Answer
$-1,3$
Work Step by Step
Use Remainder and Factor Theorem:
$P(-2)=3\times16+8-21\times4+11\times2+6=0$
$P(\frac{1}{3})=\frac{3}{81}-\frac{1}{27}-\frac{21}{9}-\frac{11}{3}+6=0$
which proves that $-2,\frac{1}{3}$ are zeros of P.
Use synthetic division, we can reduce P to quadratic form
$P(x)=(x+2)(3x-1)(x^2-2x-3)=(x+2)(3x-1)(x+1)(x-3)$
so the other two zeros are $-1,3$
