# How do you factor  x^4 + x^3 + x^2 + x + 1?

Feb 20, 2016

${x}^{4} + {x}^{3} + {x}^{2} + x + 1$

$= \left({x}^{2} + \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) x + 1\right) \left({x}^{2} + \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) x + 1\right)$

#### Explanation:

This quartic has four zeros, which are the non-Real Complex $5$th roots of $1$, as we can see from:

$\left(x - 1\right) \left({x}^{4} + {x}^{3} + {x}^{2} + x + 1\right) = {x}^{5} - 1$

So if we wanted to factor this polynomial as a product of linear factors with Complex coefficients then we could write:

${x}^{4} + {x}^{3} + {x}^{2} + x + 1$

$= \left(x - \left(\cos \left(\frac{2 \pi}{5}\right) + i \sin \left(\frac{2 \pi}{5}\right)\right)\right) \cdot \left(x - \left(\cos \left(\frac{4 \pi}{5}\right) + i \sin \left(\frac{4 \pi}{5}\right)\right)\right) \cdot \left(x - \left(\cos \left(\frac{6 \pi}{5}\right) + i \sin \left(\frac{6 \pi}{5}\right)\right)\right) \cdot \left(x - \left(\cos \left(\frac{8 \pi}{5}\right) + i \sin \left(\frac{8 \pi}{5}\right)\right)\right)$

A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if $x = r$ is a zero of ${x}^{4} + {x}^{3} + {x}^{2} + x + 1$, then $x = \frac{1}{r}$ is also a zero.

Hence there is a factorisation in the form:

${x}^{4} + {x}^{3} + {x}^{2} + x + 1$

$= \left(x - {r}_{1}\right) \left(x - \frac{1}{r} _ 1\right) \left(x - {r}_{2}\right) \left(x - \frac{1}{r} _ 2\right)$

$= \left({x}^{2} - \left({r}_{1} + \frac{1}{r} _ 1\right) x + 1\right) \left({x}^{2} - \left({r}_{2} + \frac{1}{r} _ 2\right) x + 1\right)$

So let's look for a factorisation:

${x}^{4} + {x}^{3} + {x}^{2} + x + 1$

$= \left({x}^{2} + a x + 1\right) \left({x}^{2} + b x + 1\right)$

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

Equating coefficients we find:

$a + b = 1$

$2 + a b = 1$, so $a b = - 1$ and $b = - \frac{1}{a}$

Substituting $b = - \frac{1}{a}$ in $a + b = 1$ we get:

$a - \frac{1}{a} = 1$

Hence:

${a}^{2} - a - 1 = 0$

Using the quadratic formula, we can deduce:

$a = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$

Since our derivation was symmetric in $a$ and $b$, one of these roots can be used for $a$ and the other for $b$, to find:

${x}^{4} + {x}^{3} + {x}^{2} + x + 1$

$= \left({x}^{2} + \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) x + 1\right) \left({x}^{2} + \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) x + 1\right)$

If we want to factor further, use the quadratic formula on each of these quadratic factors to find the linear factors with Complex coefficients.