<s>[INST] Splitting $\Phi_{15}$ in irreducible factors over $\mathbb{F}_7$ I have to split $\Phi_{15}$ in irreducible factors over the field $\mathbb{F}_7$. It has been a while that I did this kind of stuff, and to be the honest, I've never really understood this matter. I'd really be grateful if someone could show me how this works. I don't ask you to do this whole job. It would be better if I would get some subtle advice instead.

I use the following definition:
$$
\Phi_n(X) \quad = \quad \prod_{ord(\zeta) = n} (X - \zeta)
$$
 [/INST] Every such $\zeta$ can be written uniquely as $\zeta_1\zeta_2$ where $\mathrm{ord}(\zeta_1)=3$ and $\mathrm{ord}(\zeta_2)=5$. The cases where $\mathrm{ord}(\zeta_1)=3$ are $2,4$.
The $\zeta_2$ values are roots of $1+x+x^2+x^3+x^4=(x-\zeta_5)(x-\zeta_5^2)(x-\zeta_5^3)(x-\zeta_5^4)$. So 
$$\begin{align}(x-2\zeta_5)(x-2\zeta_5^2)(x-2\zeta_5^3)(x-2\zeta_5^4)& = 2^4+2^3x+2^2x^2+2x^3+x^4 \\&= 2+x+4x^2+2x^3+x^4\end{align}$$
$$\begin{align}(x-4\zeta_5)(x-4\zeta_5^2)(x-4\zeta_5^3)(x-4\zeta_5^4)& = 4^4+4^3x+4^2x^2+4x^3+x^4 \\&= 4+x+2x^2+4x^3+x^4\end{align}$$
So, the products of these are $\Phi_{15}(x)$.
There are no roots for these, so the only possible factors can be quadratics. 
The only possible option is: $(x-2\zeta_5)(x-2\zeta_5^4)=x^2-2(\zeta_5+\zeta_5^4)x+4$ and $(x-2\zeta_5^2)(x-2\zeta_5^3)$ for the first, and the related factoring for the second. But that only works if $\zeta_5+\zeta_5^4\in\mathbb F_7$, which we know algebraically requires $\sqrt{5}\in\mathbb F_7$, which is not the case. So:
$$\Phi_{15}(x)=(2+x+4x^2+2x^3+x^4)(4+x+2x^2+4x^3+x^4)$$
</s>