How do you factor #4x^3-108#?

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Answer 1 · 2016-05-25T22:59:38

#4(x-3)(x^2 + 3x + 9)#

To factor #4x^3-108#

Begin by factoring out a #4#

#4(x^3-27)#

#(x^3-27)# is a binomial of perfect cubes.

This factors as

#a^3-b^3#

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

The signs follow SOAP
Same Sign first
Opposite Sign next
then Always Positive

The cube of #root(3)x^3# = x
The cube of #root(3)27 =3#

#(x-3)(x^2 + 3x + 9)#

The complete factorization is

#4(x-3)(x^2 + 3x + 9)#