I have the following problem: Prove that $P(x)=x^k+5x^{k-1}+3$ with an arbitrary integer $k>1$ cannot be factored into two polynomials $Q$ and $R$ with integer coefficients and degree at least 1, i.e. so that $P(x)=Q(x)\cdot R(x) \forall x \in \mathbb{R}$.
My idea is as follows: I know about Eisenstein's criterion whose conclusion is exactly what we want to prove. Sadly, we cannot use it right away. But I read that you could also apply it if you look at the polynomial modulo an arbitrary integer (if there was a factorisation before, then that one still holds modulo that integer). So I tried looking at $P(x)$ modulo 5 and the outcome is the polynomial $x^k+3$. If we then choose $p=3$ to apply Eisenstein's criterion, everything works perfectly and by the aforementioned criterion, there are no $Q$ and $R$ as mentioned in the problem modulo 5, hence not in the whole integers. Is this right?
Similarly, is it right to consider $x^3-2x^2+x-8$ modulo 5 again to get $x^3-2x^2-4x+2$ and then choose $p=2$ and apply Eisenstein again to conclude that $x^3-2x^2-4x+2$ and hence $x^3-2x^2+x-8$ cannot be factored either?
That whole area is new to me so advice would be much approved.
Use this extension of the Eisenstein's Criterion (unfortunately I have not found a reference, however the proof is more or less the same of the original Eisenstein's Criterion where $k=n$).
Let $P(x)=\sum_{k=0}^n a_kx^k\in\mathbb{Z}[x]$. If there exist a prime number $p$ and an integer $1\leq k\leq n$, such that $$p\mid a_0,a_1,\dots,a_{k-1},\quad p\nmid a_{k}\quad\mbox{and}\quad p^2 \nmid a_0,$$ then $P$ has an irreducible factor of degree $d\geq k$.
For $P(x)=x^n+5x^{n-1}+3$ with $n>1$, take $p=3$ and $k=n-1$ then by the above criterion $P$ has an irreducible factor of degree $d\geq n-1$. The case $d=n-1$ is impossible because $P$ has no factors in $\mathbb{Z}[x]$ of degree $1$ ($P$ ha no integer roots). Hence $d=n$ and $P$ is irreducible in $\mathbb{Z}[x]$.