Answer
$-\frac{1}{2},5$
Work Step by Step
Use the Remainder and Factor Theorem:
$P(-1)=2+13+7-37+15=0$
$P(3)=2\times81-13\times27+7\times9+37\times3+15=0$
Use synthetic division, we have
$P(x)=(x+1)(x-3)(2x^2-9x-5)=(x+1)(x-3)(2x+1)(x-5)$
so the other two zeros are $-\frac{1}{2},5$
