Let $R$ be UFD. Given the polynomial $f(x)=a_0+a_1x+\dots+a_mx^m$ in $R[x]$, then the content of $f(x)$ is defined to be gcd of $a_0,a_1,\dots,a_n$. It is uniqie within units of $R$. We shall denote the content of $f(x)$ by $c(f)$. A polynomial in $R[x]$ is said to be primitive if its content is 1 (that is, is a unit in $R$). Given any polunomial $f(x)\in R[x]$, we can write $f(x)=af_1(x)$ where $a=c(f)$ and where $f_1(x)\in R[x]$ is primitive (it is not so difficult to prove!).
Except for multiplication by units of $R$ this decomposition of $f(x)$, as an element of $R$ by a primitive polynomial in $R[x]$, is unique!
More presicely, the last paragraph is equivalent to the following:
Claim: Let $R$ - UFD and $f(x)=a_0+a_1x+\dots+a_mx^m$ in $R[x]$.
If $f(x)=ap(x)$ and $f(x)=bq(x)$ where $a,b\in R$ and $p(x),q(x)$ - primitive then $a\sim b$, $p(x)\sim q(x)$.
I proved this when $f(x)\neq 0$ and the proof is not so difficult.
But I guess that this is incorrect if $f(x)=0$. For example, let's take $R=\mathbb{Z}$ and consider $f(x)=0$ with $a=b=0, p(x)=x+1, q(x)=x^2+1$. We see that $a\sim b$ but $p(x)\nsim q(x)$.
Am I right?
P.S. This excerpt from Herstein's book (page 163).