# How do you factor 3x^3-24?

Feb 14, 2016

$3 {x}^{3} - 24 = 3 \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$
$= 3 \left(x - 2\right) \left(x + 1 + \sqrt{3} i\right) \left(x + i - \sqrt{3} i\right)$

#### Explanation:

Using the difference of cubes formula ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
we have

$3 {x}^{3} - 24 = 3 \left({x}^{3} - 8\right)$

$= 3 \left({x}^{3} - {2}^{3}\right)$

$= 3 \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

If we only allow real numbers, we are done. If we allow complex numbers, we can use the quadratic formula to factor ${x}^{2} + 2 x + 4$ by finding its roots.

${x}^{2} + 2 x + 4 = 0$
$\iff x = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(4\right)}}{2 \left(1\right)}$

$= \frac{- 2 \pm \sqrt{- 12}}{2}$

$= - 1 \pm \sqrt{- 3}$

$= - 1 \pm \sqrt{3} i$

Thus

$3 {x}^{3} - 24 = 3 \left(x - 2\right) \left(x + 1 + \sqrt{3} i\right) \left(x + i - \sqrt{3} i\right)$