Question #28428

1 Answer
Aug 5, 2017

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 :)