$x=2$ or $x=-3$ or $x=6$
Work Step by Step
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 $x-6=0\\x=6$
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