Answer
$3, -1\pm i$
Work Step by Step
We have $p=\pm1,\pm2,\pm3,\pm6,q=\pm1$
Possible rational zeros are $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6$
Use the synthetic division to find one zero and the quadratic form as
$P(x)=(x-3)(x^2+2x+2)$
Solve the quadratic part as $x=\frac{-2\pm \sqrt {4-8}}{2}=-1\pm i$
So the zeros are $3, -1\pm i$
