# How do you factor x^3 + x^2 -9x -9 by grouping?

Jun 30, 2016

(x+1)(x-3)(x+3)

#### Explanation:

Group the terms into 'pairs' as follows.

$\left[{x}^{3} + {x}^{2}\right] + \left[- 9 x - 9\right]$

now factorise each pair.

$\textcolor{red}{{x}^{2}} \left(x + 1\right) - \textcolor{red}{9} \left(x + 1\right)$

We now have a common factor of (x+1) which can be 'taken out'

$\left(x + 1\right) \left(\textcolor{red}{{x}^{2} - 9}\right) \ldots \ldots . . \left(A\right)$

$\textcolor{red}{{x}^{2} - 9} \text{ is a " color(blue)"difference of squares}$ and is factorised in general as follows.

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

${x}^{2} = {\left(x\right)}^{2} \text{ and " 9=(3)^2rArra=x" and } b = 3$

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

Substitute these factors into (A)

$\Rightarrow \left(x + 1\right) \left({x}^{2} - 9\right) = \left(x + 1\right) \left(x - 3\right) \left(x + 3\right)$

Thus ${x}^{3} + {x}^{2} - 9 x - 9 = \left(x + 1\right) \left(x - 3\right) \left(x + 3\right)$