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

Jul 19, 2018

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

#### Explanation:

First, we see that there's a factor of $x$ we can easily cancel:
${x}^{3} - {x}^{2} - x + x = {x}^{3} - {x}^{2}$

From there, we can also factor out ${x}^{2}$:
${x}^{3} - {x}^{2} = {x}^{2} \left(x - 1\right)$

Since each of these is fully factored, we are done.

Jul 19, 2018

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

#### Explanation:

This expression can be simplified to

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

From here, we notice that both terms have an ${x}^{2}$ in common, which we can factor out. At this point, we are essentially dividing.

We get

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

Hope this helps!