Why can't $x^k+5x^{k-1}+3$ be factored? I have a polynomial $P(x)=x^k+5x^{k-1}+3$, where $k\in\mathbb{Z}$ and $k>1$. Now I have to show that you can't factor $P(x)$ into two polynomials with degree $\ge1$ and only integer coefficients. How can I show this?  
 A: Assume $k\ge2$.
By the rational root theorem, only $\pm1$ and $\pm3$ are candidates for rational roots - and by inspection are not roots. Therefore,
if $P(x)=Q(x)R(x)$ with $q:=\deg Q>0$, $r:=\deg R>0$, we conclude that $q\ge2$ and $r\ge 2$. Modulo $3$ we have $(x-1)x^{k-1}=x^k-x^{k-1}$, hence wlog. $Q(x)\equiv (x-1)x^{q-1}\pmod 3$, $R(x)\equiv x^r\pmod 3$.This implies that the constant terms of both $Q$ ans $R$ are multiples of $3$, hence the constant term of $P$ would have to be a multiple of $9$ - contradiction
A: I will mimic the proof of Perron's irreducibility criterion which says that if $f(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0\in\mathbb Z[x]$ and $|a_{n-1}|\gt 1+|a_0|+|a_1|+\ldots+|a_{n-2}|$ then $f$ is irreducible.
Do you know Rouché's theorem? 
If $|x|=1\Longrightarrow|5x^{k-1}|>|x^k+3|$. Therefore $P(x)$ has exactly $k-1$ roots inside the unit circle.
Suppose now that $P$ is reducible with $P(x)=Q(x)R(x)$ and $|Q(0)|=1, |R(0)|=3$. All the roots of $Q$ or $R$ are inside the unit circle and the product of those roots is $Q(0)$ or $R(0)$ respectively. Contradiction. 
