Let $f(x)$ an polinomial with degree $n$ in $\mathbb{Z} [x]$. If $\frac{a}{b}$ is a root of $f(x)$ with $(a,b)=1$, then $\forall c \in \mathbb{Z}; (bc-a) \mid f(c).$
Proof:
If $\frac{a}{b}$ is a root of $f(x) \rightarrow(x-\frac{a}{b}) \mid f(x)$.
By definition of divisibility $\exists h(x) \in \mathbb{Q}[x]$ such that
$f(x)=(x-\frac{a}{b}) h(x) $
$bf(x)=b(x-\frac{a}{b}) h(x) $
$bf(x)=(bx-a) h(x) $
By definition of divisibility, $(bx-a) \mid bf(x)$, but $(bx-a)$ is irredutible on $\mathbb{Q}[x]$ then $(bx-a) \mid b \lor (bx-a) \mid f(x) $
$(bx-a) \nmid b$ because $degree(bx-a)>degree(b)$ then $(bx-a) \mid f(x)$.
Evaluating $c$ in $f(x)$
$(bc-a) \mid f(c)$
Is the demonstration correct?
One needs to appeal to Gauss's Lemma at some stage. A primitive polynomial is a polynomial $f(x)=a_0+a_1x+\cdots+a_nx^n$ over the integers with $\gcd(a_0,a_1,\ldots,a_n)=1$. One form of Gauss's lemma states that the product of two primitive polynomials is primitive.
More generally the content of a polynomial $f(x)=a_0+a_1x+\cdots+a_nx^n$ over the integers is $c_f=\gcd(a_0,a_1,\ldots,a_n)$. Another form of Gauss's Lemma states that $c_{fg}=c_fc_g$.
In your example you have $$bf(x)=(bx-a)h(x).$$ All we know is that $h(x)$ has rational coefficients. I claim that $h(x)$ has integer coefficients, and has content divisible by $b$. We cab write $h(x)=m^{-1}h_1(x)$ with $h_1$ having integer coefficients, and $m\in\Bbb N$. Then $$mbf(x)=(bx-a)h_1(x).$$ Using Gauss's Lemma, and the fact that the content of $bx-a$ is one, we get $c_{h_1}=mb c_f$; the coefficients of $h_1$ are divisible by $mb$. So the coefficients of $h$ are integers divisible by $b$. Then $$f(x)=(bx-a)h_2(x)$$ where $h_2(x)$ has integer coefficients. For each integer $c$, then $$f(c)=(bc-a)h_2(c)$$ etc.
In your argument you in effect are saying that $f(x)=(bx-a)h_2(x)$ and therefore $f(c)$ is divisible by $bc-a$. What is lacking there is a proof that $h_2$ has integer coefficients, rather than just rational coefficients.