# How do you factor: y= m^2-4m-21 ?

Dec 10, 2015

$y = \left(m + 3\right) \left(m - 7\right)$

#### Explanation:

The things to note is that:
$\left(+ 3\right) \times \left(- 7\right) = - 21$
$- 7 + 3 = - 4$

$\textcolor{b l u e}{\text{Considering the RHS:}}$

There are two condition to give ${m}^{2}$ and they are $\left(- m\right) \times \left(- m\right) \mathmr{and} \left(+ m\right) \times \left(+ m\right)$

Lets try the $+ m$

So we have: (m+?)(m+?) " giving " m^2 + "something"

Considering the numbers we look at first lets try:
$\left(m + 3\right) \left(m - 7\right) \to {m}^{2} - 7 m + 3 m - 21$

But $- 7 m + 3 m = - 4 m$

$\textcolor{b l u e}{\text{Putting it all together:}}$

$y = \left(m + 3\right) \left(m - 7\right) = {m}^{2} - 4 m - 21$