Answer
$x=-2$ or $x=2$ or $x=-\displaystyle \frac{2}{3}$
Work Step by Step
We solve by factoring with grouping:
$3x^{3}+2x^{2}-12x-8=0$
$x^{2}(3x+2)-4(3x+2)=0$
$(3x+2)(x^{2}-4)=0$
$(3x+2)(x+2)(x-2)=0$
$3x+2=0$ or $x+2=0$ or $x-2=0$
$x=-\displaystyle \frac{2}{3}$ or $x=-2$ or $x=2$
