Question

How do you factor `x³+x²-x-1`?

Answer 1

The complete factorization is `(x+1)^2(x−1)` or `(x+1)(x+1)(x−1)`.

Explanation:

I used synthetic division to solve this. (Do you need further explanation?)

Answer 2

You can factor by grouping to find:

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

Then use the difference of squares identity to find:

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

Explanation:

First factor by grouping:

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

Then notice that `x^2-1 = x^2-1^2` is a difference of squares, so we can use the difference of squares identity [ `a^2-b^2 = (a-b)(a+b)` ] to find:

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

Alternatively, notice that the sum of the coefficients (`1+1-1-1`) is `0`, so `x=1` is a zero of this cubic polynomial and `(x-1)` is a factor.

Divide `x^3+x^2-x-1` by `(x-1)` to get `x^2+2x+1` :

Then recognise that `x^2+2x+1 = (x+1)^2` is a perfect square trinomial. One little trick to spot this one is that `11^2 = 121`, the `1,2,1` matching the coefficients of the quadratic and `1,1` matching the coefficients of the linear factor.