# Irreducible polynomial proof check

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 $$f(-2)<0 we can conclude that $$\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 $$\mathbb Q[x]$$.

• Must a reducible polynomial have a linear factor? E.g. is $(x^2 + 1)(x^2 + 2)$ reducible in $\mathbb{Q}$? Apr 1, 2020 at 3:23

This proof is not correct since it only establishes that $$x^n+x+3$$ has no root in $$\mathbb{Z}$$, which is equivalent to proving it has no linear factor (a factor of degree $$1$$). In order to prove irreducibility, you need to rule out also factors of higher degrees $$2,3,\dots,\lfloor n/2 \rfloor$$. For example, try your approach on $$x^5+x+1$$. Clearly derivative $$5x^4+1$$ is positive so it can have only one real root as well, yet $$x^5+x+1=(x^2+x+1)(x^3-x^2+1).$$
By the way, you can show non-existence of rational root by using rational root theorem. If $$f(x)=x^n+x+3$$ had a rational root $$p/q$$, then $$p \mid 3$$ and $$q\mid 1$$, so $$p/q \in \{\pm 1,\pm 3\}$$. Since clearly none of the $$f(-1),f(1),f(3),f(-3)$$ can be zero , we are done.
Just for completeness, see correct proofs in Show that $x^n + x + 3$ is irreducible for all $n \geq 2.$