How do you factor completely $7x^3-56$ ?

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Answer 1 · 2016-09-25T20:59:12

$7x^3-56 = 7(x-2)(x^2+2x+4)$

The difference of cubes identity can be written:

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

We use this with $a=x$ and $b=2$ below.

First note that both terms are divisible by $7$ , so we can separate that out as a factor first:

$7x^3-56 = 7(x^3-8)$

$color(white)(7x^3-56) = 7(x^3-2^3)$

$color(white)(7x^3-56) = 7(x-2)(x^2+2x+4)$

This is as far as we can go with Real coefficients. If we allow Complex coefficients, then this factors further as:

$color(white)(7x^3-56) = 7(x-2)(x-2omega)(x-2omega^2)$

where $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ .

How do you factor completely 7x^3-56 7x^3-56 ?