How do you find the zeroes of p(x)= x^3+x^2-x-1?

May 11, 2015

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

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

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

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

$= \left(x - 1\right) \left(x + 1\right) \left(x + 1\right)$

This is zero for $x = - 1$ (twice) and $x = 1$.

The process I used here is called factoring by grouping.