## Precalculus (6th Edition) Blitzer

$(x+5)(x^{2}-2)$
$x^{3}+5x^{2}-2x-10=\quad$ Factor in pairs; factor out $x^{2}$ from the first pair, and $-2$ from the second $=x^{2}(x+5)-2(x+5)$ $=(x+5)(x^{2}-2)$ We could treat the second parentheses as a difference of squares, using $(\sqrt{2})^{2}=2$, but factoring implies integer coefficients, so we stop here.