 Suppose we're expanding a product of sums or a difference. The partial products are formed by choosing one term from each factor and multiplying. It's sort of like ordering off a menu. We'll pick one item from the first factor, one from the second, and one from the third. So maybe we'll choose x squared, 2x, and x, which will give us a 2x to the fourth term in the expansion. Or maybe we'll choose 3x, 2x, and negative 8, which will give us negative 48x squared. Or maybe we'll choose 5, 2x, and negative 8 for a negative 80x term. Now if they have the same degree, we can add the like terms and get a single term. So let's find all partial products that include x cubed in the expansion of this product. So our partial products are formed by choosing one term from each factor and so they'll be found by multiplying two terms together. So the partial products that include x cubed are going to be, well if we want an x cubed term, we can choose the x squared term from the first factor and the 8x term from the second factor and multiply them together. And so in our product, we'll have an 8x cubed term. Could we get another x cubed term? Well what if we choose the minus 2x term from the first factor? Then to get an x cubed term, we have to choose the x squared term from the second factor. And so we'll get a partial product minus 2x times x squared or minus 2x cubed. Could we get another one? Well the only thing we haven't chosen from the first factor is 7. So what if we chose the 7 term from the first factor and then our choices of terms from the second factor give us a problem. There's no choice of term from the second factor that will give us an x cubed. And so that means the partial products that include x cubed are only these two. Now if the partial products are like terms, we'll add them together when we simplify. So way back in the day when you did things like 2x plus 7 times 3x plus 5, you did this thing where you multiplied a whole bunch of things together. You said, oh look these two middle terms here are both degree 1 terms. We can add them together and get our final product. So going back to this problem, find the x cubed term in the expansion. Well we already found the partial products that include the x cubed terms. There will be no other x cubed terms, so when we simplify these like terms will be added. And this gives us the x cubed term in our expansion. Well let's try that again. Find the degree 2 term in the expansion of this mess. Now the important thing to remember is we have to make every possible choice. So if we want the degree 2 term, we could choose the 2x term from the first factor, the x term from the second factor, and that is going to give us an x squared, so we have to choose the minus 7 term from the third factor. Now we could have also chosen the 2x term from the first factor, the 3 term from the second factor, but that only gives us an x term, so we must choose the 5x term from the third factor in order to give us an x squared term. We've been picking that 2x term, well let's pick the 5 term. We can pick the 5 term from the first factor, and since we want a degree 2 term, we have to pick the x terms from the other two factors. So we must pick the x term from the second factor and the 5x term from the third factor. Now it might not be obvious why we have to do that, why can't we just pick those terms? Well let's see what happens. If we pick the 5 term from the first factor and the 3 term from the second factor, we might take the 5x term from the third factor, but even then we'd only get 5 times 3 times 5x, that's 75x, which is not a second degree term, and those are the ones we're looking for. And in fact there are no other partial products of the second degree, and so the second degree term and the expansion will be the sum of these partial products. 41x squared.