How many different roots does $p(x) = x^{12}+2x^{6}+1 \in \mathbb{F}_3[x]$ have?

Question

I want to find out how many different roots does $p(x) = x^{12}+2x^{6}+1 \in \mathbb{F}_3[x]$ have (not necessarily in $\mathbb{F}_3$, but in an body that has a subgroup isomorphic to $\mathbb{F}_3$), where $\mathbb{F}_3$ is the finite body with $3$ elements ($\cong \mathbb{Z}_3$).
$p'(x)=12x^{11}+12x^{5} \equiv 0$ so $p(x)$ has roots of multiplicity $\geq 2$. I actually know that the roots with multuplicity $\geq 2$ are the zeroes of g.c.d$(p(x), p'(x))$ = g.c.d$(p(x), 0)=p(x)$. Therefore $p(x)$ only has roots with multiplicity $\geq 2$.
$p(x)$ doesn't have any roots in $\mathbb{F}_3$ so I have to build a body where $p(x)$ has at least a root. For example $$\mathbb{F}_3[T]/\langle T^{12}+2T^6+1\rangle$$ I know that here $\overline{T}$ is a root. Using Frobenius' homomosphism I get that $\overline{T}^3, \overline{T}^9, \overline{T}^{27} = 2\overline{T}^9$ are roots. If I use Frobenius once again, I go back to $\overline{T}^9$. Does this mean I've found all the roots? How can I be sure there aren't any more roots?

Answer 1

HINT: In $\mathbb{F}_3[x]$ $$p(x) = x^{12}+2x^{6}+1 = (x^{6} + 1)^2$$ How many roots do you expect to find in a splitting field, counting multiplicity?