Answer
$(2a-1)(3a-4)$
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 $6\times 4=24$ that add to $-11$ are $-8$ and $-3.$
$6a^{2}-11a+4=6a^{2}-3a-8a+4$
$=\left(6a^{2}-3a\right)+(-8a+4)$
$=3a(2a-1)-4(2a-1)$
$=(2a-1)(3a-4)$
