 One important skill in algebra is the ability to multiply two polynomials. And it's very helpful to rephrase an algebra problem as a geometry problem. And in this particular case, well you might remember from geometry, if I want to find the area of a rectangle, I can find that area by multiplying the length and the width. So the area is the product of length and width. And in algebra, if I want to find a product, well product is product of length and width is the area of a rectangle. I can find a product by finding the area of a rectangle. And what makes this work is that area is what's called a conserved quantity. And the idea is I can break a figure up into several regions, compute the area of each of the pieces, and then add those areas together. And the areas of the individual regions may be easier to figure out than the area of the whole figure. So let's take a look at that. So let's say I want to multiply 3x minus 7 by 2x plus 1. So I have this nice algebraic product. And what I can do is I can think about this product as corresponding to the area of a rectangle, where one side is 3x minus 7 and the other side is 2x plus 1. Let's go ahead and draw that. Now you defeat the purpose if you find the area of this rectangle by multiplying 3x minus 7 by 2x plus 1. What makes this work is I can partition this rectangle into a bunch of pieces, each of which has an easier area to compute. So to do that, what I'm going to do is I'm going to cut this rectangle into pieces. And I'm going to cut it in such a way that each term of each factor gives me part of one side. So I'll split it like this. And now I have a bunch of smaller rectangles, but each rectangle has a very easy area to compute. So for example, this first rectangle, this is a rectangle that's 3x wide by 2x in height. Here we have a minus 7 by 2x rectangle. Here's a minus 7 by 1 and so on. So I'll find the area of each of these sub rectangles. And so this first one, 3x by 2x, that area with times height 6x squared. This rectangle, 3x by 1, so the area of this is going to be 3x times 1. That'll be 3x. This is minus 7. We're in algebra. Negative numbers don't bother us. Negative 7 by 2x, that's width by height, that's negative 14x. And finally, negative 7 by 1, this last rectangle has area minus 7. And because area is a conserved quantity, if I want to find the area of the whole thing, I can just add the pieces together. Now if we've set everything up as efficiently as possible, the terms along what we might loosely call a diagonal will be of the same degree. And we can add those terms and write our sum along the outside. So here, our first what is not really a diagonal, but sort of looks like one, this 6x squared. There's no other square term, so this is going to be 6x squared by itself. Now we have a true diagonal here. This is minus 14x plus 3x. And so that adds to negative 11x. And then our last, not really a diagonal, minus 7. And so when I add all of these together, what I get is 6x squared minus x minus 7. And there's my product. Now you might think about, well, let's see, what if I multiply something that looks like this? Well, you might have been taught how to multiply two polynomials using something called FOIL, which is an acronym for first or it's, I forget. But the problem is that FOIL only works on problems that are specifically constructed to use FOIL. And what that means is that most polynomial products can't be done using FOIL. Forget FOIL. FOIL is useless for almost every polynomial product. So we cannot, we can, however, find products this way. So again, the product is going to be the area of a rectangle where I know the two sides. One of them is this, the other one is this one. And so my rectangle, x squared, 2x minus 5, there's my first factor. The other one, x squared minus 5x plus 2, there's my second factor. And I'm going to cut this rectangle apart so that each term of each factor is one side of a rectangle. So there's my partitioning. And again, I can go through and find the areas of each of these individual rectangles. So this first one, this is x squared by x squared. So this first rectangle has area x to the fourth. Next one, this is x squared by minus 5x, there's my side. x squared by minus 5x, that's minus 5x cubed. This one, x squared by 2, that's 2x squared. 2x by x squared, that's 2x cubed. 2x by minus 5x, that's minus 10x squared. 2x by 2, that's going to be 4x, minus 5 by x squared, minus 5x squared. Minus 5 by minus 5x, that's 25x, minus 5 by 2, that's minus 10. And again, if we've set everything up as efficiently as possible, the terms along the diagonal should have the same degree and I should be able to add them. So let's see, our first diagonal is just this x to the fourth. Our next diagonal, 2x cubed minus 5x cubed, that gives us minus 3x cubed. Our next diagonal, which is a true diagonal, here's minus 5x squared, minus 10x squared, plus 2x squared, that's minus 15 plus 2, that's minus 13x squared. Our next diagonal, 25x plus 4x, that's 29x. And our last diagonal, minus 10, and adding those terms together, I get my product, which I'll write along the outside. And so my product, x squared plus 2x minus 5, x squared minus 5x plus 2, is going to be x to the fourth, minus 3x cubed, minus 13x squared, plus 29x minus 10, and there's my product.