 So let's take a look at something called the rational root theorem and this is based on the following idea. So what we wanted to in general is we want to find the roots of a polynomial. We want to find the solutions to polynomial equal to zero. Now if I have a quadratic polynomial then I can use the quadratic formula to find the roots. If I have a third degree or a fourth degree polynomial there are also formulas for finding them but they get really horrifically complicated. But we also have one other thing. In general what we're most interested in are rational roots, roots that can be expressed as a quotient of two integers. And so we have what's called the rational root theorem and that's the following. Suppose I have a rational solution and assume that I've reduced that to whatever it can be. I have a rational solution to some polynomial equations equal to zero. Then the denominator, the denominator of my rational solution is going to be a divisor of my coefficient of the highest degree term. And the numerator is going to be a divisor of my constant term. And what that says is given any polynomial equation I can tell you what the rational roots might be because they're a divisor of the constant term divided by a divisor of the leading coefficient. And because positive and negative numbers also serve as divisors they're really plus or minus the quotient of the two. Now once I have those roots I could then apply something called the factor theorem. And what this says is that suppose I find a solution to this polynomial equal to zero then as soon as I find that solution I also have a factor. And you might have been taught in previous courses that when you're given a polynomial equation you factor to find the solutions. And that's really the worst way of doing it because what we're really looking at is the factor theorem tells us that you solve the equation to find the factors. You don't factor to find the solutions because that in general is going to be too difficult. How do you find the solutions? Well to find the solutions we use the rational root theorem to see what could be a solution. Well how does knowing one factor help us find the factorization of the polynomial? Well once I have one factor remember the problem of factoring is I'd like to write this as a product of two things. If I know one of the things is this then I can find the other factor by dividing. I can take this and I can divide it by x minus a and where this all comes together is that I can do that very easily and quickly using synthetic division. Now in general when I say I want to factor a polynomial we'd like to limit the coefficients of the factors to rational numbers. And so any non-rational roots while there's still roots we don't ordinarily use them in our factorization process. So for example let's try a simple problem factor 9xq plus 5x minus 2. And so to begin with so we can find this factorization by trying to solve the equation 9xq plus 5x minus 2 equals 0. So I've taken my polynomial expression I make an equation by setting it equal to 0 and I try to solve it. So here's the worst way of trying to do this factor the left hand side to find the solutions. Well you don't want to do that if you could factor the left hand side you would have factored it back here. So again what makes this work is the rational root theorem guarantees that any rational root is going to be a divisor of the constant term that's 2 divided by the coefficient of the highest degree term that's going to be 9. So my divisors my potential divisors well things that divide 2 are going to be 1 and 2 things that divide 9 are going to be 1, 3 and 9 and a quotient of one of these with one of these is a possible rational root because those divisors can be plus or minus I also want to take my quotients as being plus or minus. So that suggests that I have the potential roots as follows. So I can take 1 divided by 1 I can take 1 divided by 3 I can take 1 divided by 9 I could take 2 divided by 1 I could take 2 divided by 3 and I could take 2 divided by 9 and so here are my potential roots of this polynomial and so potential factors are going to be x minus one of these. Well which one? Well I'll randomly pick one because any randomly selected number is the correct answer then if I pick any one of these I'll probably have the root the factor I actually we don't actually know which if any of these are actually roots so the only thing we can do is we can check each possibility until we find one or if we do work our way all the way to the end here and don't find a rational root then we know that there are no rational roots and there is no factorization possible beyond what we already have. So since we want to do lots and lots of tedious arithmetic we want to do the easy things first so let's pick out the whole number of possibilities first and if we get lucky in these work then we'll have something that will have done some easy arithmetic and if we are unlucky and they don't work the arithmetic isn't too tedious so I'll try x equals positive one so I'll apply my synthetic division algorithm and and I'll try x equals one so again if I apply the synthetic division algorithm and I use one here what I get as my last term is the same as evaluating this polynomial at x equals one and if the polynomial evaluates to zero I know that x equals one is a root and that x minus one is going to be a factor so I'll apply that synthetic division algorithm drop the nine down multiply and add multiply nine by one and add multiply 14 by one and add and at x equals one I find the polynomial evaluates to 12 which means that x equals one is not a root and x minus one is not a factor and since our first attempt to find a root failed we should give up and join the french for a legion well we should try the next possibility before we do something quite so drastic so let's see well that is a plus or minus one so I might try x equals negative one and see how that works so I'll apply my synthetic division algorithm I'll drop the nine down multiply add multiply negative one by negative nine add multiply 14 by negative one is negative 14 and add and I get negative 16 and that tells me that x equals negative one the polynomial evaluates to negative 16 so negative one is not a root and x minus negative one is not a factor well the next whole number possibility is two so again I'll apply my synthetic division algorithm drop the nine multiply add multiply add multiply add and again I find x equals two is not a root again that's a plus or minus two so I'll try negative two see if that gets me anywhere I'll drop the nine down multiply add multiply add multiply add and once again x equals negative two is not a root well you know it was coming sooner or later we have those fractions in there we're going to have to try the fractional solutions and so we'll try again we'll go starting from the start x equals one third and so again applying the synthetic division algorithm I'll drop it down one third times nine is three add one third times three is one add one third times six is two add and I get a zero at the end there and so what happens with that zero and x equals one third the polynomial evaluates to zero so x minus a third is a factor and the other factor is going to be the quotient that I get from the synthetic division so let's take a look at that x equals one third is a root so I know this polynomial is x minus a third times some other polynomial which works out to be nine x squared plus three x plus six and I want to see if I can factor that so now the other term this term is quadratic so that means it's a good idea to waste time using trial and error factorization well it's quadratic let's apply the quadratic formula to find the rational roots to see if they exist so I'll substitute into my quadratic formula a is nine b is three c is six so that's negative b plus or minus b squared minus four ac all over two a and I'll let the arithmetic dust settle and at this point should be obvious that when I do this next step here I'm going to be taking the square root of a negative number and that means my remaining roots are going to be complex numbers and there are going to be no additional rational factors so here is the complete factorization as far as we're able to take it