If $R$ is a UFD, $p(x)\in R[x]$ and $a/b$ is a root of $p(x)$ in the fraction field, then we have $p(x)=(bx-a)q(x)$ for $q(x)\in R[x]$. Suppose $R$ is a UFD and $p(x)\in R[x]$ a polynomial of degree $\ge 1$.  Suppose $\frac ab$ is in  the fraction field $K$ of $R$, with $a$ and $b\in R$ and $\text{gcd}(a,b)=1$, and such it is a root of $p(x)$, so $p(\frac ab)=0$. 
Question: Is it true that then we have $p(x)=(bx-a)q(x)$ with $q(x)\in R[x]$?
It is clear that $q(x)\in K[X]$ exists and it is unique. Since the highest degree coefficient of $p(X)$ is divisible by $b$ and the constant term is divisible by $a$ (e.g. the rational root theorem ) one can easily show that the highest degree coefficient and the constant term of $q(x)$ are in $R$. This shows the case $p(x)$ has degree 2. 
A suspect it should a consequence of the so called Gauss Lemma, but I was not able to find a convincing argument. 
 A: Note: $\,p(x) = (bx\!-\!a)q(x) = (bx-a)\frac{c}d\bar q(x)\,$ by Factor Theorem in $K[x],\,$ $\color{#0a0}{{\rm primitive} \ \bar q}\in R[x]$ 
thus $\, d p(x) = (bx\!-\!a)\, c\,\color{#0a0}{\bar q(x)},\,$ so taking $C = $ content of this
$\Rightarrow\,  d\,C(p(x))\  =\  1\cdot c\cdot\color{#0a0} 1,\, $ i.e. $\,d\mid c\,$ in $\,R,\,$ so $\,q = \frac{c}d \bar q\in R[x].\ \ $ QED

Or by nonmonic Division: $\, b^k p(x) = (bx\!-\!a) q(x) + r,\ r\in R,\,$ so $\,r = 0\,$ by eval at $\,x = a/b.\,$ Note $\color{#c00}{(b,bx\!-\!a)}=(b,a)=\color{#c00}1,\,$ so $\,\color{#c00}{b\mid (bx\!-\!a)}q(x)\,\overset{\rm Euclid}\Longrightarrow\,b\mid q(x),\,$ thus $\,b^k\mid q(x)\,$ by induction [or we can use localization or the AC method as described in the link].
A: By the Factor Theorem, since $f(a/b)=0$, we have
$$f(x)=\left(x-{\small{\frac{a}{b}}}\right)v(x)$$
for some $v\in K[x]$.

Letting $g(x)=v(x)/b$, we have 
$$f(x)=(bx-a)g(x)$$
where $g\in K[x]$.

Our goal is to show $g\in R[x]$.

Letting $d\in R$ be the least common multiple of the denominators of the fractionally reduced coefficients of $g$, it follows that 
$$g(x)=G(x)/d$$
where $G\in R[x]$ is primitive.

Then from $f(x)=(bx-a)g(x)$, we get $d{\,\cdot}f(x)=(bx-a)G(x)$.

By Gauss' lemma, since $bx-a$ and $G$ are primitive in $R[x]$, the product $(bx-a)G(x)$ is primitive in $R[x]$.

Hence $d{\,\cdot}f(x)$ is primitive in $R[x]$, so $d$ is a unit of $R$.

Finally, since $G\in R[x]$ and $d$ is a unit of $R$, the equation $g(x)=G(x)/d$ implies $g\in R[x]$, as was to be shown.
