## Algebra 2 (1st Edition)

$x=2$ or $x=-3$ or $x=6$
From the earlier parts: $f(x)=x^3−5x^2−12x+36=(x−2)(x+3)(x−6)$. Thus $f(x)=(x−2)(x+3)(x−6)=0$ Thus $x-2=0\\x=2$ or $x+3=0\\x=-3$ or $x-6=0\\x=6$