Showing the irreducibility of $x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$ in $\mathbb{Q}[x]$ I would like to show the irreducibility of  $x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$ and $x^8 - 120 x^6 + 4360 x^4 - 45600 x^2 + 15376$ in $\mathbb{Q}[x]$. In both cases Eisenstein criterion fails. I also attempted some linear changes of variables but nothing seems to work. Any help?
 A: Let $f(x)=x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$. 
$f$ has degree $8$ and assumes prime values at these $18 > 2 \cdot 8$ points and so must be irreducible: 
$$
\begin{array}{rl}
n & f(n) \\
\pm 1 & 2137 \\
\pm 3 & -4583 \\
\pm 5 & -8039 \\
\pm 7 & 1117657 \\
\pm 13 & 557943577 \\
\pm 15 & 1936431961 \\
\pm 33 & 1330287723097 \\
\pm 37 & 3360699226777 \\
\pm 55 & 82083690591961 \\
\end{array}
$$
A: Both these polynomials have irreducible remainders in $\mathcal{F}_2[x]$, field of integers modulo 2, when we try to divide each by $x^2+x+1$.
The reference is to Artin's Algebra, Proposition 12.4.3

Let $f(x) = a_n x^n + \dots + a_0$ be an integer polynomial, and let $p$ be a prime integer that does not divide the leading coefficient $a_n$. If the residue $\bar{f}$ of $f$ modulo $p$ is an irreducible element of $\mathcal{F}_p[x]$, then $f$ is an irreducible element of $\mathcal{Q}[x]$.

This means if we divide our polynomials in field $\mathcal{F}_2[x]$, and we are lucky (= this often works) to get a remainder which is an irreducible polynomial in this field, then the original polynomial is irreducible. 
In field $\mathcal{F}_2[x]$ the following polynomials are irreducible, $x^2+x+1$ and $x+1$.
If we divide the original polynomials by $x^2+x+1$, the remainders are
$$16575 + 8959 x \equiv 1+x \mod 2$$
$$60855 + 49959 x \equiv 1+x \mod 2$$
Note
If any of these two polynomials had proper divisors in $\mathcal{Q}[x]$, then it had proper divisors in $\mathcal{Z}[x]$ as well, Artin 12.3.6.
This might suggest that if any of these polynomials had proper divisors in $\mathcal{Z}[x]$, then divisors would be monic polynomials with limited choice of the last term, ie $8836=2^2 \times 47^2$. But I do not have a deep insight to pursue it further.
