# How do you factor (x^5) - (x^4) + (x^3) - (x^2) - 12x + 12?

Jun 16, 2015

$f \left(x\right) = {x}^{5} - {x}^{4} + {x}^{3} - {x}^{2} - 12 x + 12$

$= \left(x - 1\right) \left({x}^{4} + {x}^{2} - 12\right)$

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

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

#### Explanation:

The sum of the coefficients is zero, that is:

$1 - 1 + 1 - 1 - 12 + 12 = 0$

So $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor of $f \left(x\right)$

The remaining quartic factor is actually quadratic in ${x}^{2}$, so factors just as easily into two quadratic factors:

${x}^{4} + {x}^{2} - 12 = {\left({x}^{2}\right)}^{2} + \left({x}^{2}\right) - 12$

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

The first of these can be viewed as a difference of squares:

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

...using the difference of squares identity:

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

The remaining quadratic factor has no linear factors with real coefficients, since ${x}^{2} + 4 \ge 4 > 0$ for all $x \in \mathbb{R}$.