 We can divide polynomials using an area model if we remember products correspond to areas, and the area of a figure is the sum of the areas of its components. Moreover, if our figure is drawn correctly, the terms along the bishop diagonals have the same degree, and this will be incredibly useful for us. For example, let's divide x cubed minus 4x squared minus 7x minus 8 by x minus 5. We can represent the problem by using a rectangle with height x minus 5. So our height is broken into two segments, and we'll break our width into some number of pieces, and our bishop diagonals will add 2x cubed minus 4x squared minus 7x and minus 8, and we'll leave some space at the end here to allow for the possibility of a remainder. This first bishop diagonal has one rectangle with area x cubed, and so the area with times height must be x cubed, but since its height is x, the width must be x squared. Now the first column has a width of x squared, so we can find the remaining areas. So this rectangle must have area width times height, but we know the width is x squared and the height is negative 5, and so the area will be negative 5x squared. The next bishop diagonal has to have a total area of negative 4x squared, and so negative 4x squared must be the area of these two rectangles put together. So negative 4x squared is negative 5x squared plus some unknown area. Solving gives us... and since the height of this rectangle is x, the width will be x as well. But now the second column has width x, and so the remaining rectangle will have area negative 5x. The next bishop diagonal has total area negative 7x, but we already know the areas of some of the rectangles, this one in particular, so we only have one rectangle whose area we don't know and we find, but area is the product of height and width, and we already know the height is x, and so the width must be negative 2. The width of the third column is negative 2, so the area of the lower rectangle must be... the next bishop diagonal must add to negative 8, and so negative 8 must be 10 plus the area, and so the area will be negative 18. Now, while we could form a rectangle with area negative 18 and height x, the width would be a fractional quantity, and we haven't had expressions like this as widths. So instead we could read our figure as showing the product, x minus 5 times x squared plus x minus 2, that's this portion of the figure, plus negative 18, that's this portion of the figure, is going to be our original amount, x cubed minus 4x squared minus 7x minus 8, or we can rearrange if we divide everything by x minus 5, and equivalently we can read this as a quotient, x cubed minus 4x squared minus 7x minus 8 divided by x minus 5 gives us x squared plus x minus 2, with remainder negative 18. And the problem doesn't change significantly if we have a quadratic divisor. So here we'll set out a rectangle with height x squared minus 4x minus 7. Notice that since our divisor has three terms, the height will be partitioned into three parts, x squared, negative 4x, and negative 7. And we'll partition our width into some number of pieces, make sure we have enough, and in this case our bishop diagonals will add to x to the fourth, 8x cubed, minus 12x squared, 13x, and 35. Now our first bishop diagonal only has one rectangle, and again we know the area must be x to the fourth, we know the height x squared, and so that tells us the width is x squared. But now that we know the width, we know the heights of the remaining rectangles, and so the remaining rectangles in the first column will have area. The next bishop diagonal must have total area 8x cubed, and so we know that part of that area, negative 4x cubed, and then the area of this rectangle, which will be, and since area is width times height, and we know the height we find, the width is 12x. And again we know the width of this second column of rectangles, and we know the heights of the two remaining rectangles in the column, and so the remaining rectangles have areas. The next bishop diagonal has areas that add to negative 12x squared, so that's this diagonal, but we already know the areas of two of these rectangles, and so there's only one whose area we don't know. Well, actually we do know it because we find, and since area is product of width and height, and we know the height, we find the width is, the remaining rectangles in the third column, now that we know the width will be, since at this point our partial quotient is a real number, any further divisions will lead to fractional answers, so we're done. Or are we? Remember that among integers you can continue long division to get a decimal answer. Continuing past this point would give us a polynomial equivalent of a decimal. Well, no one does that, but we could. We do still need to finish out the problem, because we know that the next two bishop diagonals have to add to 13x and 35, and this means we need two additional areas, the area that will make this bishop diagonal 13x, which will be, and the area that will make this bishop diagonal 35. Now, while we do have two remaining rectangles, since we're done with the problem, it doesn't make a difference how we split up that remaining area, if we put it all into one of those rectangles. And the one rectangle here must have area, so we can read our figure as showing x squared minus 4x minus 7, that's height times width, x squared plus 12x plus 43, gives us this figure, and we add 269x plus 336, we get our entire figure, x to the fourth minus 12x squared plus 13x minus 35. Equivalently, if we divide by one of our factors, we get, or we can express this as a quotient width remainder.