Answer
$(2h+3)(4h-7)$
Work Step by Step
When factoring $ax^{2}+bx+c$, we search for factors of $ac$ whose sum is $b,$
and, if we find them, we rewrite $bx$ and proceed to factor in groups.
Here, factors of $8\times(-21)=-168$ that add to $-2$ are $-14$ and $+12.$
$8h^{2}-2h-21=8h^{2}+12h-14h-21$
$=\left(8h^{2}+12h\right)+(-14h-21)$
$=4h(2h+3)-7(2h+3)$
$=(2h+3)(4h-7)$
