As commented below, I had made a false claim. So here is a modification. Let $f(x)=4x^2+9x+4$. The roots are $(1/8)(-9\pm \sqrt{17})$. Note that $\mathbb{Q}(\sqrt{17})$ has class number 1. If we factor $2=p p'$ then the roots looks like $p^2/p'^2, p'^2/p^2$, up to units. In particular, the extensions of the maps $\varphi_{\alpha}$ treat $x$ as an element of valuation $2$. So, any element in the kernel of the extended map (much less a purported generator) must have constant term which has valuation $2$. Thus the constant term is divisible by $4$.
This means that the constant term of $\prod_{\nu}f_{\nu}(x)$ is divisible by $16$, and thus cannot (even up to units) equal $f(x)$.