Answer
Zeros:$\displaystyle \quad\frac{3}{2},\quad\pm 1$
$P(x)=(2x-3)(x+1)(x-1)$
Work Step by Step
Factor by grouping:
$P(x)=x^{2}(2x-3)-1(2x-3)$
$=(2x-3)(x^{2}-1)$
...the second parentheses have a difference of squares,
$P(x)=(2x-3)(x+1)(x-1)$
and the zeros are $\displaystyle \quad\frac{3}{2},\quad\pm 1$