Answer
$b(2a-1)(3a+2)$
Work Step by Step
$ 6a^{2}b+ab-2b\qquad$... factor out the common term, $b$
$=b(6a^{2}+a-2)$
... Searching for two factors of $ac=-12$ whose sum is $b=1,$
we find$\qquad-3$ and $4.$
Rewrite the middle term and factor in pairs:
$=b(6a^{2}-3a+4a-2)=$
$=b[3a(2a-1)+2(2a-1)]$
= $b(2a-1)(3a+2)$
