Question

How do you factor completely `x^3+x^2-x-1`?

Answer 1

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

Explanation:

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

First notice that the sum of the coefficients is `0` so `f(1) = 1+1-1-1 = 0` and `x-1` is a factor of `f(x)`.

Divide by `(x-1)`...

Then factor the remaining quadratic factor - recognisable as a perfect square trinomial.

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

A little trick to spot that `x^2+2x+1` is `(x+1)^2` is given by the pattern of the coefficients: `1, 2, 1` is the same pattern of digits as `121 = 11^2`.