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

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Answer 1 · 2016-04-07T18:21:49

$(z-4)^3 - 4(z-4)=(z-4)(z-6)(z-2)$

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

$a^2-b^2 = (a-b)(a+b)$

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

$(z-4)^3 - 4(z-4)$

$=(z-4)((z-4)^2-4)$

$=(z-4)((z-4)^2-2^2)$

$=(z-4)((z-4)-2)((z-4)+2)$

$=(z-4)(z-6)(z-2)$

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