Question

Question #28428

Answer 1

`x(x^3+2x^2+x-1)`

Explanation:

Let

`f(x)=x^4+2x^3+x^2-x`

First, to find factors I would try to solve:

`f(a)=0`

for small values of `a`.

E.g. `f(-1)` or `f(-2)`

If `f(a)=0`, then `(x-a)` is a factor.

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

`f(x)=x(x^3+2x^2+x-1)`

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 :)