# How do you factor x^(12) -1?

Dec 26, 2015

Break the problem down to the exponents you can handle. Ideally, Difference of Squares or sum and difference of cubes can be used. An explanation is given below on how to go about it.

#### Explanation:

Factoring ${x}^{12} - 1$ can be tricky but can be broken down and worked it out step by step.

Let us recollect some rules.

1. Difference of squares
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

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

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

Now let us come to our problem.

${x}^{12} - 1$
Let us make x^12 into something which we can recognize easily.

${x}^{12} = {x}^{6 \cdot 2} = {\left({x}^{6}\right)}^{2}$ We have used ${a}^{m n} = {\left({a}^{m}\right)}^{n}$

${\left({x}^{6}\right)}^{2} - {1}^{2}$ Now it is of form ${a}^{2} - {b}^{2}$

$\left({x}^{6} - 1\right) \left({x}^{6} + 1\right)$

Let us factorize ${x}^{6} - 1$

${x}^{6} - 1 = {\left({x}^{3}\right)}^{2} - {1}^{2}$ Rewriting ${x}^{6}$ to get into ${a}^{2} - {b}^{2}$ form
${x}^{6} - 1 = \left({x}^{3} - 1\right) \left({x}^{3} + 1\right)$
${x}^{6} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right) \left(x + 1\right) \left({x}^{2} - x + 1\right)$

Now to factorize $\left({x}^{6} + 1\right)$
${x}^{6} + 1 = {\left({x}^{2}\right)}^{3} + {1}^{3}$ Rewriting ${x}^{6}$ to get it into ${a}^{3} + {b}^{3}$ form
${x}^{6} + 1 = \left({x}^{2} + 1\right) \left({x}^{4} - {x}^{2} + 1\right)$

Factors of $\left({x}^{6} - 1\right) \left({x}^{6} + 1\right)$
$= \left(x - 1\right) \left({x}^{2} + x + 1\right) \left(x + 1\right) \left({x}^{2} - x + 1\right) \left({x}^{2} + 1\right) \left({x}^{4} - {x}^{2} + 1\right)$

Rearranging
$\left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right) \left({x}^{2} - x + 1\right) \left({x}^{2} + x + 1\right) \left({x}^{4} - {x}^{2} + 1\right)$ Answer.

Dec 26, 2015

Use Complex arithmetic, then match up conjugate pairs to find:

${x}^{12} - 1 = \left(x - 1\right) \left({x}^{2} - \sqrt{3} x + 1\right) \left({x}^{2} - x + 1\right) \left({x}^{2} + 1\right) \left({x}^{2} + x + 1\right) \left({x}^{2} + \sqrt{3} x + 1\right) \left(x + 1\right)$

#### Explanation:

Here's an alternative method using Complex arithmetic to produce a result with Real coefficients...

Note that if ${a}^{12} = 1$ then ${a}^{12} - 1 = 0$ and $\left(x - a\right)$ is a factor of ${x}^{12} - 1$.

The twelve Complex roots of $1$ are shown in this picture: I have marked the Complex conjugate pairs with the same colour.

For example: $\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} i$

So, picking out the roots from right to left, then multiplying up Complex conjugate pairs we find:

${x}^{12} - 1$

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

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