# How do you factor (z-4)^3-4(z-4) ?

Apr 7, 2016

${\left(z - 4\right)}^{3} - 4 \left(z - 4\right) = \left(z - 4\right) \left(z - 6\right) \left(z - 2\right)$

#### Explanation:

Separate out the common $\left(z - 4\right)$ factor then use the difference of squares identity:

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

with $a = \left(z - 4\right)$ and $b = 2$ as follows:

${\left(z - 4\right)}^{3} - 4 \left(z - 4\right)$

$= \left(z - 4\right) \left({\left(z - 4\right)}^{2} - 4\right)$

$= \left(z - 4\right) \left({\left(z - 4\right)}^{2} - {2}^{2}\right)$

$= \left(z - 4\right) \left(\left(z - 4\right) - 2\right) \left(\left(z - 4\right) + 2\right)$

$= \left(z - 4\right) \left(z - 6\right) \left(z - 2\right)$