# How do you factor n^3+5n^2-4n-20 by grouping?

May 16, 2016

(n + 5)(n-2)(n+2)

#### Explanation:

'Group' the terms as follows:

$\left[{n}^{3} + 5 {n}^{2}\right] + \left[- 4 n - 20\right]$

now factorise each group.

$\Rightarrow {n}^{2} \left(n + 5\right) - 4 \left(n + 5\right)$

There is now a common factor of (n +5) which can be taken out.

$\Rightarrow \left(n + 5\right) \left({n}^{2} - 4\right)$

now ${n}^{2} - 4 \text{ is a difference of squares}$

which $\textcolor{b l u e}{\text{ In general is factorised as follows }}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\Rightarrow {n}^{2} - 4 = \left(n - 2\right) \left(n + 2\right)$

$\Rightarrow {n}^{3} + 5 {n}^{2} - 4 n - 20 = \left(n + 5\right) \left(n - 2\right) \left(n + 2\right)$