Solve the given polynomial, use the rational 0 theorem, Discarte's rule of signs, and possibly the graph of the polynomial function shown by a graphing utility as an aid and obtaining the first root. So the first thing is the rational 0 theorem says all of the factors of the constant divided by all of the factors of the leading coefficient. So we know it's going to be positive -1 positive -2, positive -4 and positive -8 and then positive negative 113th positive -2 thirteenths, positive -4 thirteenths, positive -8 thirteenths. I didn't write it out because I'm limited on room, but remember it's positive negative 12481, thirteen 213 four 13813. Then we want to look at Descartes rule of signs. So I'm going to put in. I'm going to count how many sign changes in the original. So how many times did it change to plus to minus or minus to plus and I get 2. Then I'm going to put in negative X throughout the original equation and simplify and count the sign changes again. This time I get 3. The degree was 5, so I know that that had to add up to a total of 5. So if I look at my positives and negatives, that leaves me zero for my eyes. So 2-3 and zero. But I can take pairs because of that conjugate idea from any of the positives or negatives. So if I have two or higher, I'm going to take subtract 2 at a time from one location. So if I took the two positives and put them as the eyes, I'd have 032 or I could have 212 or 014. The fact that there's always a negative means I'm going to check the negatives because the positives may or may not occur, but the negatives are always going to have something. So I'm going to look at the graph, and when I graph this on my graphing calculator, it appears to me that it's going to perhaps touch at -2. So I'm going to use synthetic division here. I've written out my coefficients, and I'm going to put -2 in the box. We bring down the first number. We multiply, we add, we multiply, we add, we multiply and we add. We multiply and we add, We multiply and we add. So the fact that I have a zero in the remainder spot tells me it is indeed a factor. So we could think about this original equation now as X - -2 IE one of our solutions times 13 X to the 4th plus 25X cubed -28 X squared minus fifty, X + 4, all equaling 0. So now we're going to look at this new polynomial and figure out what will divide out of it. If we did the whole thing again, we would realize that in the Discartes rule assigned, we no longer have the 8 as a possibility. But having looked at the graph, I think -2 is a double root. So I'm going to take out -2 again and see what happens. If it's a double root, I should get a remainder of 0 with this new polynomial. So we bring down the first number, we multiply, we add, we multiply, we add, we multiply, we add, we multiply, we add. So now I have -2 as a root a second time, so I'd actually get X - -2 again, or I get X + 2 quantity squared thirteen X ^3 -, X ^2 - 26, X +2 equaling 0. So now I think I'm going to go to a new page where I have more room to write. OK, if we look at this our new polynomial here thirteen X ^3 -, X ^2 - 20 six X + 2 = 0. If we think about the rational 0 theorem here, we would see that we could have positive -1 or positive -2 over positive -1 or positive -13. So we've eliminated how many possibilities we could have. Now if we did our positive negatives and I's, we would have 1-2 sign changes because the I's always have to come in pairs and there were only three. I actually know that there would have to be one negative. If I don't see that, I can put in that negative X and compute it, so I'd get negative thirteen X ^3 -, X ^2 + 26 X +2. So the other possibility for positive negatives and I's would be 01/2. If I grab my calculator and graph this, I'm going to see, well, if I grab my calculator and graph it correctly, I'm going to know it's a cubic. So we've got to cross somewhere. If I have thirteen X ^3 minus X ^2 -20 six X + 2, it looks to me that they're gonna both be fractions. Actually, it looks like all three of them are fractions. I'm gonna try, I don't know, I'm just going to start trying I guess. It looks like I have two that are positive and one that's negative based off the graph. So I might try 113th and see if that works. So 13 negative one, negative 26 and two if I bring down the first number, add, multiply, add multiply. Hey, that one worked. So I'd have X + 2 ^2 from before X -, 1 thirteenth times thirteen, X ^2 - 26 all equals zero. I can pull a 13 out of the 2nd polynomial. No third polynomial, sorry. If I pull out a 13, I get X ^2 -, 2 left. So then if I multiply it through, I get X + 2 ^2 13 X -1 and X ^2 -, 2. We could think of as the difference of two squares, where we have root 2 being the second part, so we'd have X minus root 2 and X plus root 2. So our solutions would be -2 113th root 2 and negative root 2. All of them would have multiplicity of one other than this -2, which would have a multiplicity of two.