Given the phrase "enter all answers including repititions", I wonder if you're supposed to find all possible rational zeros? If so, they're all found by putting the factors of the constant (32) over the factors of the leading coefficient (2). 32 has the factors 1, 2, 4, 8, 16, and 32, and 2 has the factors 1 and 2. Take both positive and negative values of each: ±{1/1, 1/2, 2/1, 2/2, 4/1, 4/2, 8/1, 8/2, 16/1, 16/2, 32/1, 32/2}. Obviously, there are a lot of repititions here, like 1/1=2/2 and 4/2=2/1. Your teacher seems to want those.
If you don't need all possible zeros, but only actual zeros, you can use synthetic division, plug values in, or graph the function to see which of these possibilities are actually solutions. We wind up with three distinct values, and a "double root" at x=4. So x=-2, 1/2, 4, 4.
I hope that helps, and let me know if you have any questions!
Precalc rocks!
Liz Z.
