Sep 18, 2015

# (x+1)(x−1)(x+2)(x+4)

#### Explanation:

First thing to always remember is a bi quadratic has 4 roots whether they are distinct or not.

Start of with the basic

$\pm 1$

We find that

$p \left(1\right) , p \left(- 1\right) = 0$
$\implies \left(x - 1\right) \left(x + 1\right)$are factors of $p \left(x\right)$

Lets divide

$\frac{p \left(x\right)}{\left(x - 1\right) \left(x + 1\right)}$ = ${x}^{2} + 6 x + 8$

$\implies \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 6 x + 8\right) = p \left(x\right)$

Now factor ${x}^{2} + 6 x + 8$

$= \left(x + 2\right) \left(x + 4\right)$

Now
$\implies \left(x - 1\right) \left(x + 1\right) \left(x + 2\right) \left(x + 4\right) = p \left(x\right)$

We have found the four roots and hence we can be certain we have factored the given polynomial completely