# How can you factor x^2+x+1 completely ?

Oct 2, 2017

Given, ${x}^{2} + x + 1$

$\Rightarrow {x}^{2} + 2 x + 1 - x$

$\Rightarrow {\left(x + 1\right)}^{2} - {\left(\sqrt{x}\right)}^{2}$[ formula a^2+2ab+b^2 = (a+b)^2]

$\Rightarrow \left(x + 1 + \sqrt{x}\right) \left(x + 1 - \sqrt{x}\right)$
[ applying formula ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$]

Oct 2, 2017

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

#### Explanation:

Note that:

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

is in the standard form:

$a {x}^{2} + b x + c$

with $a = 1$, $b = 1$ and $c = 1$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = \textcolor{b l u e}{1} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{1}\right) = - 3$

Since $\Delta < 0$ this quadratic has no real zeros and no linear factors with real coefficients.

We can still factor it, but we need to use non-real Complex coefficients.

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

We can complete the square then use this with $A = \left(x + \frac{1}{2}\right)$ and $B = \frac{\sqrt{3}}{2} i$ as follows:

${x}^{2} + x + 1 = {x}^{2} + x + \frac{1}{4} + \frac{3}{4}$

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

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

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

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

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

Bonus

Note that:

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

So the zeros $- \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ of $\left({x}^{2} + x + 1\right)$ that we found above are also zeros of $\left({x}^{3} - 1\right)$.

That is, they are cube roots of $1$.

They are often denoted:

$\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i \text{ }$ and $\text{ } {\omega}^{2} = \overline{\omega} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

$\omega$ is called the primitive complex cube root of $1$ and crops up a lot when solving cubic equations lacking simple roots.