Proving $f(x)=x^4+4x^3+3x^2+7x-4$ is irreducible

Question

Here's my attempt, I'm almost there but I'm stuck:

Using a hint, I wrote the modular reduction:

Reducing the coefficients modulo $2$ gives: $\left [ f \right ]_2=x^4+x^2+x=x(x^3+x+1)$.

Reducing the coefficients modulo $3$ gives: $\left [ f \right ]_3=x^4+x^3+x-1$

If $f$ were reducible then either it decomposes as $(x-x_0)g(x)$ (in which case $x_0$ is a root of $f$, therefore $\left [ x_0 \right ]_3$ is a root of $\left [ f \right ]_3$ but it is easy to verify $\left [ f \right ]_3$ has not roots so this is impossible) or it decomposes as $p(x)q(x)$, in which case $\left [ f \right ]_2=\left [ p \right ]_2\left [ q \right ]_2$. But I don't know how to find a contradiction here.

Thanks in advance.

Answer 1

Hint: Write out what it means for the product of two $2$nd degree polynomials to have product equal to $f$. Get an system of equations in the coefficients. Check whether the system is consistent.

Answer 2

The polynomial is irreducible modulo $17$. This is (a little bit) easier than to check over the integers. First, there is no root modulo $17$, and then writing the polynomial as $(x^2+ax+b)(x^2+cx+d)$ quickly gives a contradiction modulo $17$.

Substituting $x$ by $x-1$ we obtain $f=x^4-3x^2+9x-11$, which is a bit easier to handle. First we have $c=-a$ and then $bd=-11=6$. This means, we have $(b, d)=(1,6),(2,3), (3,2), (4,10), (5,8), (6,1), (7,13), (8,5),\ldots $, which is enough to solve the last two equations for $a$.

Answer 3

HINT.-You have $$f(1)=11\\f(-1)=-11\\f(3)=233\\f(5)=1231\\f(-9)=3821\\f(-17)=64613\\f(25)=455171$$ Then there is seven integers $n=1,-1,3,5,-9,-17,25$ such that $f(n)$ is prime. This gives $3$ integers more than the degree of the polynomial.

Note that just one prime $f(n)$ gives a strong enough probability that $f(x)$ be irreducible.

If I remember correctly a result of mine established that if there are two integers more than the degree of the polynomial giving $ f (n) $ prime then $ f (x) $ is irreducible. It could be that instead of $ 2 $ it has been $ 3 $ but that is exhibited here, with $ 3 $ more integers. If you want, take this just as a comment.