 So let's take a look at factoring quadratic expressions. Before we go too much further, we'll start out by observing that factoring is the hardest easy problem in mathematics. It's easy to explain what the problem is, but often it's very difficult to actually solve the problem. Now, your prior experience with factoring suggests that this might not be true, but the reality is it only seems easy, because we've only ever asked to solve easy problems. So you might have been asked to do something like factor 12, and you've never been asked to do something much harder, like factor 3,891. And that means that in general, trying to factor something is hard, and so we'll try to make it a little bit easier. To keep things simple, we'll limit ourselves to rational factors, polynomial factors whose coefficients are rational numbers. So if we write x squared minus 4 equals x minus 2 times x plus 2, this is a factorization, and all of its factors have rational coefficients. Likewise, 6x squared minus 5x plus 1 can be written as 6 times x minus 1-3 times x minus 1-2, this is a factorization, and all of our factors have rational coefficients. x squared minus 2 can be written as x minus square root 2 times x plus square root 2. Well, this is a factorization, but the factors do not have rational coefficients. If a polynomial can be written as a product of rational factors, we say it is factorable. So these two are factorable. Otherwise, we say that the polynomial is unfactorable. So this is a factorization, but it's not a rational factorization, and so this is unfactorable. Well, maybe. Maybe we just haven't found the factorization. So let's think about this a little bit. Suppose we could write a quadratic expression as a product of two linear factors. So x squared plus bx plus c, maybe I can write that as x minus p times x minus q. With that in mind, let's consider the equation x squared plus bx plus c equals 0. How would we solve this type of equation? Well, equals means replaceable. If x squared plus bx plus c is equal to x minus p times x minus q, then every place I see x squared plus bx plus c, I can replace it with x minus p times x minus q. So I can change my equation into a different equation, and now the equation is in the form product equals 0. And I know that if I have two things that multiply to 0, one of them has to be 0. So either x minus p is 0 or x minus q is 0. Well, I can solve both equations, and I get the solutions x equals p, x equals q. And the thing to notice here is that our solutions, x equals p, x equals q, correspond to the factors x minus p, x minus q. And so this is an example of what's known as the root theorem. Let x be any polynomial in x. If x equals a is a solution to polynomial equals 0, then x minus a is a factor of our polynomial. So let's say we want to factor x squared minus x minus 12. So the root theorem guarantees that if I can solve polynomial equal to 0, I can factor. So to find the factors, we'll solve the equation x squared minus x minus 12 equals 0. If only we had a good way to solve any quadratic equation we were confronted with, oh, wait a minute. Since this is a quadratic equation, we can use the quadratic formula. So we'll substitute the values of our coefficients into the quadratic formula. And when we evaluate the quadratic formula, we'll have our solutions. x equals 4 or x equals negative 3. If it's not written down, it didn't happen. So let's summarize our results. The solutions to x squared minus x minus 12 equals 0 are x equals 4 or x equals negative 3. So two factors are x minus 4 or x minus negative 3. Or maybe we don't like that minus a negative. We could say the factors are x minus 4 or x plus 3. Now in general, the root theorem will give us some factors. But we want all factors. So let's think about that a little bit more. In x squared minus x minus 12, we found two factors, x minus 4 and x plus 3. Since x squared minus x minus 12 is a second degree polynomial, it can only have two linear factors. And what this suggests is that x squared minus x minus 12 has to be x minus 4 times x plus 3. And if we multiply out the right-hand side, we can verify that this is in fact true. And we have factored x squared minus x minus 12. And this gives us a good way to factor quadratic expressions. So let's factor x squared minus 9x plus 20. To find the factors, we'll solve the equation x squared minus 9x plus 20 equals 0. So we'll use the quadratic formula, which gives us the solutions 5 or 4. And so the factors of x squared minus 9x plus 20 are x minus 5 and x minus 4. And so we can write our polynomial as a product. There's one very important idea to keep in mind, which appears with the factorization of something like 6x squared minus 5x minus 6. So we find our equation 6x squared minus 5x minus 6 equals 0 has solutions, x equals 3 halves, or x equals minus 2 thirds. And as before, we might write down our factors, x minus 3 halves, x minus negative x plus 2 thirds. But this is wrong. The problem is that when we expand x minus 3 halves times x plus 2 thirds, the very first thing we get is an x squared, but we want to get a 6x squared. So what can we do? Well, here it's useful to keep in mind an important idea. It doesn't matter what you write down first, fix things as you go along. And so the problem here is that I get an x squared, but I want a 6x squared. And so we can fix this by including a factor of 6. So in our factorization, we'll want to include that factor of 6, and that fixes our problem. Now, of course, I've had many years of experience in doing factorization, so I was able to come up with this factor of 6 very quickly and very easily, if only there are some easy and obvious ways of figuring out where that factor of 6 comes from. Well, actually, there is. When we find our factors this way, we always get factors in the form x minus something or x plus something, which means that when we expand, we always get an x squared as our leading coefficient. So if we want an x squared as our leading coefficient, we're done. If we want anything else, we'll have to include a factor that's equal to our leading coefficient. So let's say I want to factor 2x squared minus 15x plus 18. We'll begin by solving 2x squared minus 15x plus 18 equals 0. So we'll drop that into the quadratic formula. We'll get solutions of 6 or 3 halves, which says that we have factors x minus 6x minus 3 halves. But since I want a leading coefficient of 2, I'll need to include a factor of 2 as well.