Answer
$2(2x-y)(3x+4y)$
Work Step by Step
$ 12x^{2}+10xy-8y^{2}\qquad$...factor out the common term, $2$.
$=2(6x^{2}+5xy-4y^{2})$
... Searching for two factors of $ac=-24$ whose sum is $b=+5,$
we find$\qquad-3$ and $8.$
Rewrite the middle term and factor in pairs:
$=2(6x^{2}-3xy+8xy-4y^{2})=$
$=2[3x(2x-y)+4y(2x-y)]$
= $2(2x-y)(3x+4y)$
