Let $\mathbb{F}$ be a field. Suppose that we have a polynomial $f(x) \in \mathbb{F}[x]$ of degree $k$. Its reciprocal will be $f^*(x)=x^{k}f(x^{-1})$, and we say $f$ is self-reciprocal if $f=f^*$.
Suppose that $f(x)=p(x)q(x)$, where $p$ and $q$ are irreducible, and $f$ is self-reciprocal. Then either
$p(x)=\pm p^*(x)$ and $q(x)=\pm q^*(x)$;
$p(x)=a q^*(x)$ and $q(x)=a^{-1} p^*(x)$, for some $a \in \mathbb{F}$.
My attempt:
Notice that, since $f=f^*$, we have that $f^*(x)=x^{k}f(x^{-1})=x^{k}p(x^{-1})q(x^{-1})=p^*(x)q^*(x)$. Thus, $pq=p^*q^*$. Since $\mathbb{F}$ is a field, $\mathbb{F}[x]$ is an Euclidean Domain, so the concept of irreducible and prime is the same, so either $q(x)|p^*(x)$ or $q(x)|q^*(x)$.
If $q(x)|q^*(x)$, since $\deg(q)=\deg(q^*)$, we have that $q^*(x)=c q(x)$ for some constant $c \in \mathbb{F}$. But then $q(x)=c^{-1} q^*(x)$, so $q^*(x)|q(x)$, and thus $q(x)=\pm q^*(x)$. With this we also get $p(x)=\pm p^*(x)$.
If $q(x)|p^*(x)$, I'm stuck. How can I show that one is a constant multiple of the other?
