# How do you factor x³+x²-x-1?

Nov 13, 2015

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?)

Nov 13, 2015

You can factor by grouping to find:

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

Then use the difference of squares identity to find:

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

#### Explanation:

First factor by grouping:

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

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

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

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 $\left(x - 1\right)$ is a factor.

Divide ${x}^{3} + {x}^{2} - x - 1$ by $\left(x - 1\right)$ to get ${x}^{2} + 2 x + 1$ :

Then recognise that ${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$ 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.