# How do you factor by grouping x^3 + x^2 - x - 1?

May 25, 2015

${x}^{3} + {x}^{2} - x - 1$

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

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

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

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

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

...using the difference of squares identity

$\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)$

Dec 31, 2017

Find a useful grouping, then factor.

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

#### Explanation:

Factor   x^3+x^2−x−1

The idea of grouping ${x}^{2}$ with -$1$ looks really tempting
because it's the Difference of Two Squares..

1) Find a useful grouping
$\left({x}^{3} - x\right) + \left({x}^{2} - 1\right)$

2) Factor each group
$x \left({x}^{2} - 1\right) + 1 \left({x}^{2} - 1\right)$

3) Factor out $\left({x}^{2} - 1\right)$ from each group
$\left({x}^{2} - 1\right) \left(x + 1\right)$

4) Factor $\left({x}^{2} - 1\right)$ as the Difference of Two Squares
$\left(x - 1\right) \left(x + 1\right) \left(x + 1\right)$ $\leftarrow$ answer

5) You can write it this way if you want:
${\left(x + 1\right)}^{2} \left(x - 1\right)$ $\leftarrow$ same answer