 Once we have the quadratic formula, we can use it in order to factor quadratic expressions. So this emerges from an important result in mathematics known as the factor theorem. Suppose I have a solution to polynomial equal to zero. So I have some polynomial and it's an equation where that polynomial is equal to zero. And suppose a solution falls out of the sky and hits me on the head. Then one of the things I know is that x minus a is going to be a factor of that polynomial. So if I can find a solution, then I can factor. And we also use the term root. When we talk about the polynomial itself, we say that x equals a is a root. And what we mean is that x equals a is a solution to the equation polynomial equals zero. Now one of the things this means is that I can factor a polynomial by finding the roots. And this is a very important. You never, never, never, never, never try to solve a polynomial equation by trying to factor it. Instead what you tend to do is you find the solutions and use those to find the factors. So the quadratic expression is a particularly useful example because we have this nice quadratic formula that gives us all of the roots to the polynomial, all the solutions. And so that means the quadratic formula will also tell us how to factor any quadratic expression. So for example let's take a look at x squared minus x minus 12. If you've been practicing, practicing, practicing, practicing, factoring for years and years and years and years you might be able to look at this and say, oh I know how to factor that. But let's say that you want to factor it without having to go through the years and years and years of practice. Well to find the factors we'll look for the roots and that means we'll try to solve the equation polynomial quadratic expression equal to zero. So how do we solve something like this? Well we don't want to factor it. That's the problem we're trying to deal with in the first place. We don't want to factor this but that's okay. It's a polynomial. It's a quadratic expression and we have this very nice quadratic formula that will tell us what the solutions are. So x equals, well I know a is 1, b is negative 1, c is negative 12. So I can drop those into the quadratic formula and after all the arithmetic does settles I'll end up with both solutions x equals 4 or x equals negative 3. And again the factor theorem tells us that once I know the solutions to quadratic equals 0 I also know the factors x minus 4, x minus negative 3. And so that tells me I can factor this polynomial as x minus 4 times x minus negative 3 and I'll do a little bit of cleanup that minus and negative becomes plus 3. So here's another example, factor 6x squared plus x minus 12. And if you've been doing factorization for years and years and years and done hundreds of examples you're going to look at one of these and say, ew, I don't like this type. These are the ones that are really hard and painful because they require lots and lots and lots and lots of trial and error because 6 has lots of factors, 12 has lots of factors. And this is going to be something that's going to take us 20 minutes to figure out. Well we can actually solve it much more rapidly if we remember the root theorem. So again I want to factor this. I'm going to solve the equation quadratic equal to 0. And again I have this nice quadratic formula. So I'll drop these coefficients into the quadratic formula. A equals 6, B equals 1, C equals negative 12. And my quadratic formula gives me this and after all the arithmetic thus settles I do need to know that the square root of 289 is a whole number. That's actually the square root of 289 is 17. But I can then find my factor 16 over 12, negative 18 over 12, or I'll reduce that to 4 thirds or negative 3 halves. And so I have the solutions, 4 thirds, negative 3 halves. And that tells me that this is the product x minus 4 thirds, x minus negative 3 halves. Now there's one limitation to this particular approach. I know that x minus 4 thirds is a factor. I know that x minus negative 3 halves is a factor. There may be other factors that don't give us solutions. In particular there may be a constant factor that I'm missing here. So let's think about that. Now if I expand the right hand side I'm going to get x times x. I'm going to get an x squared over on the right hand side. But I want a 6x squared over on the left hand side. So I have this really complicated problem to figure out how I get a 6x squared. And what I need is an additional factor of 6 to complete de-factorization. So I'll drop in that factor of 6 and there's my factorization. Now as a factorization of this polynomial expression this is a great way of leaving it. This is a beautiful expression for what is the factorization of this. Well it's 6 times this times this and there's our factorization. Now if you've done polynomial factorization for years and years and years and years and done hundreds of examples you might look at this and say I've never come across anything like this. And the reason is that we like to have integer coefficients in our factors. We don't really need them but we like to have them. So what we can do is we can clean this up a little bit. We can eliminate these fractions by breaking this 6 apart into 3 times 2 and then distributing it among the different factors. So 6 I'm going to replace that with 3 times 2 and I'll rewrite this x minus negative as a plus and that becomes 3 and a 2 here. I can rearrange the product anyway I want to because of associativity and commutativity of multiplication and they can distribute the 3 into here, the 2 into there and that'll get rid of my fractions. But again this is a perfectly beautiful, perfectly legitimate, perfectly correct factorization of 6 x plus x minus 12. Everything you do after this point is in the nature of cleanup.