How to factor polynomial $\:f(x)=1+x^2+x^4,\:\:f\in \:Z_3\left [x\right]$

Question

I have been trying to solve the following polynomial factoring: $$ f(x) = 1 + x^2 + x^4,\quad f∈Z_3[x] $$ But I am stuck with: $$ \begin{align} f(x) &= 1 - 2x^2 + x^4 \\ &= (x^2 -1)^2 \\ &= x^2 -1 \\ &= (x + 1)(x - 1) \end{align} $$

How should I progress pass this?

Answer 1

We have that $(1+x^2+x^4)(1-x^2)=(1-x^6)$ hence $$ f(x) = (1-x+x^2)(1+x+x^2) = \Phi_3(x)\,\Phi_6(x) $$ and over $\mathbb{F}_3$ $$ f(x) = (1+2x+x^2)(1-2x+x^2) = (1+x)^2 (1-x)^2 $$ completely splits. If $p$ is a prime $\geq 5$ and $h$ is the least integer such that $6\mid (p^h-1)$,
$f(x)$ completely splits over $\mathbb{F}_{p^h}$.

Answer 2

Given that we are working over $\Bbb{Z}/3\Bbb{Z}$, clearly $1$ and $-1$ are roots of the polynomial, so long division by $(x-1)(x+1)=x^2-1$ shows that $$x^4+x^2+1=(x-1)(x+1)(x^2+2)=(x-1)(x+1)(x^2-1)=(x-1)^2(x+1)^2.$$