The bishop diagonals contain terms of the same degree, and a fourth of our two items is that I can't count. Note that the total area of the figure must be 6x squared plus x minus 35, where our first bishop diagonal adds to 6x squared, this second bishop diagonal has terms of add to x, and this third bishop diagonal, just this last square, has terms of add to negative 35. What's important here is that we can reverse the process. Suppose we didn't know the areas of these smaller rectangles, and we didn't know the width of the big rectangle. If we still know the area, we'll be able to find the length. So let's consider that. This first bishop diagonal, which must add to 6x squared, only has one rectangle, so this rectangle must have area 6x squared. And since the height is known, it's 3x, and the area is 6x squared, then our width, because it's a rectangle, must solve width times height equals 6x squared. We know the height, 3x, and that tells us the width is 2x. But wait, now we know the width of the first column, so we can find the area of the other rectangles of the column. There's only one rectangle, but the area of that bottom rectangle must be width times height. We know the width, we know the height, and so we know the area, negative 14x. Now let's take a look at that second bishop diagonal. The two terms along that second bishop diagonal have to add to x. So that means negative 14x, plus this area, we'll call it area 1, has to be x. And so that tells us area 1 is, and since the area is 15x, and the height is 3x, then we know that the width times the height is 15x, and so that tells us the width is 5. But wait, now we know the width of the second column, so we can find the areas of the remaining rectangles. The area of the bottom rectangle area is width times height, which will be the third bishop diagonal should add to negative 35, and it does, so we're done. Now the important thing here is figuring out what we found. The figure tells us several things. For example, it tells us that the product 3x minus 7 by 2x plus 5 is 6x squared plus x minus 35, and so that answers, that's not the problem we're trying to solve. Or is it? For example, we can rearrange by dividing both sides by 2x plus 5, and rearranging gives us 3x minus 7 equals 6x squared plus x minus 35, divided by 2x plus 5, and so this is the answer to a different question. This is still not the problem we're trying to solve, but we could divide by 3x minus 7 instead and get, and here's a useful thing to notice. The same figure can be used for a product and two different quotients. We're in a really unusual situation in that we have a by 1 get 2 free. So let's see if we can find other quotients. Let's try an example where we don't already know the answer. Let's divide 4x squared minus 4x plus 7 by 2x plus 7. We'll set up a rectangle with height 2x plus 7, so our height will be divided into two portions, and we'll divide the width into some number of partitions. We don't know how many, so let's just add a bunch. Now the bishop diagonals will add to the terms of our dividend. So the first bishop diagonal will add to 4x squared, the next one to negative 4x, the next one to 7, and we have no more terms in our dividend, so we won't worry about the remaining bishop diagonals, at least not yet. Since the first bishop diagonal only has a single rectangle, the first rectangle must have the entire area 4x squared. Since the height is 2x, the width will also be 2x, and the rectangle below has area 14x. The next bishop diagonal has to add to negative 4x. Now we already have part of that bishop diagonal, this 14x, and so we know that negative 4x must be 14x plus the area of this other rectangle, which will be, and the area of this rectangle, well that's the product of width and height, so we can find its width, which will be. Again, we know the width of this next column, so we can find the area of the rectangle below, which will be, and the last bishop diagonal must add to 7, and so that means 7 is the sum of negative 63 plus the area of this rectangle, which must be, and at this point, because the degree of this term is lower than the degree of the height, we don't want to continue because then we'll get fractional answers, and so our figure shows several things. First of all, this part is the product 2x plus 7 times 2x minus 9, but then if we add 70, we get our original dividend, or we can divide by either factor and get a statement about a quotient with remainder.