I want to prove that
$f = x^5 - 6x^3 +2x^2 - 4x +5$ is irreducible in $\mathbb Q[x]$.
Since it is primitive, it is irreducible over $\mathbb Q$ iff it is irreducible over $\mathbb Z$.
Eisensteins criterion obviously doesn't work and reduction to $\mathbb Z/p \mathbb Z$ didn't work for me either.
So my attempt is to assume it factorized as two polynomials $f= gh$ with $\deg f, g \geq 1$ and solve a system of equations. I'm sure that leads to a solution/contradiction (does it?) but in algebra there always is a short or more neat solution, so I would appreciate any hint. I am only allowed to use the most basic criterions.
First note that $f$ has no rational roots by the Rational Root Test: the possible candidates are $\pm 1$ and $\pm 5$ and none of those is a root. Thus the only possible factorization is $f = gh$ where $\deg(g) = 3$ and $\deg(h) = 2$.
Now observe that mod $2$, the factorization of $\overline{f}$ into irreducibles is $$ \overline{f} = (x+1)(x^4 + x^3 + x^2 + x + 1) \, . $$ (To see that the quartic factor is irreducible: first, note that it has no roots and then check that $x^2 + x + 1$, the only degree $2$ irreducible mod $2$, is not a factor.) This shows that the factorization above into a cubic and a quadratic is impossible, since the polynomial could only factor further mod $2$.
Addendum: Since the reduction mod $p$ map is a homomorphism, then we have $\overline{f} = \overline{g} \overline{h}$, where $\overline{f}$ denotes its reduction mod $2$. An irreducible factor $\pi$ of $\overline{f}$ must be an irreducible factor of $\overline{g}$ or $\overline{h}$, and since $\overline{g}$ and $\overline{h}$ have degrees $\leq 3$, then $\pi$ must have degree $\leq 3$, too. But we have found an irreducible factor with degree $4$, which is a contradiction. Thus the factorization $f = gh$ is impossible, so $f$ is irreducible.