On this problem, it specifies that you can use a graphing utility as an aid. So I started with typing in the equation into my graph to figure out what it was going to look like and where it was going to cross the X axis, which would be my zeros. It also said to use the rational 0 theorem and Discarte's rule of sign. So I started with looking at Discarte's rule of sign. I counted how many sign changes and in the original there were two sign changes. I stuck in negative X and I counted the sign changes again and there were three. So one possibility is 2 rationals, 3 negatives, and 0 imaginary. A different possibility would be 0 positives, 3 negatives and two imaginary. Another possibility would be two positives, one negative and two imaginary, and the last one would be 0 positives, one negative and four imaginary. So those are each possibilities. So those would go together. Each column has to add up to the number of solutions, and in this case we should expect 5 answers, so each of those really do add up to five. The next thing I'm going to do is I'm going to look at what my possible rational zeros might be. So it's going to be the constant factors all divided by the leading coefficient factors. So positive, negative, 124510 and 20 all over positive -1 and positive -5. Looking at my graphing utility, I see that it crosses at -2. So I'm going to put -2 in. I'm going to do the synthetic division and I'm going to get out of 0. So X + 2 was a factor, and that left me five X ^2 + 9 five X to the 4th plus nine X ^3 - 20 seven X ^2 - 40 five X + 10 = 0. Now doing our constant divided by our leading coefficient again, 10 is now smaller than 20. So I've gotten rid of some of the possible numbers that I should try. So now it's positive, negative 125 and 10 all over positive -1 and five. So looking at my graphing calculator, I see that one 5th works, SO59 negative 27, negative 45 and 10. Doing my synthetic Division, I get a zero out, so now it's X + 2 * X - 1/5 * 5 X cubed plus ten X ^2 - 25 X -10. Now all of these terms in this third polynomial that's being multiplied together have A5 in common. So I can take out a 5. If I take a 5 out of this term, then I can multiply the five through in the term right in front of it and get rid of the fraction. So five X -, 1. Now this one is going to get a -2 out of it because if I look at my graph at -2, it's actually going to bounce on the graph. So I knew that -2 is going to have an even multiplicity. So if I do my synthetic division again, I get X + 2, five X - 1, another X + 2, and X ^2 - 5. So I get an X + 2 quantity squared five X - 1 and X ^2 - 5. I could think of as the difference of squares X minus root 5X plus root 5. So our final answer is -2 one fifth root 5 and negative root 5. That -2 occurred twice. It's called a double root, but there really are five solutions there. It's just that -2 occurs twice.