# How do you factor 9-(k+3)^2?

May 2, 2017

$- k \left(6 + k\right)$

#### Explanation:

This expression is a $\textcolor{b l u e}{\text{difference of squares}}$ which factorises in general as.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{for } 9 - {\left(k + 3\right)}^{2}$

$a = 3 \text{ and } b = k + 3$

$\Rightarrow 9 - {\left(k + 3\right)}^{2} = \left(3 - \left(k + 3\right)\right) \left(3 + \left(k + 3\right)\right)$

$\textcolor{w h i t e}{\Rightarrow 9 - {\left(k + 3\right)}^{2}} = \left(3 - k - 3\right) \left(6 + k\right)$

$\textcolor{w h i t e}{\Rightarrow 9 - {\left(k + 3\right)}^{2}} = - k \left(6 + k\right)$