 So let's put our root and remainder theorems together and look at solving polynomial equations. And we'll approach it this way. We can use the rational root theorem and the zero factor property to solve polynomial equations. First, we'll use the rational root theorem to find a rational root. We'll use the factor theorem to rewrite the polynomial as a product. And we'll lather rinse repeat. And at the end, we'll use the zero factor property to solve our equation. So let's consider our polynomial equation, xq plus 4x squared plus 9x plus 18 equals zero. The left-hand side is a polynomial with integer coefficients, and so the rational root theorem guarantees that a rational root will be among the divisors of 18 over the divisors of one. And so that means that the divisors of 18 are 1, 2, 3, 6, 9, and 18. The only divisor of 1 is 1, so the rational roots, if they exist, have to be among plus or minus 1, 2, 3, 6, 9, or 18. And again, the important thing to remember is that the rational root theorem only gives possibilities for the roots. You have to check to see which, if any of these, are in fact roots. And that means we'll have to check each of these to see if it's a root. Since we want to do this in the hardest way possible, maybe not. Since we want to do this the easiest way we can, we'll start by checking the easiest numbers. So we'll check to see if 1 is a root. And an easy way to do that is to use the remainder theorem. Remember, if we want to find the value of a polynomial at x equals a, we can find the remainder when the polynomial is divided by x minus a. Of course, unless we have a fast and easy way to divide, we shouldn't use the remainder theorem. Fortunately, we do have a fast and easy way to divide by x minus a. We can use synthetic division. So let's see if x equals 1 is a root. So remember, this means we want to know if x equals 1 makes the polynomial 0. And we can do that by finding the remainder when we divide by x minus 1. So let's divide our polynomial by x minus 1 using synthetic division and see what our remainder is. So we'll prepare our synthetic division table, then divide by x minus 1. So remember, this tells us that our dividend can be written as x minus 1 times a polynomial plus a remainder. But that means that x minus 1 is not a root. Because if x equals 1, the polynomial will evaluate to the remainder 32. We'll check to see if minus 1 is a root. We'll divide. So again, we know our polynomial is x minus negative 1 times something plus 12, which means that negative 1 is not a root. We'll check to see if 2 is a root. And again, since the remainder isn't 0, this is not a root. So we'll try negative 2. And remainder equals 8 means that this is not a root. So we'll try x equals 3. Nope. x equals negative 3. Finally, and we see that x cubed plus 4x squared plus 9x plus 18 factors as x plus 3 times x squared plus x plus 6. Now our equation is written in the form product equals 0. So by the 0 product property, we know that one of the factors has to be 0. So either x plus 3 is equal to 0, which we solve, or x squared plus x plus 6 is equal to 0, which we'll solve using the quadratic formula. And so we have our three solutions, two of which are complex numbers. How about this polynomial equation? The divisors of 6 are 1, 2, 3, and 6, while the divisors of 2 are 1 and 2. So the rational roots, if they exist, have to be of the form divisor of 6 over divisor of 2. And so this gives us a lot of possibilities. And again, the rational root theorem only gives possibilities. You have to check to see which, if any, are the roots. And that means we have to check through all of these, or at least as many of these as we can until we find a root. And if we do so after a little bit of effort, we find that x equals 1 half is a root. And so our polynomial factors is x minus 1 half times. And we'll make our life a little bit easier by factoring a 2 out of this second polynomial. So now we have product equal to 0. So we know that 1 of the factors equals 0. So either x minus 1 half is 0, which gives us a solution. 2 equals 0. Well, that doesn't happen. So we can ignore this case. Or x squared plus 8x plus 6 is equal to 0. So we'll run this through the quadratic formula and get a couple more solutions.