# How do you factor x^3 + 6x^2 - 9x - 54 by grouping?

Jul 23, 2016

(x+6)(x-3)(x+3)

#### Explanation:

Group the terms into 'pairs' as follows.

$\left[{x}^{3} + 6 {x}^{2}\right] + \left[- 9 x - 54\right]$

Now, factorise each 'pair'

$\textcolor{red}{{x}^{2}} \left(x + 6\right) \textcolor{red}{- 9} \left(x + 6\right)$

Take out the common factor of (x +6)

$\left(x + 6\right) \textcolor{red}{\left({x}^{2} - 9\right)}$

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

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

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

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

Pulling this altogether we get.

${x}^{3} + 6 {x}^{2} - 9 x - 54 = \left(x + 6\right) \left(x - 3\right) \left(x + 3\right)$