Irreducibility of $x^5 - 6x^3 +2x^2 - 4x +5$ in $\mathbb Q[x]$ 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.
 A: COMMENT.-There is no linear factor. If $f(x)=x^5 - 6x^3 +2x^2 - 4x +5$ and $f$ is reducible then $f(x)=g(x)h(x)$ with, say, degree of $g$ and $h$ respectivement equal to $3$ and $2$. It follows $f(n)$ is prime iff $g(n)$ or $h(n)$ is equal to $\pm1$. We can expose nine prime values for $f(n)$
$$f(0)=5\\f(1)=-2\\f(-2)=37\\f(2)=-11\\f(-4)=-587\\f(4)=661\\
f(-5)=6379\\f(-8)=-29531\\f(12)=2387\\$$ What else?
A: 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.
A: The polynomial is irreducible modulo $19$. So this does work. But I would also use an attempt to write it as $f(x)=g(x)h(x)$ and solve it modulo $19$. By the rational root test, there are no linear factors.
A: Hint: use the rarional root test: give you the potential solutions. Verify that these (4) candidates are not solutions.
https://en.wikipedia.org/wiki/Rational_root_theorem
