The polynomial $f(x) = 2x^6 + 6x^5 + 4x^4 + 5x^3 + 3x +1 $ is irreducible in $\mathbb{Z}[x]$ How can I prove that the polynomial $f(x) = 2x^6 + 6x^5 + 4x^4 + 5x^3 + 3x +1 $ is irreducible in $\mathbb{Z}[x]$?
 A: Using Cohn's irreducibility criterion (see also Corollary 1 in article ON AN IRREDUCIBILITY THEOREM OF A. COHN) if we find integer $b\geq 2$ such that $0 \leq a_k \leq b-1$ and $\sum_{k=0}^{n}a_k b^k$ is a prime, then $f(x)=\sum_{k=0}^{n}a_k x^k$ is irreducible in $\mathbb{Z}[x]$.
In this case $b=14$ satisfies conditions of the criterion since $f(14)=18453443$ is a prime and so the $f(x)$ is irreducible in $\mathbb{Z}[x]$. Not as pretty as Eisenstein but works.
A: As others pointed out we can use modular factorization. I think that a quinella of modulo two and five gives us a winning betting slip with the least amount of paper-and-pencil work.
Modulo $2$ we have 
$$
f(x)\equiv x^3+x+1
$$
which is well known to be irreducible. Like quasi (+1) we conclude that if we have a factorization $f(x)=g(x)h(x)$ in $\Bbb{Z}[x]$ then we must have $g(x)\equiv x^3+x+1$ and
$h(x)\equiv 1$. Here the leading coefficient of $h$ must be even, so $g$ must have an odd leading coefficient. Therefore $\deg g(x)=3=\deg h(x)$.
Let's look at it modulo $5$.
$$
f(x)\equiv2x^6+x^5-x^4-2x+1.
$$
We see that $f(-2)\equiv0\pmod5$ so in $\Bbb{Z}_5[x]$ $f$ is divisible by $(x+2)$. By polynomial division we get
$$
f(x)=2(x+2)(x^5+x^4-1)\in\Bbb{Z}_5[x].
$$
That degree five factor, denote it by $m(x)$, is actually irreducible in $\Bbb{Z}_5[x]$. There are many ways to see this. The simplest may be 
to observe that the reciprocal of $m(x)$ is
$$
\tilde{m}(x)=x^5m(\frac1x)=-(x^5-x-1).
$$
This falls under the umbrella of Artin-Schreier polynomials. 
A standard exercise is to show that whenever $a\not\equiv 0\pmod p$, $p$ a prime, the polynomial
$$
x^p-x-a
$$
is irreducible in $\Bbb{Z}_p[x]$. See here for a variety of arguments with varying degrees of sophistication proving this result. With $\tilde{m}(x)$ known to be irreducible we can conclude that $m(x)$ is irreducible also.
Anyway, with the knowledge that modulo five $f$ has an irreducible factor of degree five, we have run into a contradiction with the modulo two conclusion that the only possible factorization is the product of two cubics. Therefore $f$ is irreducible.
A: Using the Berlekamp algorithm for polynomials $f\in \mathbb{F}_p[x]$, we see that $f(x)$ is irreducible for $p=29, 31, 47$, etc., see Bemte's comment. Hence $f(x)$ is irreducible in $\mathbb{Z}[x]$. 
A: $f$ has degree $6$ and assumes prime values for these $13 > 2 \cdot 6$ points and so must be irreducible: 
$$
\begin{array}{rl}
n & f(n) \\
2 & 431 \\
14 & 18453443 \\
17 & 57152981 \\
21 & 196861141 \\
30 & 1607175091 \\
39 & 7588383193 \\
89 & 1027721432693 \\
105 & 2757260165941 \\
122 & 6757665621431 \\
131 & 10340492467001 \\
161 & 35484310770581 \\
182 & 73889824920731 \\
201 & 133863373175221 \\
\end{array}
$$
