# Question #28428

Aug 5, 2017

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

#### Explanation:

Let

$f \left(x\right) = {x}^{4} + 2 {x}^{3} + {x}^{2} - x$

First, to find factors I would try to solve:

$f \left(a\right) = 0$

for small values of $a$.

E.g. $f \left(- 1\right)$ or $f \left(- 2\right)$

If $f \left(a\right) = 0$, then $\left(x - a\right)$ is a factor.

Unfortunately, no values of $a$ will work so all you can do is take out the common factor of $x$:

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

So, can we factorise the cubic part in brackets further?

Well, if we graph the cubic part, you can see that there is only one x-intercept, which means that the cubic function can't be factorised any further (more factors means more x-intercepts).

graph{x^3+2x^2+x-1 [-10, 10, -5, 5]}

So because the cubic part can't be factorised, taking out $x$ is as far as you can go without getting into trouble with the cops :)