$f(x)=x^n+5x^{n-1}+3$ can't be expressed as the product of two polynomials 
Let $f(x)=x^n+5x^{n-1}+3$ where $n\geq1$ is an integer. Prove that $f(x)$ can't be expressed as the product of two polynomials each of which has all its coefficients integers and degree $\geq1$.

If the condition that each polynomial must have all its coefficients integers was not there, then I needed to show only that $f(x)$ is irreducible over real numbers. But since this condition is given,therefore, we can't exclude the case when $f(x)$ is reducible over real numbers but not with polynomials of integer coefficients. Anyone with idea how to proceed? Thanks in advance!!
 A: Hint $\ $ This is a minor variant of Eisenstein's criterion. Mod $3$ it factors as $\rm\:x^{n-1}(x+5)\:$ so by uniqueness of factorization if $\rm\:f = gh\:$ then, mod $3,\,$ $\rm\:g = x^j,\, $ $\rm\,h = x^k(x+5),\,$ $\rm\:j\!+\!k = n\!-\!1.\,$ But not $\rm\,j,k > 0\,$ else $\rm\:3\:|\:g(0),h(0)\:$ $\Rightarrow$ $\rm\:9\:|\:g(0)h(0)=f(0).\:$ Hence either $\rm\:j=0,\:$ so $\rm\:f\:$ is irreducible, or $\rm\:k=0,\:$ and $\rm\:f\:$ has a linear factor $\rm\:g,\:$ which is easily ruled out. $\ \ $ QED
A: The notation $[x^t]P(x)$ denotes the coefficient of $x^t$ in polynomial $P(x)$.
Suppose that $f(x)=g(x)h(x)$, where $g(x),h(x)$ are monic polynomials, $g(x),h(x)\in\Bbb Z[x]$ and $\deg g>0$, $\deg h>0$, $\deg g+\deg h=n$. Let $g(x)=\sum_k\alpha_kx^k$, $h(x)=\sum_k\beta_kx^k$, where $\alpha_k,\beta_k$ is integer whenever $k\ge0$.
First, we have $\alpha_0\beta_0=3$, so $|\alpha_0|=3$ and $|\beta_0|=1$ or $|\alpha_0|=1$ and $|\beta_0|=3$. WLOG, suppose that $|\alpha_0|=3$ and $|\beta_0|=1$. Let $m$ is the smallest positive integer such that $3\nmid\alpha_m$ (such $m$ exists because $g(x)$ is monic). Now we have
$$[z^m]f(x)=\alpha_0\beta_m+\alpha_1\beta_{m-1}+\cdots+\alpha_m\beta_0\equiv\alpha_m\beta_0\not\equiv0\pmod3$$
So $m\ge n-1$, and $\deg g\ge n-1$, thus $\deg h\le 1$, hence $\deg h=1$, and $h(x)=x+\beta_0$, where $\beta_0=\pm1$, so $h(\beta_0)=0$, and $f(\beta_0)=0$, but it's obvious that $f(\pm1)\neq0$, Q.E.D.
