Answer
$P(x)=(2x-3)(x-1)(x+1)$
zeros: $\displaystyle \pm 1, \frac{3}{2}$
Work Step by Step
Factor in pairs:
$2x^{3}-3x^{2}-2x+3=x^{2}(2x-3)-(2x-3)=(2x-3)(x^{2}-1)$
... recognize a difference of squares.
$P(x)=(2x-3)(x-1)(x+1)$
zeros: $\displaystyle \pm 1, \frac{3}{2}$
