# How do you factor the expression c^4 + c^3 - 12c - 12?

Apr 13, 2016

${c}^{4} + {c}^{3} - 12 c - 12 = \left(c - \sqrt[3]{12}\right) \left({c}^{2} + \sqrt[3]{12} c + \sqrt[3]{144}\right) \left(c + 1\right)$

#### Explanation:

We can factor this quartic using grouping then the difference of cubes identity, which may be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

with $a = c$ and $b = \sqrt[3]{12}$ as follows:

${c}^{4} + {c}^{3} - 12 c - 12$

$= \left({c}^{4} + {c}^{3}\right) - \left(12 c + 12\right)$

$= {c}^{3} \left(c + 1\right) - 12 \left(c + 1\right)$

$= \left({c}^{3} - 12\right) \left(c + 1\right)$

$= \left({c}^{3} - {\left(\sqrt[3]{12}\right)}^{3}\right) \left(c + 1\right)$

$= \left(c - \sqrt[3]{12}\right) \left({c}^{2} + \sqrt[3]{12} c + \sqrt[3]{144}\right) \left(c + 1\right)$

If we allow Complex coefficients then this can be factored further as:

$= \left(c - \sqrt[3]{12}\right) \left(c - \omega \sqrt[3]{12}\right) \left(c - {\omega}^{2} \sqrt[3]{12}\right) \left(c + 1\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.