Answer
$(6a-m)(5a+m)$
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 $(30)\times(-1)=-30$ that add to $+1$ are ....$+6$ and $-5$
$30a^{2}+am-m^{2}=30a^{2}-5am+6am-m^{2}$
$=\left(30a^{2}-5am\right)+\left(6am-m^{2}\right)$
$=5a(6a-m)+m(6a-m)$
$=(6a-m)(5a+m)$
