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.
Best Answer
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.