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

Nov 29, 2015

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

Explanation:

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

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

Divide by $\left(x - 1\right)$...

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

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

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