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#.