Irreducibility of $x^5-10x^4+55x^3-110x^2+184x-60$ in $\mathbb{Q}[x]$ Let $g(x)$ be the above polynomial, I know that this polynomial is indeed irreducible, but I'm not entirely sure how to show it since the reduction mod $p$ test gives $g(k) \neq 0 \enspace \forall \enspace  k \in \mathbb{Z_{11}}$ and mostly $g(k) = 0$  for some $k \in \mathbb{Z_{p}}$ for primes $p=\{2,3,5,7,11\}$. Is it enough to say that it's reducible in $\mathbb{Z_{11}}$ that it is irreducible in $\mathbb{Q}[x]$? Is there any way to get to 11 without brute force calculations?
 A: If the polynomial is reducible in $\Bbb{Z}[x]$ then $g=fh$ for some nonconstant polynomials $f,h\in\Bbb{Z}[x]$ and so $\overline{g}\equiv\overline{f}\overline{h}\pmod{p}$, meaning that $\overline{g}$ is reducible in $\Bbb{F}_p[x]$. It follows that if $\overline{g}$ is irreducible in $\Bbb{F}_p[x]$ for some $p$, then it is irreducible in $\Bbb{Z}[x]$.
However, to show that $\overline{g}$ is irreducible in $\Bbb{F}_p[x]$ it does not suffice to show that $\overline{g}(k)\neq0$ for all $k\in\Bbb{F}_p$. This only shows that $\overline{g}$ has no linear factor; it might still be the product of an irreducible quadratic and cubic.
You have already checked that $g$ is reducible mod $2$, $3$, $5$ and $7$, so the smallest possible prime that can show $g$ is irreducible is $11$. You have already checked that $g$ has no linear factors, so by the argument above it remains to check that it has no quadratic factors. You can check by brute force that $g$ is not the product of an irreducible quadratic and cubic, either showing that
$$g=(x^3+ax^2+bx+c)(x^2+dx+e),$$
has no solution $a,b,c,d,e\in\Bbb{F}_{11}$, or alternatively, by showing that
$$g(a+bi)=0,$$
has no solution $a,b\in\Bbb{F}_{11}$, where $i^2=-1$.
A: The polynomial is irreducible over $\Bbb F_{11}$ by Berlekamp, or by a direct computation, i.e., $f=x^5+x^4+8x+6$ cannot be written as a quadratic polynomial times a cubic one, by considering the linear equations over $\Bbb F_{11}$ arising by comparing coefficients. This is very easy and not a "brute force" calculation.
Hence $x^5-10x^4+55x^3-110x^2+184x-60$ is irreducible over $\Bbb Q$.
