 So we've talked about in this lecture already why finding factorizations of a polynomial is critical? Finding the factorization is how we solve polynomial equations. So we have to look for the roots. We want to test them using synthetic division. But how do we know which roots to look for or which one's not to? Well, first of all, we have to know there's a limit on how many roots we're going to take. A polynomial function cannot have more real roots than its degree. And in fact, I could replace that word with complex numbers, which we'll talk about that in a future lecture, right? But it can't have any more roots than its degree. So if you have a degree 4 polynomial, it must have most 4 roots. If you have a degree 3 polynomial, at most 3, if it's a degree 17 polynomial, it can have the most 17 roots. So there's a limit. This process will eventually have to terminate. But okay, so I know I don't need to be searching for 100 roots if I have a degree 3 polynomial. But what do I try? What are what are the ones to check? And this is going to be one of our favorite theorems in this section here. This is commonly referred to as the rational roots theorem. Let F be a polynomial function of degree 1 or higher. So it's not a constant polynomial. And so it's going to look something like the following. Just here's the generic polynomial. The leading coefficient is not 0. And it's not a constant polynomial. It does actually have some variable in play right here. And suppose that all of the coefficients here are integers, whole numbers. Positive or negative, 0. All of that's okay. So looking at the leading term in the constant term. What we're going to do is we're going to look for divisors of the leading term. And we're going to look for divisors of the constant term. All right. So the rational roots theorem then claims that all of the rational roots of this polynomial will be of the form p over q. So it's a fraction. So it looks like p over q. Where p is a factor of the constant term. And q is a factor of the leading term. So we can look at the divisors of the first and last coefficient and compare them to find all the possible rational roots of a polynomial. This is a theorem where it's best to illustrate via an example. Take the polynomial f of x equals 2x cubed plus 11x squared minus 7x minus 6. So if we look at the constant term, that's a negative 6. So negative 6 here, this is going to give us all of our possible p's. So p here could be one of the factors of negative 6. You could get 1, you could get 2, you could get 3, and you could get 6. Those are the factors of 6. But we could also take negatives. I mean, that doesn't have to be positive, right? So we could get like plus or minus 1, both of those divide negative 6. Plus or minus 2, plus or minus 3, and plus or minus 6. Those are all the possible divisors of negative 6. On the other hand, if you take the leading coefficient, which is 2, the possible divisors of 2 are going to be plus or minus 1 and plus or minus 2. 2 is a prime number, so there's not a lot of divisors there. And so the rational roots theorem says that every possible rational root of f of x will be a combination of these numbers here, p divided by q. And so let's look at all the possibilities one could get here. We could get 1 divided by 1 plus or minus, plus or minus. Well, I'm going to come back to that issue. We could get 1 divided by 1. We could get 2 divided by 1. We could get 3 divided by 1. We could get 6 divided by 1. We could get negative 1 divided by 1. So I'm going to write up plus or minus right here. We could get negative 2 divided by 1, negative 3 divided by 1, negative 6 divided by 1. Okay, so we got those. We could also get one divided by negative one, we get two divided by negative one, we get three divided by negative one, we get six divided by negative one. But we could also get negative one divided by negative one, negative two divided by negative one, negative three divided by negative one, and negative six divided by negative one. Now you'll notice that if you look at these ones right here, plus or minus one divided by negative one, that's going to give you either a negative one or a positive one. This one right here, this also will give us a positive one or a negative one. So in some of these, some of these results are actually redundant. Dividing by negative here actually gives us the same possibilities that we did before. So we don't need those terms right there. Okay, so you don't have to worry about, so when it comes to denominator, just always make the denominator be positive because since the numerator could be positive or negative, having a positive or negative denominator is kind of redundant. So just, you don't have to worry about the negative denominators. So I'm actually just going to get rid of these things that we wrote here a moment ago. We don't need them because they're redundant. Oops, I do want that one. The next thing to consider is we could take plus or minus one divided by two. We could take plus or minus two divided by two. We get a plus or minus three divided by two. And we could take plus or minus six divided by two. So we took all the possibilities here where we took the divisors of P divided by the divisors of Q. And like I said, we don't have to worry about negative signs for the divisors on the Q side of things. We get the following, but there's still some redundancy here. Two divided by two is the same thing as one. So we actually don't need that. And the same thing here, six divided by two is the same thing as three. So let's clean this thing up a little bit. The divisors we really want to look about, the rational root theorem tells us we have plus or minus one, plus or minus two, plus or minus three, plus or minus six, plus or minus one-half, and plus or minus three-halves. These are the possible rational roots of this polynomial. And that gives us a list of 12, right? So there's one, two, three, four, five, six terms here, but there's plus or minus, so each of those is doubled. So we get actually 12 possibilities. Now that might seem like, oh no, that's a lot of possibilities. Well, that's a whole lot better than infinity that we had a moment ago, right? Before the rational roots theorem, it could be that any number under the sun was a root of the polynomial. So the fact that we can list it as a finite list is much, much better than infinity. And now I want to show you how you can actually utilize this rational roots observation here. Let's take the exact same polynomial, 2x cubed plus 11x squared minus 7x minus six. And so the rational roots theorem told us, I'll just copy it down, that the potential rational roots were plus or minus one, two, three, six, whoops, plus or minus six, one-half, and three-halves. And so what we want to do is determine what are the roots of this polynomial. And so we could try, we could try various roots. So let's say we just start off with wanting to take a random guess, let's pick something like, say, 2. Like, what if we want to try 2? So 2 is our candidate here. We're going to run synthetic division using 2. So we're going to write down the coefficients of a polynomial in descending order, 2, 11, negative 7, negative 6, and go through the calculation, bring down the 2. 2 times 2 is 4, plus 11 is 15, times 2 is 30, minus 7 will be 23, times 2, you get 46, plus, or minus 6, you're going to get 40 right here. We can actually, we actually were able to predict a long time ago, these numbers just kind of getting bigger and bigger and bigger. Didn't look like we were going to lead to a remainder. So this tells us that positive 2 is not an option here. It didn't turn out to be a root. We could try something else like, what if we did negative 2? Same basic thing here. 2, 11, negative 7, negative 6, like so. Bring down the 2. 2 times negative 2 is negative 4, plus 11 is going to give us their positive 7, times negative 2, we get negative 14, minus 7 is negative 21, times negative 2 gives us 42, minus 6 is going to give us 36, which still is not zero. So it turns out that neither plus or minus 2 turn out. So we can kind of take it off the tally there. And so we can try this trial by error, trial by error, try until we find a root that works. This next one, I'm going to try 1. Honestly, 1 is the one I usually start with. It's, I just like simple arithmetic number 1 there. You're going to see in a moment why I hesitated to do 1. So if you take do 1, we're going to bring down the 2. 2 times 1 is 2, plus 11 is 13, times 1 is 13, minus 7 is 6, times 1 is 6, minus 6 is 0. So the reason I didn't start off with 1 is because it actually turns out to be a divisor right here, right? So this tells us like, yay, our remainder was 0. We have a factor, joint of the world. And so this actually gives us a factorization. Whenever you find a root of the polynomial, you get a factor. So we know that x minus 1 is a factor of the polynomial, but we also know these guys right here give us a factor. And so we start off with a degree 3 polynomial, so it goes down by 1. We get 2x squared minus plus 13x, plus 6. And so now whenever you find a factor, you no longer look at the original polynomial. We no longer want 2x cubed plus 11x squared minus 7x minus 6. We want to look at this depressed polynomial. And it's not depressed because it's sad. It's just, it's smaller than we've started off with. It's now a degree 2 polynomial. And the fact that it's degree 2 means I'm actually going to factor using the techniques I learned in the previous unit in this lecture. So how do you factor a quadratic? Look at the leading terms, the leading constant term, right? 2 times 6, that's going to give us 12. We need factors of 12 that add up to be 13. That's going to be 12 times 1. And so if we factor this thing by groups, we're going to get 2x squared plus 12x. That's my first group. And then my second group, I'm going to just take x plus 6. And so factoring this thing right here, we would pull out a 2x from the first group, leaving behind x plus 6. From the second group, if you think you can take out nothing, that actually means you can take out a 1. You just can't take anything bigger than 1 there. And so this factor is s2x plus 1 and x plus 6. So when you put all this together, we found now our complete factorization of f of x. We're going to get x minus 6 we had from before. We're going to have a 2x plus 1, and we're going to have an x plus 6. Now let's zoom out a little bit. So this was our final answer. Put a big box around it so we can find it. And then look at the rational roots theorem that we had previously, right? It's this business right here. So the rational roots theorem says our roots, we're going to be amongst the list plus or minus 1, plus or minus 2, plus or minus 3, plus or minus 6, plus or minus 1 half, plus or minus 3 halves. And so when you look at this, we get a 1, x equals 1. We got the second one, 2x plus 1 gives us a negative 1 half and then x plus 6 tells us negative 6. So those numbers were in that list. And so we would have found them eventually if we kept on going. But it's better whenever you find a root to use the depressed polynomial because it'll speed up the polynomial factorization process dramatically. So the rational roots theorem is a critical tool to help us solve these polynomial equations, be able to factor them. Now, it turns out that this rational roots theorem is a critical tool because we have a list of 12 numbers. We just have to try all of them. So worst case scenario, it's 12 synthetic divisions. But it turns out that we can hone in on the roots a lot faster than just this list. And so turns out that finding out that 2 was a failure actually led me to think that 1 was going to be successful. And we'll talk about this more in the next lecture, how we can go through this process even faster and we wouldn't have to try all 12 possibilities.