If $\alpha$ = $\beta^q - \beta$ where both $\alpha$ , $\beta $ belongs to $F_q^n$ which is extension of $F_q$ Clearly $\beta$ is a root of $f(x) = x^q - x - \alpha$ and the other roots are its conjugates w.r.t $F_q$ so $f(x)$ splits in $F_q^n$ . But the degree is q so there are q distinct roots and my problem is how to represent it.
 A: The $F_q$-conjugates of $\beta$ are $\beta^{q^i}$, $i\ge0$. We get
$$
\begin{aligned}
\beta^q&=\beta+\alpha\\
\beta^{q^2}&=\beta^q+\alpha^q=\beta+\alpha+\alpha^q\\
\beta^{q^3}&=\cdots =\beta+\alpha+\alpha^q+\alpha^{q^2}\\
\vdots&=\ddots\\
\beta^{q^n}&=\cdots=\beta+\alpha+\alpha^q+\cdots+\alpha^{q^{n-1}}=\beta+tr^{q^n}_q(\alpha)=\beta,
\end{aligned}
$$
because it was assumed that the (relative) trace $tr^{q^n}_q(\alpha)$ vanishes.
Thus we can conclude that the number of $F_q$-conjugates of $\beta$ is a factor of $n$. I am not sure whether we can say anything more about it.
If you are interested in the factorization of $f(x)=x^q-x-\alpha$, then that it easy. If $\gamma\in F_q$ is an arbitrary element, then
$$
\begin{aligned}
f(\beta+\gamma)&=(\beta+\gamma)^q-(\beta+\gamma)-\alpha\\
&=\beta^q+\gamma^q-\beta-\gamma-\alpha\\
&=f(\beta)+\gamma^q-\gamma=0+0=0.
\end{aligned}
$$
So all the zeros of $f(x)$ are of this form, and we have a factorization
$$
f(x)=\prod_{\gamma\in F_q}(x-\beta-\gamma).
$$
Note that because the coefficients of $f(x)$ are not in $F_q$, its relation to the minimal polynomial of $\beta$ over $F_q$ is unclear.
