 An important problem in mathematics is known as the factoring problem. An important property of real numbers is the distributive property. A times the quantity B plus C is AB plus AC. Now, because this is an equality, we can either go from the left-hand side to the right-hand side, or from the right-hand side to the left-hand side. And so we use the following terms. We expand by going from left to right, from A times B plus C into AB plus AC. And it's important to note that this produces a sum. On the other hand, if we go from right to left, we say that we factor AB plus AC into A times quantity B plus C. And it's important to note here that this produces a product. So let's try to factor 5x plus 5y. So again, both factoring and expanding are based on the distributive property, so let's bring that back. So factoring starts with the sum and removes, outside of set of parentheses, a factor that is common to all the terms of the sum. So it helps to read out what these two terms are. The first term is 5 times x, and the second term is 5 times y. And so a factor common to both terms is 5. And so we'll remove that common factor outside the set of parentheses, and inside we'll leave the other factors of the two terms, x and y. Now factoring in general is a very hard problem, so we have to break it down into a number of steps. And so we typically begin as follows. If we want to factor an expression, we begin by looking for a monomial factor, a factor common to all terms in the expression. For example, if we have 7 plus 14 plus 35, we might make the following observation. 7 plus 14 plus 35, we'll let 7 times 1 plus 7 times 2 plus 7 times 5. And since all of our terms have a factor of 7, we can remove 7 as a common factor. Likewise, in 8x plus 5x plus 7x, we observe that that's 8 times x plus 5 times x plus 7 times x. And every one of our terms has a factor of x, so we can remove that as a common factor. How about something like 8x plus 6? So the first thing that we notice is that 8x is 2 times 4x, and 6 is 2 times 3. And that means our expression, 8x plus 6, can be rewritten as 2 times 4x plus 2 times 3. But now both of our terms have this factor of 2. So the distributive property says we can remove that common factor and get our factorization 2 times 4x plus 3. Now the important thing to remember about factoring is that factoring is a very hard problem. And that means that in general factoring is so hard you'll have no idea of where to begin. And so it's important to remember a small step is better than no step. So here we might begin by noticing anything at all about our two terms. And maybe we'll begin by noticing that both our coefficients are even, and so both of them are 2 times something. And so 12x cubed y squared, well that's 2 times 6x cubed y squared. And 8xy cubed is 2 times. And that means both of our terms have a common factor of 2, and we can remove that and get our first step in the factorization. Now if you're a politician, you can say, look, I did something, I solved the problem. But if you're a mathematician or a good human being, you'll ask yourself, have I really solved the problem? And so we still have this sum inside the parentheses 6x cubed y squared plus 4xy cubed. And again we might also notice that both coefficients are even, so both of them have a factor of 2. And so I can rewrite them. And so my expression 6x cubed y squared plus 4xy cubed can be rewritten by removing a factor of 2. And remember that equals means that wherever we see the one thing, we can replace it with the other. So I see a 6x cubed y squared plus 4xy cubed here. I can replace it with 2 times 3x cubed y squared plus 2xy cubed. But wait, there's still more. I have the sum 3x cubed y squared plus 2xy cubed. And because it's a sum, I might be able to factor it further. And we notice that 3x cubed y squared and 2xy cubed both have a factor of x. 3x cubed y squared is x times 3x squared y squared. And 2xy cubed is x times 2y cubed. And so this sum can be rewritten where both terms are products of x and something. And I can use the distributive property. Again equals means replaceable. So I see 3x cubed y squared plus 2xy cubed. And I'll replace it with x times 3x squared y squared plus 2y cubed. But wait, there's still more. We have a sum. So it's possible we might be able to remove some additional common factors. So let's take a look at the two terms of our sum. And we see that both of these can be rewritten as y squared times something else. And so that means our sum can be factored and equals means replaceable. So 3x squared y squared plus 2y cubed can be replaced with y squared times 3x squared plus 2y. Now we still have a sum 3x squared plus 2y. So it's possible we might be able to factor this further. But if we look closely, we see that these two terms do not have any common factors. So it looks like we're done. However, it's traditional to multiply the constants together. So if it's not too much trouble, we can multiply this 2 times 2 to get our final form of the factorization. Now you might take a look at this and say, wow, that's a lot of steps to get to the answer. And the truth is, it is. And we don't necessarily have to go through this every time. Ideally, you'd notice right from the start that 12x cubed y squared is 4xy squared times 3x squared. And that 8xy cubed is also 4xy squared times 2y. And then we could do our factorization in one step. And in fact, this is exactly how you run a marathon. You get up one morning and say, hey, I think I'd like to run 26 miles a day. And you get out and run. Well, actually you typically train for a marathon by taking small short runs over a long period of time. And the same thing with factoring. Eventually you build yourself up to this simple one step factorization. But in the meantime, keep in mind that a small step is better than no step. Well, let's train up for the marathon and take slightly longer runs here. So let's say I want to factor 8x cubed plus 20x squared plus 4x to the fifth. And we might begin by noticing the coefficients 8, 20, and 4 all have a common factor of 4. Meanwhile, the variable portions x cubed, x squared, and x to the fifth all have a common factor of x squared. And so we can remove these common factors. And so our sum is 4x squared times 2x plus 4x squared times 5 plus 4x squared times x cubed. And I can remove that common factor and obtain my factorization.