Answer
$2(3y+4)(y-2)$
Work Step by Step
Since $2$ is common to all the terms of the equation, we take it out as a common factor:
$6y^{2}-4y-16$
=$2(3y^{2}-2y-8)$
In accordance with the rules of factoring trinomials, we need to find two integers whose product is $-24$ and whose sum is $-2$. A little searching determines that these numbers are $4$ and $-6$. Therefore, we can express the middle term $-2y$ as $(+4y-6y)$ and proceed to factoring by grouping:
$2(3y^{2}-2y-8)$
=$2(3y^{2}+4y-6y-8)$
=$2[y(3y+4)-2(3y+4)]$
=$2[(3y+4)(y-2)]$
=$2(3y+4)(y-2)$
