# How do you factor completely  x^4 + 2x^3 - 3x - 6?

Apr 7, 2016

${x}^{4} + 2 {x}^{3} - 3 x - 6 = \left(x - \sqrt[3]{3}\right) \left({x}^{2} + \sqrt[3]{3} x + \sqrt[3]{9}\right) \left(x + 2\right)$

#### Explanation:

Notice that the ratio between the 1st and 2nd terms is the same as that between the 3rd and 4th terms so this will factor by grouping.

The remaining cubic term can be expressed as a difference of cubes, so we can use the identity:

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

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

${x}^{4} + 2 {x}^{3} - 3 x - 6$

$= \left({x}^{4} + 2 {x}^{3}\right) - \left(3 x + 6\right)$

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

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

$= \left({x}^{3} - {\left(\sqrt[3]{3}\right)}^{3}\right) \left(x + 2\right)$

$= \left(x - \sqrt[3]{3}\right) \left({x}^{2} + \sqrt[3]{3} x + \sqrt[3]{9}\right) \left(x + 2\right)$

This is as far as we can go with Real coefficients, but if you allow Complex coefficients then this can be factored as:

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

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