# How do you multiply (m - 1) (m^2 + 2m + 6)?

Apr 13, 2015

First you multiply the $m$ with all the others in the second group, and then the $- 1$

$= m \cdot \left({m}^{2} + 2 m + 6\right) + \left(- 1\right) \cdot \left({m}^{2} + 2 m + 6\right)$

$= \left(m . {m}^{2} + m \cdot 2 m + m \cdot 6\right) + \left(- 1 {m}^{2} - 2 m - 6\right)$

$= {m}^{3} + 2 {m}^{2} + 6 m - 1 {m}^{2} - 2 m - 6$. now add like powers:

$= {m}^{3} + \left(2 - 1\right) {m}^{2} + \left(6 - 2\right) m - 6$

$= {m}^{3} + {m}^{2} + 4 m - 6$