We're trying to find the solution set, so looking at Discartes rule of sign we talk about positive negatives and imaginaries. So 040022004. If we looked at the sign changes in the original, there weren't any, so 0 positives all the way along. If I stick in negative XI get X to the 4th minus eight, X ^3 + 8, X squared minus eight, X + 7 equaling 0. So there are four sign changes. Now we know imaginary numbers always have to come in pairs. So if I have 040, I could take two away from the negatives and put them in imaginary 02/2, or I could take or also I can take another two away from the negative and get 004. So I'm going to just start with the possible rational roots. It's the possible factors of the of the constant divided by the possible factors of the leading coefficient. So in this case, positive -1, positive -7 divided by positive -1 I realize that all of the coefficients are positive 18887. I also know from Discarte's rule of sign that I shouldn't expect any positive rational roots. So I'm going to stick -1 in as my first chance, and I bring down the one. If I'm below the line, we multiply by the number in the box and put it above the next column, above the line. In the next column, if I'm above the line, I'm going to add. Doing this procedure, I can see that this does divide evenly because it has a remainder of 0. So I get X + 1 * X ^3 plus seven X ^2 + X + 7 = 0. I'm going to try -1 again and see if it happens to be a double root. So I'm going to use the new coefficients 1717. I'm actually going to do the Discarte's rule of signs again. When I look at that, I realize there's no positive possibilities. When I stick in negative XI get negative X ^3 plus seven X ^2 -, X + 7 and there are three sign changes. So 030 are my possibilities for positive negative imaginary. Anytime I have a number two or greater, I can take things out in multiples of two. So I also have a possibility of 01/2. So putting in -1 I don't get a remainder of 0. I get a remainder of 12 or a point on the graph of -112. So I know that that one cannot work. So I'm going to try the -7 next because the possible roots are still positive -1 and positive -7. I know I don't have any positives, so I'm going to try -1 and -7 negative 7 does give me out a zero as a factor, so it has to be a factor. A zero as a remainder, so it has to be a factor. So we end up with X ^2 + 1. So X + 1 * X + 7 * X ^2 + 1. If we solve that, we'd get X equal -1 negative 7 positive negative I. If we wanted to think about what the graph of this might be, we have -1 negative 7. We see that it's a fourth degree. So as we go out to Infinity, for the X negative Infinity, the Y values are either going to be both positive or both negative. The leading coefficient is a +1 so that we know it's going to be a positive. We come down here at this zero. We look at the multiplicity. The multiplicity is odd. So if the Y values on one side are positive, the Y values on the other side have to be negative. We come to the next 0. It also has an odd multiplicity, so if the Y values are negative on one side, they've got to be positive on the other. Granted, very rough sketch, but that's basically what the sketch would look like.