Show that $f=x^n+x+3$ is irreducible in $\mathbb Q[x]$ for all $n\geq 2$.
I am aware that this question has already been posted to the website but I just want to know if my attempt at proving it is correct. Here is my proof:
First suppose that $n$ is odd. Then $f$ certainly has at least one real root. Since $f'(x)$ is strictly positive for all $x\in\mathbb{R}$ this means $f$ is a strictly increasing function. Therefore it has exactly one real root $\alpha$. This root has multiplicity $1$ since $f'(\alpha)\neq 0$. We can write $f(x)=(x-\alpha)g(x)$ where $g$ is an irreducible element of $\mathbb{R}[x]$. In order to show $f$ is irreducible in $\mathbb{Q}[x]$, which by Gauss's lemma is equivalent to $f$ being irreducible in $\mathbb Z[x]$, it is enough to show that $\alpha \not\in \mathbb Z$. Since $f(-2)<0<f(-1)$ we can conclude that $\alpha \in (-2,-1)$ so $f$ is irreducible.
Now suppose $n$ is even. Since $f'=nx^{n-1}+1$ has exactly one root at $\beta= -(\sqrt[n-1]{n})^{-1}$ we can see that $f(\beta)$ is the global minimum of $f(x)$. Since $-1<\beta<0$ we have $f(\beta)=\beta^n+\beta+3>0-1+3=2$ so $f$ has no real roots which means it is irreducible in $\mathbb R[x]$ and therefore irreducible in $\mathbb Q[x]$.
