Answer
$(3s-t)(4s+5t)$
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 $(12)\times(-5)=-60$ that add to $11$ are ....$+15$ and $-4.$
$12s^{2}+11st-5t^{2}=12s^{2}-4st+15st-5t^{2}$
$=\left(12s^{2}-4st\right)+\left(15st-5t^{2}\right)$
$=4s(3s-t)+5t(3s-t)$
$=(3s-t)(4s+5t)$
