Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible. This is what I'm trying to prove:

Let $f(x)\in\mathbb{Q}[x]$ and $\deg(f(x))>1$. Prove that $f(x)$ is irreducible in $\mathbb{Q}[x]$ iff its reciprocal polynomial $f^*(x)$ is irreducible in $\mathbb{Q}[x]$. 

Note: The reciprocal polynomial $f^*(x)=x^nf(1/x) \in \mathbb{Q}[x]$, where $n=\deg f$. 
So my thoughts are to prove the contrapositive, i.e. $f^*(x)$ is reducible iff $f(x)$ is reducible. 
So to prove the ($\Rightarrow$) direction I assume that $f^*(x)=g(x)h(x)$, so I get that $g(x)h(x)=x^nf(1/x)$ which implies $f(1/x)=(1/x^n)g(x)h(x)$. I want to say this somehow makes $f(x)$ reducible but I am unable to proceed as $f(1/x)\notin \mathbb{Q}[x] $. I thought about substituting $y=1/x$, which gives $f(y)=y^ng(1/y)h(1/y)$, but I am unable to show that this is irreducible in $\mathbb{Q}[y]$. If I could, then it would be irreducible in $\mathbb{Q}[x]$ since $\mathbb{Q}[y]\cong \mathbb{Q}[x]$ 
Any suggestions or hints will be appreciated.
 A: 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.
A: 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.
