Do the grouping $2x^{3}+3x^{2}-8x-12=\left(2x^{3}+3x^{2}\right)+\left(-8x-12\right)$, and factor out $x^{2}$ in the first and $-4$ in the second group.
$$x^{2}\left(2x+3\right)-4\left(2x+3\right)$$
Factor out common term $2x+3$ by using distributive property.
$$\left(2x+3\right)\left(x^{2}-4\right)$$
Consider $x^{2}-4$. Rewrite $x^{2}-4$ as $x^{2}-2^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
