# Thread: How to factor (x^3 + 8) ?

1. ## How to factor (x^3 + 8) ?

How do I factor (x^3 + 8) ?

Any help would be greatly appreciated!

2. Originally Posted by s3a
How do I factor (x^3 + 8) ?

Any help would be greatly appreciated!
Notice that

$\displaystyle x^3 + 8 = x^3 + 2^3$, a sum of two cubes.

So it can be factorised using the sum of two cubes rule...

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

Therefore

$\displaystyle x^3 + 8 = (x + 2)(x^2 - 2x + 4)$.

3. Let $\displaystyle f(x)=x^{3}+8$

Since $\displaystyle f(-2)=0$, by factor theorem, $\displaystyle (x+2)$ is a factor.

So $\displaystyle x^{3}+8=(x+2)(ax^{2}+bx+c)$

Comparing coeff. of $\displaystyle x^{3}$, $\displaystyle a=1$

Comparing constants, $\displaystyle c=4$

Comparing coeff. of $\displaystyle x^{2}$, $\displaystyle b+2a=0$
$\displaystyle b=-2$

Note: you can find the quadratic factor by long division also.

Thus
$\displaystyle f(x)=(x+2)(x^{2}-2x+4)$

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