Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible.
Observe that $(f^*)^*=f$ if and only if $f$ has nonzero constant term, and $(gh)^*=g^*h^*$ for all $g,h$.
Assume $f$ is irreducible but $f^*=gh$ is not, where $g,h$ are nonconstant. Then since $f^*$ has nonzero constant term, so do $g$ and $h$, hence $g^*,h^*$ are both nonconstant, and since $f=(f^*)^*=g^*h^*$ we have a contradiction.
The other implication actually isn't true! For a counterexample, let $h(x)$ be irreducible with nonzero constant term and let $f(x)=xh(x)$. Then $(f^*)^*=h(x)$ and $((f^*)^*)^*=f^*$, so $f^*$ is irreducible even though $f$ is not.
Note that $f^{*} = 0$ if and only if $f = 0$ and that for $f \neq 0$ we have:
- $\deg f^{*} \leq \deg f$.
- $\deg f = \deg f^{*}$ if and only if $f(0) \neq 0$.
- If $f(0) \neq 0$ then $\left( f^{*} \right)^{*} = f$.
In addition, we have $$ (gh)^{*}(x) = x^{\deg (gh)} (gh) \left( \frac{1}{x} \right) = x^{\deg g} g \left( \frac{1}{x} \right) x^{\deg h} h \left( \frac{1}{x} \right) = g^{*}(x) h^{*}(x). $$
Assume that $f$ is irreducible. Since $\deg f > 1$, we must have $f(0) \neq 0$. Write $f^{*} = gh$ and apply $^{*}$ to obtain $f = \left( f^{*} \right)^{*} = g^{*} h^{*}$. Since $f$ is irreducible and non-zero, we must have $\deg g^{*} = 0$ or $\deg h^{*} = 0$. If $\deg g^{*} = 0$ then we have
$$ 1 < \deg f =\deg g^{*} + \deg h^{*} = \deg h^{*} = \deg f^{*} = \deg g + \deg h$$
which implies that $\deg h = 0$. Similarly, if $\deg h^{*} = 0$ then $\deg g = 0$. Similarly, for the case $f^{*}$ is irreducible.