Question

(bm2+3am2-4b-12a)/(2b+bm+3 am+6a)

Answer 1

$\frac{bm^2+3am^2-4b-12a}{2b+bm+3am+6a}=\\\\\frac{m^2(b+3a)-4(b+3a)}{b(2+m)+3a(m+2)}=\frac{(m^2-4)(b+3a)}{(b+3a)(m+2)}=\\\\\frac{m^2-2^2}{m+2}=\frac{(m-2)(m+2)}{m+2}=m-2$