Factor the expression by grouping. First, the expression needs to be rewritten as $3x^{2}+ax+bx-8$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-10$$ $$ab=3\left(-8\right)=-24$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-24$.
$$1,-24$$ $$2,-12$$ $$3,-8$$ $$4,-6$$
Calculate the sum for each pair.
$$1-24=-23$$ $$2-12=-10$$ $$3-8=-5$$ $$4-6=-2$$
The solution is the pair that gives sum $-10$.
$$a=-12$$ $$b=2$$
Rewrite $3x^{2}-10x-8$ as $\left(3x^{2}-12x\right)+\left(2x-8\right)$.
$$\left(3x^{2}-12x\right)+\left(2x-8\right)$$
Factor out $3x$ in the first and $2$ in the second group.
$$3x\left(x-4\right)+2\left(x-4\right)$$
Factor out common term $x-4$ by using distributive property.
