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: $$ \Phi_n(X) \quad = \quad \prod_{ord(\zeta) = n} (X - \zeta) $$
On
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:
$$\Phi_{15}(x)=(2+x+4x^2+2x^3+x^4)(4+x+2x^2+4x^3+x^4)$$
On
+1 to the question and both answers.
Koenraad, a different way of reaching the conclusion in Jack's answer is to use Galois theory. You hopefully remember that the Galois groups of finite fields are generated by the Frobenius map - in this case $F:a\mapsto a^7$. So if $\zeta$ is a fixed primitive fifteenth root of unity in whatever extension field $L$ of $K=\Bbb{F}_7$ it resides, its conjugates are $$ \zeta^7,\quad \zeta^{49}=\zeta^{3\cdot15+4}=\zeta^4,\quad\zeta^{4\cdot7}=\zeta^{15+13}=\zeta^{13},\quad\zeta^{7\cdot13}=\zeta^{6\cdot15+1}=\zeta. $$ So by iterating the Frobenius map we generated the listed four conjugates. Therefore the polynomial $$ p(x)=(x-\zeta)(x-\zeta^7)(x-\zeta^4)(x-\zeta^{13}) $$ is the minimal polynomial of $\zeta$ over $K$. The nice trick in Thomas' answer is based on the observation that we can write these conjugates in the form $$ \begin{aligned} \zeta&=\zeta^{10}\zeta^6\\ \zeta^7&=\zeta^{10}\zeta^{12}\\ \zeta^4&=\zeta^{10}\zeta^9\\ \zeta^{13}&=\zeta^{10}\zeta^3, \end{aligned} $$ where the recurring factor $\zeta^{10}$ on the right hand side is of order three (and can be identified as an element of $K$), and the other factors are exactly the roots of unity of order five.
Don't know if this helps. Ask, if you need more details.
Consider that $x^3-1$ completely splits over $\mathbb{F}_7$ as: $$ x^3-1 = (x-2)(x-1)(x+3) \tag{1}$$ while $x^5-1$ splits as: $$ x^5-1 = (x-1)(x^4+x^3+x^2+x+1),\tag{2}$$ since the smallest field with characteristic $7$ in which there is a primitive fifth root of unity if $\mathbb{F}_{7^4}$, because $k=4$ is the first natural number such that $5\mid 7^k-1$. For the same reason, the splitting field of $x^{15}-1$ over $\mathbb{F}_7$ is $\mathbb{F}_{7^4}$. Since $\deg \Phi_{15} = 8$, we have that $\Phi_{15}$ splits as the product of two irreducible polynomials of degree four over $\mathbb{F}_7$.
The key lemma is the following: if $x^l-1$ completely splits over $\mathbb{F}_{7^a}$, then $l$ must divide $|\mathbb{F}_{7^a}^*|=7^a-1$ by the Lagrange's theorem. Since the multiplicative group of a finite field is cyclic, this gives also a sufficient condition for $\mathbb{F}_{7^a}$ to be the splitting field of $x^l-1$ over $\mathbb{F}_7$.
Back to the original question, how to recover the two factors of $\Phi_{15}$? Just look at the Jyrki Lahtonen's answer, that shows how the Frobenius automorphism of a finite field $\mathbb{F}_{p^4}\simeq \mathbb{F}_p[x]/(q(x))$ gives that the roots of $q(x)$ are of the form $\xi,\xi^p,\xi^{p^2},\xi^{p^3}$.