Answer
$4(2a-b)(2a-3b)$
Work Step by Step
$ 16a^{2}-32ab+12b^{2}\qquad$... factor out the common term, $4$
$=4(4a^{2}-8ab+3b^{2})$
... Searching for two factors of $ac=12$ whose sum is $b=-8,$
we find$\qquad-2$ and $-6.$
Rewrite the middle term and factor in pairs:
$=4(4a^{2}-2ab-6ab+3b^{2})=$
$=4[2a(2a-b)-3b(2a-b)]$
= $4(2a-b)(2a-3b)$
