Answer
(a) $\pm1,\pm2,\pm3,\pm6,\pm9,\pm18$
(b) 2 or 0 positive and 3 or 1 negative real zeros.
Work Step by Step
(a) List all possible rational zeros (without testing to see whether they actually are zeros).
$\pm1,\pm2,\pm3,\pm6,\pm9,\pm18$
(b) Determine the possible number of positive and negative real zeros using Descartes’ Rule of Signs.
P(x) has 2 sign changes, so there may be 2 or 0 positive real zeros,
$P(-x)=-x^5+6x^3-x^2-2x+18$ and there are 3 sign changes, which means possibly there are 3 or 1 negative zeros. In total, it is possible that there are 5,3,or 1 real zeros.
