Answer
$x=3$ or $x=2$ or $x=-2$
Work Step by Step
We solve by factoring with grouping:
$x^{3}-3x^{2}-4x+12=0$
$x^{2}(x-3)-4(x-3)=0$
$(x-3)(x^{2}-4)=0$
$(x-3)(x-2)(x+2)=0$
$x-3=0$ or $x-2=0$ or $x+2=0$
$x=3$ or $x=2$ or $x=-2$
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