Peter wrote: Let $ f(x)=x^{n}+5x^{n-1}+3$, where $ n>1$ is an integer. Prove that $ f(x)$ cannot be expressed as the product of two nonconstant polynomials with integer coefficients. First, it's an easy job to show that $ f(x)$ has no linear factor because we have a good weapon: Rational Root Theorem Let $ P(x)=p_{n}x^{n}+\cdots+p_{0}\in Z[x]$ and $ p_{n}\neq 0$. If $ r$ and $ s \neq 0$ are relatively prime integers such that $ P \left( \frac{r}{s}\right)=0$, then we have $ r \vert p_{0}$ and $ s \vert p_{n}$. If $ f(x)=x^{n}+5x^{n-1}+3$ has a linear factor in $ Z[x]$, then $ f$ has a rational root. By applying RRT, we find that $ 1$, $ -1$, $ 3$, and $ -3$ are only candidates. However, one can easily check that none of $ f(1)$, $ f(1)$, $ f(3)$ and $ f(-3)$ can be zero. This is a contradiction. Now, we establish that $ f$ has no factor in $ Z[x]$ with arbitrary degrees $ \geq 1$. Here, we present two different solutions using standard techniques: [First Solution] Assume that $ f(x)$ can be expressed as the product of two nonconstant polynomials with integer coefficients: \[ f(x)= \left( a_{k}x^{k}+\cdots+a_{0}\right)\left( b_{m}x^{m}+\cdots+b_{0}\right), \] where $ 1 \leq k,m<n$, $ k+m=n$, $ (a_{0},b_{0})=(\pm 1, \pm 3)$, and $ (a_{k},b_{m})=(\pm 1, \pm 1)$. Let $ B=\{ j \in \{0, \cdots, m\}\; \vert \; b_{j}\not\equiv 0 \; (mod \; 3) \}$. The main trick is to think of $ l=Min B$, which is well-defined because $ m \in B$. Since $ b_{0}= \pm 3$ or $ 0 \not\in B$, we get $ 1 \leq l \leq m$. Since $ b_{0}\equiv \cdots \equiv b_{l-1}\equiv 0 (mod \; 3)$, after reading the coefficient of $ x^{l}$ modulo $ 3$ in \[ x^{n}+5x^{n-1}+3=\left( a_{k}x^{k}+\cdots+a_{0}\right)\left( b_{m}x^{m}+\cdots+b_{0}\right), \] we find that the coefficient of $ x^{l}$ in $ x^{n}+5x^{n-1}+3$ is not divisible by $ 3$. Indeed, we get \[ a_{l}b_{0}+\cdots+a_{1}b_{l-1}+a_{0}b_{l}\equiv a_{0}b_{l}\not \equiv 0 (mod \; 3). \] Since, in $ x^{n}+5x^{n-1}+3$, only two terms $ x^{n}$ and $ x^{n-1}$ have the coefficient not divisible by $ 3$, we see that $ l=n$ or $ l=n-1$. It now follows that $ k=n-m \leq n-l \leq 1$. So, we get $ k=1$ and $ a_{1}=a_{k}= \pm 1$. Now, look at the polynomial factorization again: \[ x^{n}+5x^{n-1}+3=\left( a_{1}x+a_{0}\right)\left( b_{m}x^{m}+\cdots+b_{0}\right). \] Since $ a_{1}=\pm 1$ and $ a_{0}= \pm 1$, we find that $ a_{1}x+a_{0}=\pm x \pm 1$ so that $ f(x)=x^{n}+5x^{n-1}+3$ is divisible by $ x+1$ or $ x-1$. In other words, $ f(1)=0$ or $ f(-1)=0$. However, this is a contriadiction because $ f(1)=9$ and $ f(-1)=3+4(-1)^{n-1}$. This completes the proof. [Second Solution] Assume $ f(x)=F(x)G(x)$, where $ F(x), G(x) \in Z[x]$ and $ 1 \leq deg F < n$. Since $ 3=f(0)=F(0)G(0)$, we may also assume that $ (F(0), G(0))=(1,3)$ or $ (-1,-3)$. The key idea is to empoly complex numbers. By the fundamental theorem of algebra, we can write \[ F(x)=A(x-z_{1}) \cdots (x-z_{k}) \;\; \text{and}\;\; G(x)=B(x-z_{k+1}) \cdots (x-z_{n}), \] where $ z_{1}, \cdots, z_{n}\in C$ and $ A=B=\pm 1$. After looking at $ F(0)$, we find that \[ \pm 1 =F(0) = \pm A z_{1}\cdots z_{k}\;\; \text{or}\;\; \left\vert \prod_{1 \leq i \leq k}z_{i}\right\vert =1. \] The key observation is that $ z_{1}$, $ \cdots$, $ z_{k}$ are also zeros of $ f=FG$. It follows that \[ {z_{i}}^{n}+5{z_{i}}^{n-1}+3=0 \;\; \text{or}\;\;{z_{i}}^{n-1}(z_{i}+5)=-3 \] for all $ i \in \{1, \cdots, k \}$. Hence, we obtain \[ 3^{k}= \left\vert \prod_{1 \leq i \leq k}{z_{i}}^{n-1}(z_{i}+5) \right\vert = \left\vert \prod_{1 \leq i \leq k}(z_{i}+5) \right\vert =\vert F(-5) \vert. \] Now, looking at $ f(-5)$, we deduce that \[ 3 = \vert f(-5) \vert = \vert F(-5) G(-5) \vert=3^{k}\vert G(-5) \vert. \] Since $ G(-5) \in Z$, this means that $ k=1$ or $ deg F=1$. Hence, $ f(x)=x^{n}+5x^{n-1}+3$ has a linear factor $ F(x)$ with integral coefficients. Since $ F(0)=\pm 1$ and $ A=\pm 1$, we have $ F(x)= \pm x \pm 1$ so that $ F(x)$ is divisible by $ x+1$ or $ x-1$. This is impossible as in the first solution.

Rather unpretty solutions were posted above.I would like to write down another.I am deeply convinced that Extended Eisenstein criterion was already discussed here,that's why I will leave this theorem without proof. Extended Eisenstein criterion: Suppose that polynomial $ P(x)\in\mathbb{Z}[x]$ has the following form: \[ P(x) = a_nx^n + a_{n - 1}x^{n - 1} + \dots + a_1x + a_0 \] And there exist prime number $ p$,such that $ p^2\not{|} a_0$ and $ p|a_0,a_1,\dots,a_k$,but $ p\not{|} a_{k + 1},\dots,a_n$,then $ P(x)$ has an irreducible factor of a degree greater than $ k$.As a particular case if we take $ k = n - 1$ then $ P$ is irreducible. Let's return to the initial problem. By the above mentioned theorem,$ f$ has an irreducible factor of a degree $ n - 1$,therefore,it is enough to show that $ f$ has no integer roots. Suppose to the contrary that $ f$ has an integer root,say $ r$.Then $ r^{n} + 5r^{n - 1} + 3 = 0$. This equation implies that $ |r|$ is either $ 3$ or $ 1$.It is not hard to check the case $ |r| = 1$ and to observe that if $ n > 2$ then $ 9|3$ what is impossible. It means that either $ n = 2$ and $ |r| = 3$ or $ f$ has no roots. But equation $ r^2 + 5r + 3$ has no integer roots,contradiction. $ \blacksquare$.

Peter wrote: Let $f(x)=x^{n}+5x^{n-1}+3$, where $n>1$ is an integer. Prove that $f(x)$ cannot be expressed as the product of two nonconstant polynomials with integer coefficients. Remark: Follows directly by Perron's Criterion .