 In our previous video, we learned about the remainder theorem and at first glance, this method of computing remainder seems dramatically easier than the long division of polynomial polynomials we saw before. Now the remainder theorem only applied in situations where you had x minus c and so that's a situation where actually synthetic division came into play, which we saw in that video. And function evaluation might not actually necessarily be more efficient. I argued that if you're evaluating the polynomial at f of c, synthetic division can actually be faster than evaluating the polynomial in general. So the remainder theorem connects function evaluation with division. It's an important connection. And we will see shortly that the remainder theorem is essential to the study of roots of polynomials. Another theorem that's connected to this discussion is the factor theorem, which we've seen a version of this before. If f is a polynomial function, then x minus c is a factor of f of x if and only if f of c equals zero. And that's very easy to see here because first of all, if f of x did factor x minus c times some other function, we'll call it g of x, right? If f of c was a factor, then that would imply that f of c, which is equal to c minus c times g of c. What's g of c? I have no idea, but we're going to take zero times g of c, which that's definitely going to be zero. And so therefore, the function evaluated at c gives you zero. So x minus c being a factor implies that c is a root. Now, what if we go the other way around? So what if, for example, f of c equals zero? Well, by the remainder theorem, this is the remainder if you divide by x minus c. So that would imply that when you take f of x and you divide it by x minus c, there's some quotient we call the g of x, and there's some remainder r. But as that's zero in this situation, you get x minus c times g of x. So the x minus c is the divisor right there. So finding the factors of a polynomial are identical to finding the roots. This is why it's so important to set a polynomial. When you have a polynomial equation, you set it equal to zero, and then you factor the left-hand side. The zero product property, of course, combined with all of this stuff we talked about, the factor theorem, the remainder theorem. The collection of all this theory here tells us that finding the factors is equivalent to finding the roots. And if you have a polynomial f of x equals zero, you're trying to solve the equation, the solution for x is going to be its roots. Finding the roots of the polynomial there solves the equation, and finding this roots is equivalent to finding the factors. So solving polynomials equations really comes down to factoring. This is why we emphasized factoring when we solve quadratic equations. Yes, the quadratic formula did exist, and we can find the solutions of a quadratic equation using that. But using the quadratic equation, believe it or not, is essentially the same thing as factoring. In the end, to solve a polynomial equation, we have to factor. And so finding factors is going to be critical for us. Let's use the factor theorem to determine whether f of x has the following factor. So if we take f of x equals 2x cubed minus x squared plus 2x minus 3, what are some of the factors? Let's first consider x minus 1. If you want to consider whether x minus 1 is a factor or not, my idea is to suggest here is actually to do division, and we're going to do synthetic division. So we're going to write down the coefficients of the polynomial in descending order. We take 2, negative 1, 2, and negative 3. And if we're dividing by x minus 1, we're just going to record the number 1 right here. So we get 1, and then we go from there. So we're going to bring down the 2. 2 times 1 is 2, minus 1 is 1, times 1 is 1, plus 2 is 3, times 1 is 3, minus 3 is 0. So this right here is our remainder. The remainder turned out to be 0. So this tells us that the polynomial is divisible by x minus 3. In fact, what we see here is that f of x, this polynomial up here, it'll have as its factor, x minus 1. This is the term we were dividing by, but we have also the other factor, right? Division, you just throw things away, but in factorization, we have both of them. We have x minus 1, but we also have this factor right here. This is going to be 2x squared minus x plus 3, which, hey, x2x squared minus x plus 3. This is a quadratic polynomial. We potentially could factor that using techniques of factoring quadratic polynomials we've already learned about. What about x plus 3? Is that a divisor of our polynomial or not? I would again do synthetic division. You would take 2, negative 1, 2, negative 3. We're going to divide by negative 3 because you have a plus 3 in the factor right there. Go through the calculation. Bring down the 2. 2 times negative 3 is negative 6. Minus 1 is negative 7 times negative 3 is going to give us positive 21 plus 2 is 23. And then you take 23 times negative 3. That's going to give you negative 69 minus 3 gives you negative 72. This is not a factor because this remainder turns out not to be 0. Now, admittedly, if I was trying to factor this polynomial right here, the fact I found out that x minus 1 was a factor means that you actually have this depressed polynomial, a smaller polynomial from what you start off with. And so I could try to factor this by groups of some kind. 2 and 3 come together. That gives me a 6. I need factors of 6 that add up to be negative 1. Why did I write negative 1? That should be a positive 1, shouldn't it? A good thing I came back to this one. It should be a positive 1. Which turns out this polynomial doesn't have any rational factors because we can't find a magic pair. But just to illustrate here that the polynomial has as a factor x minus 1 because its remainder was 0 and it doesn't have x plus 3 as a factor because its remainder was not 0 right there. So although one could check x minus 1 or x minus c as a factor polynomial by computing f of c, the remainder theorem says you could also check this. Look at f of 1. f of 1 would have been 0. Oh, that means x minus 1 was a factor. You could also have done f of negative 3, which tells you that that was negative 72. That's not a factor. Synthetic division I think is the preferred method here because when you do find a factor, you also find the other factor, this 2x squared plus x plus 3 right there. And so that can be critical to helping us here. We want to always shrink things down and this will become more apparent in future examples. Use synthetic division to find factors and the corresponding quotient that comes with them.