# How do you factor h(x)= x^4 + 4x^3 + x^2 - 6x?

Jul 22, 2015

$h \left(x\right) = x \left(x - 1\right) \left(x + 2\right) \left(x + 3\right)$

#### Explanation:

Extract the common factor of $x$ from the terms of ${x}^{4} + 4 {x}^{3} + {x}^{2} - 6 x$
$\textcolor{w h i t e}{\text{XXXX}}$$= x \left({x}^{3} + 4 {x}^{2} + x - 6\right)$

Considering the term: ${x}^{3} + 4 x + x - 6$
Notice that the coefficients of the first 3 terms sum to the negative of the final constant
$\rightarrow \left(x - 1\right)$ is a factor of this expression.

Using synthetic division (or whatever method you choose):
$\textcolor{w h i t e}{\text{XXXX}}$$= x \left(x - 1\right) \left({x}^{2} + 5 x + 6\right)$

We can use standard AC methods to factor this final term:
$\textcolor{w h i t e}{\text{XXXX}}$$= x \left(x - 1\right) \left(x + 2\right) \left(x + 3\right)$