Show that $f = x^5 + 5x^3 + 1 \in \mathbb Q[x]$ is irreducible My solution: Because of Gauss Lemma, it is enough to show irreducibility over $\mathbb Z[x]$. The only possible integers roots are divisors of $1$, so $\pm 1$, but $f(1) \ne 0$ and $f(-1) \ne 0$, so $f$ is irreducible. 
That's my solution. Now I looked in the official solution, which is more involved. So I wanted to ask if my solution has some faults?
For completeness, the official solution: Because of Gauss Lemma, it is enough to show irredubility in $\mathbb Z[x]$. We do it with reduction modulo 2, then $\overline{f} = x^5 + x^3 + 1 \in \mathbb Z_2[x]$. Is $\overline{f}$ irreducible over $\mathbb Z_2$, so is $f$ irreducible over $\mathbb Z$. We have to show that no irreducible polynomial of degree $\le 2$ divides $\overline{f}$. Because $\overline{f}$ has no root in $\mathbb Z_2$, it has no linear factor. And the only irreducible polynomial over $\mathbb Z_2$ is $x^2 + x + 1$. Polynomial division shows that this is no divisor of $\overline{f}$. It is $\overline{f} = (x^3 + x^2 + x)(x^2 + x + 1) + x + 1$.

A polynomial in $\mathbb{Q}[X]$ may have no rational roots, but that does only mean that it has no linear factors (i.e. degree $1$) in $\mathbb{Q}[X]$. You forgot to test for factorizations into higher-degree polynomials. One of those would need to have degree not greater than $2$, and that is what the official proof checked.
