Answer
(a) $\pm1,\pm1/2,\pm1/3,\pm1/6,\pm2,\pm2/3,\pm3,\pm3/2,\pm4,\pm4/3,\pm6,\pm12$
(b) 2 or 0 positive, 2 or 0 negative real zeros.
Work Step by Step
(a) $p=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12,q=\pm1,\pm2,\pm3,\pm6$
$\frac{p}{q}=\pm1,\pm1/2,\pm1/3,\pm1/6,\pm2,\pm2/3,\pm3,\pm3/2,\pm4,\pm4/3,\pm6,\pm12$
(b) $P(x)$ has 2 sign changes, meaning possibly 2 or 0 positive real zeros.
$P(-x)=6x^{10}-2x^8+5x^3+2x^2+12$ which has 2 sign changes, meaning possibly 2 or 0 negative real zeros.
