 Welcome back to our lecture series Math 1050, College of Algebra for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. In lecture 27, we want to continue the topics we introduced in lecture 26. That is, we want to continue learning how to factor larger degree polynomials with the goal to using these techniques to solve polynomial equations. So in the previous lecture, we learned things about the remainder theorem, division algorithm, the factor theorem, and how we can use synthetic division to help us factor polynomials. It's going to be our best friend in this situation. In particular, a very favorite thing from the previous lecture was the rational roots theorem that tells us a list of numbers to test with synthetic division to try to find a factorization. Let's say g of x equals x to the fourth minus 5x cube minus 5x squared plus 23x plus 10. By the rational roots theorem, we're looking for all the possible fractions that come about by taking divisors of the constant term 10 with the leading coefficient divisors, which in this case would be 1. So when your leading coefficients are 1, you can actually ignore that because the only divisor is going to be plus or minus 1. I've told you previously that the denominator will always make it be positive here. We'll allow the numerator to vary between positive and negative signs. We look at factors of 10. We're going to get plus or minus 1, plus or minus 2, plus or minus 5, and plus or minus 10. Now, we have to try these options here. Do we try 1, 2, 4, 5, or 10? And do we try positive or negative? Personally, it's usually not the big numbers. I mean, although it's possible, it could be 10. I mean, I can't deny such a possibilities. But I'm just saying from a probability point of view, you're more likely to be successful with a smaller number like 1 and 2 compared to the bigger numbers like 10. And so it's better to try, in my opinion, the smaller numbers. I like to start with positive 1. I can't really make much of a statement about who's more likely positive or negative of roots right now. We'll actually deal with that in the next video, how to determine that. So without any preference here, we're just going to start off with 1 and try that one. So we're going to divide by 1 using synthetic division. So looking at the coefficients and descending order, we get 1, negative 5, negative 5, 23, and 10. Be very cautious. If any numbers got skipped, you do need to put a 0 in there. So we go through the process here, bring down the 1. 1 times 1 is 1, minus 5 is negative 4, times 1 is negative 4, minus 5 is negative 9, times 1 is negative 9, 23 minus 9 is 14, times 1 is 14, plus 10 is 24. All right, so it didn't work. So 1 plus 1, we can take off of our list. We could try, say, 2. 2 seems like a good choice. Let's put it over here. So if we try 2, same numbers as before, 1, negative 5, negative 5, 23, and 10. So we bring down the 1. 1 times 2 is 2 minus 3, or minus 5 is negative 3, times 2 is negative 6, minus 5 is negative 11, times 2 is negative 22, plus 23 is 1 really close. But then 1 times 2 is 2, plus 10 is 12. So that didn't quite work either. So we want to take plus 2 off the list. Don't make sure we don't take off negative 2, because that's a possibility. It's like, okay, keep on going, I guess, have an optimistic point of view here. Let's try 5. Again, we're just going down the list here. Again, I'm more likely to get a smaller number than a bigger one. But it could be 10. Let's try 5 this time. And so with this technique, there is a lot of trial and error. We're just kind of guessing the root and going from there. Bring down the 1. 1 times 5 is 5, minus 5 is 0, times 5 is 0, minus 5 is negative 5, times 5 is negative 25, plus 23 is negative 2, times 5 is negative 10. And woo-hoo, we did it. We found a root, 0 is the remainder there. And so whenever you find a root, you jump up for joy, which you can't see me doing that right now, but that's what's happening. Then we're going to consult a factorization. So now we have a factorization of g of x. Let's look at that. So first of all, since 5 worked with synthetic division, that means that x minus 5 is a factor and then 5 was a root. But then we also have the quotient right here, which is another factor of g. So writing that down, g started off as a degree 4 polynomial. So we're going to get x cubed plus 0x squared minus 5x minus 2. And we want to now factor this polynomial going forward. Looking at it real quick, it is a degree 3 polynomial. It's not a difference of cubes or sum of cubes because there's three terms. And I can't use factoring by groups because there's three terms instead of four. So I have to try factoring this thing here. But look at what we have here so far. If I were to do the rational roots test again using the depressed polynomial right here, notice that you're looking for factors of negative 2 divided by factors of 1. So your possible p's and q's, don't forget your p's and q's, right? This is going to be plus or minus 1 and plus or minus 2. So that's a much smaller list than we had before. Instead of the plus or minus 1, 2, 5, and 10, we can rule out that 5 isn't going to show up again. Neither is 10. But also I want to mention that we've already shown that plus 1 and plus 2 didn't work for g. And so all of the factors of the depressed polynomial are factors of g with the exception of 5, right? That 5 is a factor of g but not here. The depressed polynomial have fewer factors of g, but all of the factors of the depressed polynomial of the quotient will be factors of g. So the fact that 1 and 2 failed before means they'll continue to fail. So I'm not going to use plus 1 or plus 2 as I go forward. So my list is I want to try negative 1 or negative 2. And so if I try negative 1, let's just try that one. So we now do the depressed polynomial. We're going to do 1, 0. Don't forget that there was a quadratic there. Some people actually just continue on from here if you want to. 1, 0, negative 5, negative 2. We're going to try negative 1 this time. Bring down the 1. 1 times negative 1 is negative 1. Plus 0 is negative 1 times negative 1 is 1. Minus 5 is negative 4 times negative 1 is 4. Minus 2 is 2. So negative 1 didn't work. Let's try negative 2. 1, 0, negative 5, negative 2. Again, there's a little bit of guesswork here, but that's OK. Bring down the 1. 1 times negative 2 is negative 2. Plus 0 is negative 2 times negative 2 is positive 4. Minus 5 is negative 1 times negative 2 is a positive 2. And there we go. We found our remainder. That is we found the remainder to be 0. So what we now have is that g of x factors as x minus 5. We get an x plus 2. Since your root was negative 2, you're going to get x minus negative 2, which is an x plus 2. And then take this thing right here as this quotient here. You're going to get x squared minus 2x minus 1. And then we try to see if that can factor at all. x squared minus 2x minus 1. You need factors of negative 1 that add up to be negative 2, which is impossible with real numbers. So in terms of factoring this, if we're looking for real roots of the polynomial, this right here is an irreducible quadratic. We'd have to use the quadratic formula to find the roots of that thing, which we're going to see that the discriminant of such a thing, b squared minus 4ac, you're going to get 4 plus 4 inside the square root. So it turns out that you have some irrational roots looks like right there. And so we can keep on going. So let's try that out. So we're going to get a plus 2 plus or minus all over 2. Simplifying this, we get 2 plus 2 root 2. So notice you take the square root of 8, which is the square root of 4 times 2, which is 2 root 2, like we have before. This sits above 2. And so putting this all together, the roots of our polynomial are going to be 5, negative 2. And then we're going to get 1 plus or minus the square root of 2. So you have these two irrational roots plus two real roots, two whole number roots, I should say. 5 and negative 2 are rational roots. So this right here is not in violation of the rational roots test, because the rational roots test tells us what the rational roots could be. It could be that there are, of course, irrational roots. That's a possibility. 1 plus or minus the square root of 2. And so we can find the roots of the polynomial. We find them from this factorization. And so finding the factorization is critical to finding the roots of this polynomial. The two processes by the factor theorem are one and the same.