Prove that the polynomial $x^nf(1/x)$ with reverted coefficients is also irreducible polynomial over $\mathbb{Q}$

abstract-algebra polynomials irreducible-polynomials

Assume $f(x)=g(x)h(x)$ with $\deg g=k$, $\deg h=m$. Then the other polynomial is $x^nf(1/x)$ and we have $x^nf(1/x)=x^kg(1/x)\cdot x^mh(1/x)$.

Related

Deriving expression for general cubic equation solution.
How to compute this integral involving a cdf?
Show that every $n$ can be written uniquely in the form $n = ab$, with $a$ square-free and $b$ a perfect square
why $(r+I)(s+I) = rs + I$ in the quotient ring $R ?$
How many fixed points in a permutation
$6^{(n+2)} + 7^{(2n+1)}$ is divisible by $43$ for $n \ge 1$
Fermat primes relation to $2^n+1$
If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $
Evaluate $\int_0^{2\pi} \frac{\sin^2\theta}{5+4\cos\theta}\,\mathrm d\theta$
Combining error terms in Simpson's rule
What is my subnet and gateway [closed]
How do factory reset my Monitor?