 In this final example for section 4.5, let us try to find the complex roots of the polynomial function, which is the same thing as writing it factored with a complex factorization. So we have g of x equals x cubed plus 13x squared plus 57x plus 85. Some things to note here, that if we do the rational roots test, we have to look at factors of 85. 85 is 5 times 17, and therefore our rational roots would look like p over q. We're going to get plus or minus 1, plus or minus 5, plus or minus 17, plus or minus 85. I hesitate to do the bigger factors, try the smaller ones first. But also notice, look at the variation of signs here, positive, positive, positive, positive. This tells us that the variation of signs is 0. The negative variation is going to turn out to be 3, so you get 3 or 1. But in particular, if your variation is 0, this means that there are no positive roots. That is a very helpful statement. So now I'm not going to be looking for the positive roots whatsoever, because there are none. So what I'm going to do is I'm going to amend my previous list and emphasize that I'm only going to look for negative numbers. Negative 1, negative 5, negative 17, negative 85. I don't really like looking for the bigger ones, negative 17, negative 85. They're just so big and it's not as likely. So I'm going to try something in the middle. I'm going to try negative 5. It's kind of in the middle of this, not too big, not too small. Sort of the Goldilocks of my potential rational roots right now. So if we go through some negative division, 1, 13, 57, and 85. If we try negative 5, we bring down the 1. 1 times negative 5 is negative 5. Plus 13 is going to be 8 times negative 5 is a negative 40. Plus 57 gives me 17. This is helpful here. Negative 5 times 17 is negative 85, which gives us the root. So my Goldilocks guess was actually a pretty good guess right there. So g of x can then be factored as x plus 5 times x squared plus 8x plus 17. As this polynomial is now quadratic, the quotient, I'm not going to try synthetic division whatsoever. Factors of 17 that add up to be 8 is not going to happen. 17 is the prime number. So my only option was going to be 1 plus 17, which was 18. That's not enough. If there was like a extra digit of 1 right there, we'd be good, but no. We can't factor this using those techniques whatsoever. So instead we got to try the quadratic formula. Because we can't factor it using the reverse foil technique. So we're going to use the quadratic formula. x equals negative 8 plus or minus the square root of 8 squared minus 4ac all over 2a. Like so. And so let's see. 4 times 17, that's a 68. So we get negative 8 plus or minus the square root of 64 minus 68 all over 2. That gives the radicand as a negative 4, which then the square of negative 4 is going to be 2i, negative 8 plus or minus 2i over 2. Fact around a 2 from the numerator, we get negative 4 plus or minus i over 2. This 2 cancels. And so we see that the non-real roots here are actually going to be negative 4 plus or minus i. That actually shows up with what Descartes-Rouls says. Descartes-Rouls said we'll either have 3 negative roots or 1 negative root. We actually only had 1 because 2 of them turned out to be non-real. And so therefore our polynomial will factor as g of x looks like x plus 5 times x minus negative 4 plus i. We're also going to get an x minus negative 4 minus i. If you don't like the double negatives, you can distribute the negative signs right here. And this would give us g of x equals x plus 5. And then we get x plus 4 minus i and x plus 4 plus i. This would be the complex factorization of this polynomial into linear factors. And in terms of the roots, the roots turned out to be x equals negative 4 plus i and negative 4 minus i. So we can find the complex roots, we can find the factorization. Finding the roots and the factorization are one and the same thing. And combining these techniques of Descartes-Rouls signs, upper and lower bound theorems, the factor theorem, the remainder theorem, it meshes together to help us very effectively solve these polynomial equations by factoring these large polynomials.