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

May 6, 2015

Factor by grouping

${x}^{3} + {x}^{2} - x - 1 = \left[{x}^{3} + {x}^{2}\right] + \left[- x - 1\right]$

The first bracket has a common factor of ${x}^{2}$ and the second bracket has a common factor of $- 1$. take those out to get:

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

Now we have two terms, one on each side of the red $\textcolor{red}{+}$.

Each term has a factor (in brackets) of $\left[x + 1\right]$. Tlhat is a common factor, so we can factor it out:

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

$\textcolor{w h i t e}{\text{ssssssssssssssssss}}$ $= \left({x}^{2} \textcolor{red}{+} \left(- 1\right)\right) \left[x + 1\right]$

$\textcolor{w h i t e}{\text{ssssssssssssssssss}}$ $= \left({x}^{2} - 1\right) \left(x + 1\right)$

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

Are we finished or can anything be factored more?

${x}^{2} - 1$ is a difference of twp squares, so we can factor it.

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

Now we are finished.