# How do you factor x^4-x^2-x-1 completely ?

May 9, 2017

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

$\textcolor{w h i t e}{{x}^{4} - {x}^{2} - x - 1} = \left(x - 1\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

where ${x}_{1} , {x}_{2} , {x}_{3}$ are the roots found below...

#### Explanation:

Given:

${x}^{4} - {x}^{2} - x + 1$

First notice that the sum of the coefficients is $0$. That is:

$1 - 1 - 1 + 1 = 0$

Hence we can tell that $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

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

Factoring the remaining cubic is somewhat more complicated:

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Discriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 1$, $b = 1$, $c = 0$ and $d = - 1$, so we find:

$\Delta = 0 + 0 + 4 - 27 + 0 = - 23$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 27 \left({x}^{3} + {x}^{2} - 1\right)$

$= 27 {x}^{3} + 27 {x}^{2} - 27$

$= {\left(3 x + 1\right)}^{3} - 3 \left(3 x + 1\right) - 25$

$= {t}^{3} - 3 t - 25$

where $t = \left(3 x + 1\right)$

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Cardano's method

We want to solve:

${t}^{3} - 3 t - 25 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 1\right) \left(u + v\right) - 25 = 0$

Add the constraint $v = \frac{1}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{1}{u} ^ 3 - 25 = 0$

Multiply through by ${u}^{3}$ and rearrange slightly to get:

${\left({u}^{3}\right)}^{2} - 25 \left({u}^{3}\right) + 1 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{25 \pm \sqrt{{\left(- 25\right)}^{2} - 4 \left(1\right) \left(1\right)}}{2 \cdot 1}$

$= \frac{25 \pm \sqrt{625 - 4}}{2}$

$= \frac{25 \pm \sqrt{621}}{2}$

$= \frac{25 \pm 3 \sqrt{69}}{2}$

Since this is Real and the derivation is symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to find Real root:

${t}_{1} = \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}} + \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}} + {\omega}^{2} \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}} + \omega \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}}$

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

Now $x = \frac{1}{3} \left(- 1 + t\right)$. So the roots of our original cubic are:

${x}_{1} = \frac{1}{3} \left(- 1 + \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}} + \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}}\right)$

${x}_{2} = \frac{1}{3} \left(- 1 + \omega \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}} + {\omega}^{2} \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left(- 1 + {\omega}^{2} \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}} + \omega \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}}\right)$

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