How to find the roots of $x^4 + x^2 + 1$ in $\mathbb{C}$ (question about the basics of factoring)

Question

So, I'm struggling a bit with factoring some of these polynomials (I'm missing many basics) so I'm working through some problems.

The first step of the solution I have for the example I'm working on is the following:

We re-write the polynomial as:

$(x^2)^2 + x^2 + 1$ and conclude that that's equal to $(x^2-w) (x^2 - w^2)$, where $w = e^{2πi}/3$.

Can someone explain why do we re-write this and how we made that conclusion?

I know that $e^{2πi}/3$ is one of the solutions to cube roots of unity and it's equal to $-1/2 + √3/2i$, but how does that relate to our problem?

Answer 1

The equation $x^4+x^2+1$ is what we call a biquadratic equation since only even powers of $x$ are involved.

Let $y:=x^2$, then $x^4+x^2+1=0$ is equivalent to $y^2+y+1=0$. Now, it is derived from quadratic formula that $y^2+y+1=(y-w)(y-w^2)$. Recalling what is $y$, one has: $$x^4+x^2+1=(x^2-w)(x^2-w^2)=(x^2-w)(x-w)(x+w).$$ The factorization is still not complete, $x^2-w$ can still be split into a product of two linear factors.

Answer 2

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Just push your argument a little further $$ y^{3} - 1 = (y - 1) (y^{2} + y + 1). $$ Substitute $y = x^{2}$ to get $$ x^{6} - 1 = (x^{2} - 1) (x^{4} + x^{2} + 1)= (x - 1) (x + 1) (x^{4} + x^{2} + 1). $$ So the roots of your polynomial are the $6$-th roots of unity other then $\pm 1$, that is, $$ \exp\left( k \frac{2 \pi i}{6} \right) $$ for $k \in A = \Set{ 1, 2, 4, 5 }$.

Answer 3

You can observe that $(x^4+x^2+1)(x^2-1)=x^6-1$, so the roots of $x^4+x^2+1$ are the $6$-th roots of $1$, except for $1$ and $-1$, so $$ e^{i\pi/3},\quad e^{2i\pi/3},\quad e^{4i\pi/3},\quad e^{5i\pi/3} $$

Or, since you know that $e^{2i\pi/3}$ and $e^{-2i\pi/3}$ are roots of $z^2+z+1$, you get as roots for the given polynomial their square roots, that is, $$ e^{i\pi/3},\quad -e^{i\pi/3},\quad e^{-i\pi/3},\quad -e^{-i\pi/3} $$