## Algebra 1: Common Core (15th Edition)

$\frac{2m^{2}(m+2)}{(m-1)(m+4)}, with$ $m\ne-4,m\ne1$
Given expression : $(\frac{2m+4}{m}) \times (\frac{m^{3}}{m^{2}+m-12}) \times (\frac{m^{2}-m-6}{m^{2}+m-2})$ This becomes : $\frac{2(m+2)}{m} \times \frac{m^{3}}{(m-3)(m+4)} \times \frac{(m-3)(m+2)}{(m-1)(m+2)}$ $= \frac{2m^{2}(m+2)}{(m-1)(m+4)}$ (After dividing out the common factors (m-3) and (m+2))