Answer
$2(6z-5)(8z-3)$
Work Step by Step
Substitute $u=4z-3$
$6u^{2}+7u-3=...$
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 $(6)\times(-3)=-18$ that add to $+7$ are ....$+9$ and $-2$
$...=6u^{2}+9u-2u-3$
$=3u(2u+3)-(2u+3)$
$=(2u+3)(3u-1)$
... bring back z
$=[2(4z-3)+3][3(4z-3)-1]$
$=(8z-6+3)(12z-9-1)$
$=(8z-3)(12z-10)$
... factor out $2$ from the second parentheses
$=2(8z-3)(6z-5)$
