Answer
$(4x+y)(3x-y)$
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 $(12)\times(-1)=-12$ that add to $-1$ are ....$+3$ and $-4.$
$12x^{2}-xy-y^{2}=12x^{2}+3xy-4xy-y^{2}$
$=\left(12x^{2}+3xy\right)+\left(-4xy-y^{2}\right)$
$=3x(4x+y)-y(4x+y)$
$=(4x+y)(3x-y)$
