 So one of the big topics in mathematics is factoring, and this is actually by far the hardest, easy operation. It's the hardest thing to do that you can describe to somebody without a lot of background in mathematics. And fortunately, there are certain things that we can do that'll make this a little bit easier. And one of the things is we can rely on the quadratic formula. And this goes back to the factor theorem, which is the following. Suppose I have a solution to some polynomial equation So suppose I know for whatever reason it falls out of the sky and hits me on the head that x equals a is a solution to this polynomial equation. Then I also know immediately that x minus a is a factor of this polynomial. And the other term we sometimes use is we talk about the root of a polynomial. The root is going to be anything that makes this polynomial equal to 0. It's anything that's going to solve this equation. Well, that root is actually really useful because it means I can factor by finding the roots. And this is an important point to emphasize. If I know the roots, I can find the factors. I can factor by finding the roots. You never, never, never, never, never, never, never, never want to factor the polynomial to find the roots. That is generally a waste of time and effort. You find the roots first, and that gives you the factorization. And here's the connection to quadratic equations. I have a way of finding the roots of a quadratic equation. So I can use the roots, use the quadratic formula to find the roots of the quadratic equation, and I can then immediately factor. Again, you never want to factor to find the roots. You always want to find the roots in order to do the factorization. So for example, let's take the expression x squared minus x minus 12. And to find the factors, I find the roots, and the roots will be the solution to the equation x squared minus x minus 12 equal to zero. And I have the quadratic formula. This is a quadratic equation. I can apply the quadratic formula. x equals negative b plus or minus square root b squared minus four a c over two a. So again, a one b negative one c negative 12. I'll substitute those in quadratic formula. I'll simplify. I'll let the dust settle. One plus or minus seven over two, eight halves or negative six halves, that's four or negative three. And so I have the solutions for negative three. And so I know that a factor will be x minus four and another factor will be x minus negative three. And so that gives me my two factors of the polynomial. And I'll simplify this a little bit. That's x plus three. And here's the connection. This is a x squared. This is a quadratic polynomial. So I only expect two possible factors, x minus four, x plus three. Now there's one possible twist that we might run into. This gives us polynomial factors, but we might be missing a constant factor. No problem. We can include that or not as the case may be. So our factor is a should say two x squared plus 11 x plus 12. And again, my roots will be the solutions. So I'll have to solve this equation here. And again, I have the quadratic formula. So I'll use the quadratic formula to give me the solutions. So substituting those values in a is two, b is five, c is negative 12. So there's minus b plus or minus squared b squared minus four a c over two a. There's my quadratic formula. I'll let the dust settle. I'll let the dust settle. And that gives me two roots, negative five plus or minus 11 over four, otherwise known as three halves or negative four. So again, I have two roots, x equals three halves, x equals negative four. So that tells me I have two factors of x minus three halves, x minus the one root, x minus negative four, x minus the other root. And the one qualifier here is that I may have some additional constant factors. So here's the thing to notice. If I expand the right hand side, I get x times x, I get x squared. But I want a two x squared, which means I'm missing a factor of two. So I'll go ahead and throw that factor of two in. And there's my factorization. Now I might want to do a little bit of cleanup. I don't really need to. This is a perfectly beautiful factorization of two x squared plus five x minus 12. But you may want to clean that up just a little bit. Let's see. We should clean up the double negative, that's x plus four. And here's a much nicer way of writing the factorization. And if you really, really, really want to do some extra work, you can make all of our factors involve whole numbers by multiplying this two into, well, if I multiply it into this factor here, that'll get rid of that fraction. That's two x minus three. But really, the factorization is complete at this step. It's nicer at this step, and this step here is kind of unnecessary. Well, let's do a horribly messy factorization, six x squared plus x minus 12. And this is one, if you've been taught how to do trial and error factorization, you could spend the next half hour or so trying to figure out what that factor is. But I can find the roots directly because I know the quadratic formula. So again, the roots are the solutions to this equation. So I'll use the quadratic formula. And I get roots of four thirds or negative three halves. And so that tells me that my factors are going to be x minus four thirds and x minus negative three halves. And again, I look at this, if I expand the right hand side, I get an x squared. I actually want a six x squared. So that means I'm missing a factor of six. To complete my factorization. And again, beautiful factorization. I really don't have to do anything beyond this point. But if I want to get whole number of factors, I can multiply the six into these expressions, and that will eliminate the fractions. And I can split the six into three times two. And if I distribute that into three into here, two into there, that'll get rid of my factors. So here six is three times two. And the associative and commutative property of multiplication, say I can put the two over here. The associativity says I can multiply these out. I can multiply these out. And that gets me three x minus four times two x plus three.