Answer
$(7m-5r)(2m+3r)$
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 $(14)\times(-15)=-210$ that add to $+11$ are ....$+21$ and $-10.$
$14m^{2}+11mr-15r^{2}= 14m^{2}-10mr+21mr-15r^{2}$
$=\left(14m^{2}-10mr\right)+\left(21mr-15r^{2}\right)$
$=2m(7m-5r)+3r(7m-5r)$
$=(7m-5r)(2m+3r)$
