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

May 10, 2015

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

To find this, first notice that ${x}^{4} + {x}^{2} + 1 > 0$ for all (real) values of $x$. So there are no linear factors, only quadratic ones.

${x}^{4} + {x}^{2} + 1 = \left(a {x}^{2} + b x + c\right) \left({\mathrm{dx}}^{2} + e x + f\right)$

Without bothering to multiply this out fully just yet, notice that the coefficient of ${x}^{4}$ gives us $a d = 1$. We might as well let $a = 1$ and $d = 1$.

... $= \left({x}^{2} + b x + c\right) \left({x}^{2} + e x + f\right)$

Next, the coefficient of ${x}^{3}$ gives us $b + e = 0$, so $e = - b$.

... $= \left({x}^{2} + b x + c\right) \left({x}^{2} - b x + f\right)$

The constant term gives us $c f = 1$, so either $c = f = 1$ or $c = f = - 1$. Let's try $c = f = 1$.

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

Note that the coefficient of $x$ will vanish nicely when these are multiplied out.

Finally notice that the coefficient of ${x}^{2}$ is $\left(1 - {b}^{2} + 1\right) = 2 - {b}^{2}$, giving us $1 = 2 - {b}^{2}$, thus ${b}^{2} = 1$, so $b = 1$ or $b = - 1$.

Nov 15, 2017

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

#### Explanation:

This really makes a bit more sense in the complex numbers...

First note that:

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

So zeros of ${x}^{4} + {x}^{2} + 1$ are also zeros of ${x}^{6} - 1$.

What are the zeros of ${x}^{6} - 1$?

The real zeros are $1$ and $- 1$, which are zeros of ${x}^{2} - 1$, the factor we introduced. So the four zeros of ${x}^{4} + {x}^{2} + 1$ are the four complex zeros of ${x}^{6} - 1$ apart from $\pm 1$.

Here are all $6$ in the complex plane:

graph{((x-1)^2+(y-0)^2-0.002)((x-1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1)^2+(y-0)^2-0.002)((x+1/2)^2+(y+sqrt(3)/2)^2-0.002)((x-1/2)^2+(y+sqrt(3)/2)^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}

They form the vertices of a regular hexagon.

de Moivre's formula tells us that:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin \theta$

where $i$ is the imaginary unit, satisfying ${i}^{2} = - 1$

For instance we find:

${\left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)}^{6} = \cos 2 \pi + i \sin 2 \pi = 1 + 0 = 1$

That's the zero $\frac{1}{2} + \frac{\sqrt{3}}{2} i$ that we see in Q1.

If $a$ is a zero of a polynomial then $\left(x - a\right)$ is a factor.

Hence:

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

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

For example:

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

$= {x}^{2} - \left(\left(\frac{1}{2} + \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{\sqrt{3}}{2} i}}}\right) + \left(\frac{1}{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{\sqrt{3}}{2} i}}}\right)\right) x + \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) \left(\frac{1}{2} - \sqrt{\frac{3}{2}} i\right)$

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

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

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

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

Nov 16, 2017

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

#### Explanation:

Given: ${x}^{4} + {x}^{2} + 1$

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

Note that:

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

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

$\textcolor{w h i t e}{\left({x}^{2} - 1\right) \left({x}^{4} + {x}^{2} + 1\right)} = {x}^{6} - 1$

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

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

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

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

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

Dividing both ends by $\left({x}^{2} - 1\right)$ we get:

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