How do you factor $4 x^{3} - 108$ $4x^3-108$ ?

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Answer 1 · 2016-05-25T22:59:38

$4 (x - 3) (x^{2} + 3 x + 9)$ $4(x-3)(x^2 + 3x + 9)$

To factor $4 x^{3} - 108$ $4x^3-108$

Begin by factoring out a $4$ $4$

$4 (x^{3} - 27)$ $4(x^3-27)$

$(x^{3} - 27)$ $(x^3-27)$ is a binomial of perfect cubes.

This factors as

$a^{3} - b^{3}$ $a^3-b^3$

$(a - b) (a^{2} + a b + b^{2})$ $(a-b)(a^2 + ab + b^2)$

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

The cube of ${\sqrt[3]{x}}^{3}$ $root(3)x^3$ = x
The cube of $\sqrt[3]{27} = 3$ $root(3)27 =3$

$(x - 3) (x^{2} + 3 x + 9)$ $(x-3)(x^2 + 3x + 9)$

The complete factorization is

$4 (x - 3) (x^{2} + 3 x + 9)$ $4(x-3)(x^2 + 3x + 9)$

How do you factor 4x3−1084x^3-108?