Answer
$2(2y-1)(4y+1)$
Work Step by Step
$ 16y^{2}-4y-2\qquad$...factor out the GCF, $2$.
$=2(8y^{2}-2y-1)$
Two factors of $ac=-8$, whose sum is $b=-2$
are $-4$ and $2$.
Rewrite $-2y$ as $-4y+2y$ and factor in pairs.
$2(8y^{2}-2y-1)$
$=2(8y^{2}-4y+2y-1)\qquad$...factor from each group if possible.
$=2[4y(2y-1)+(2y-1)]\qquad$...factor out the common binomial factor,$2y-1$.
$=2(2y-1)(4y+1)$