 So one important thing to keep in mind about doing mathematics is not to get lost in the pushing symbols around odd paper, but to remember why you're doing what you're doing. And this is particularly true in, well, everything, but we'll just as an example take a look at this in terms of rational expressions. So again, let's consider the problem simplifying this rational expression. And so first of all, the one thing you should absolutely always do is find the forbidden values that will make the denominator equal to zero. And again, the reason that you're doing this is if the denominator is ever equal to zero, this expression is meaningless. So we'll go ahead and solve for those. And that's a nice quadratic equation. And we can solve that and you find x equals three x equals negative two are the roots that will make the denominator equal to zero. So I can factor the denominator into x minus three times x minus negative two x plus two. So our denominator factors nicely. Now our next step in general is to try and simplify the numerator by factoring it. And while you might note that this is a cubic equation x to the third, so we don't have any nice quadratic formula to solve for it. So we fall back on the rational root theorem and the rational zero theorem and our solutions are going to be divisors of 12 divided by divisors of 6. And unfortunately, that's a lot of numbers. Divisors of 12 include 1, 2, 3, 4, 6, and 12. Divisors of 6 include 1, 2, and 3. So I'm going to have lots and lots and lots and lots and lots and lots of potential factors. And that's a lot of work even though we do have a fast way of doing that. So here's an important question to ask. We could factor the numerator. Why? And the reason is that factoring is only going to be useful if we're going to be able to cancel a factor from the numerator with a factor in the denominator. And, well, there's only two factors in the denominator, which means that if I factor the numerator and I don't get at least one of these, all that effort in factoring the numerator is wasted. I can't do anything useful with it. So what we really care about, what is really important here is, is x minus 3 a factor of the numerator? Or is x plus 2 a factor of the numerator? And so I can determine not whether the denominator factors, but whether its factors include x minus 3 or x plus 2. So I can answer this question directly using the factor theorem. If x equals 3 makes the polynomial, makes our numerator equal to 0, then x minus 3 is a factor. And conversely, if x equals 3 does not make this numerator 0, then x minus 3 is not a factor, and I don't have to worry about it. I'll use synthetic division in the remainder theorem to evaluate our polynomial at x equals 3. So I'll set up my synthetic division table. So again, the value I'm testing is x equals 3, my coefficients, 6x cubed, 3x squared, minus 24x, minus 12 is my constant, and I'll apply the synthetic division algorithm. I'll drop, multiply, add, multiply, add, and, well, a little analysis goes a long way. So before I try to do this computation, I might ask myself, well, you know, 39 times 3 is what I'm going to have to multiply. But when I add that to negative 12, what I really care about is whether or not I get 0. Well, 39 times 3 added to negative 12. Well, 39 is already bigger than 12. So times 3, I'm a lot bigger than 12. So if I add it to negative 12, what I get here is not going to be 0. So who cares what that product is? It's not going to give us 0 at the end of it. I don't actually need to finish the synthetic division algorithm. What I do know is that x equals 3 does not make my polynomial equal to 0. So I know that x minus 3 is not a factor. Well, let's check the other factor of significance, this x plus 2. And so I can check to see if x plus 2 is a factor. If x equals negative 2 makes the polynomial equal to 0, then x plus 2 is a factor. Otherwise, it isn't. So I'll set up my synthetic division table. So I'll drop, multiply, add, multiply, add, multiply, add. And at the end of it, I have a 0, which tells me that x equals negative 2 makes the polynomial 0, which means that x minus negative 2, x plus 2 is one factor of this numerator. And that gives me a factor. Now the other factor is going to be what I got from the synthetic division. So I also have the other factor. And now I have a common factor I can cancel out. There it goes. And I have another expression. Now we've already determined x minus 3 is not a factor of the numerator, which means that I can't simplify this expression any further. However, just to dot all your i's and cross all your t's, you may want to indicate this just to make everybody happy. You should actually point out that you're not going any further because it's not worth it and not because you decided you wanted to quit the problem at this point. So it's worth making a point of saying that x minus 3 is not a factor of the numerator. So I can't actually simplify this expression any further.