## College Algebra 7th Edition

$P(x)=(2x-3)(x-1)(x+1)$ zeros: $\displaystyle \pm 1, \frac{3}{2}$
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: $\displaystyle \pm 1, \frac{3}{2}$