Irreducibility of a particular polynomial I've got this problem for my homework: find out whether the polynomial
$$f(x)=x(x-1)(x-2)(x-3)(x-4) - a$$
is irreducible over the rationals, where $a$ is integer which is congruent to $3$ modulo $5$. 
It is easy to verify that $f(x)$ has no integer zeros (and no rational zeros too) and, because $f(x)$ is primitive, it is irreducible over $\mathbb{Z}$ iff it is irreducible over $\mathbb{Q}$. It is also clear to me that, since $f(x)$ has no integer zeros, only way to factorize $f(x)$ over $\mathbb{Z}$ is $(x^3 + bx^2 + cx +d)(x^2 + ex +f)$, i.e. as a product of one irreducible cubic polynomial and one irreducible quadratic polynomial, both primitive. I've got pretty ugly system of $5$ equations with $5$ variables. So I decided to put the whole story in $(\mathbb{Z}/5\mathbb{Z})[x]$. Then I've got the polynomial $g(x) = x^5 + 4x +2$. Since I firmly believe that $f(x)$ is irreducible over $\mathbb{Z}$, it means that I must check whether the $g(x)$ is irreducible over $\mathbb{Z}/5\mathbb{Z}$. Is there any other way than brute force?

Note that $f(x)=-f(4-x)$ and since the factorization $f(x)=q(x)r(x)$ is unique, we conclude $q(x)=q(4-x)$ for the quadratic, $r(x)=-r(4-x)$ for the cubic factor.
Thus $e=-4$. Then from $e+b=-10$, we have $b=-6$.
Then from $f+eb+c=35$, we have $f+c = -5$. And from $fb+ec+d=-50$, we have $-6f-4c+d=-50$. This implies $d\equiv 2f\pmod 5$ and together with $df\equiv -a\equiv 2\pmod 5$: $f\equiv \pm1\pmod 5$, hence $fc\equiv -1\pmod 5$ so that from $fc+ed=24$ we have $ed\equiv 0\pmod 5$, contradiction.
