How do you factor #x^3 - 16#?

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Answer 1 · 2016-04-07T22:21:01

#x^3-16#

#=(x-2 root(3)(2))(x^2+2 root(3)(2) x + 4 root(3)(4))#

#=(x-2 root(3)(2))(x-2 root(3)(2) omega)(x-2 root(3)(2) omega^2)#

Use the difference of cubes identity:

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

with #a = x# and #b = 2root(3)(2)#

So:

#x^3-16 =x^3-(2 root(3)(2))^3#

#=(x-2 root(3)(2))(x^2+2 root(3)(2) x + 4 root(3)(4))#

This is as far as we can go with Real coefficients, but if you allow Complex coefficients then this can be further factored as:

#=(x-2 root(3)(2))(x-2 root(3)(2) omega)(x-2 root(3)(2) omega^2)#

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