Divisible module which is not injective

Consider two $\mathbb Z[x]$-modules. The first is $Y = \mathbb Z[x]$ itself, and the second is the ideal $X = (2, x) \subseteq \mathbb Z[x]$. There's a natural injection $f: X \to Y$.

Now, define a $\mathbb Z[x]$-homomorphism $g: X \to \mathbb{Q}(x)/\mathbb{Z}[x]$ by: $$ \begin{array}{rcl} g(2) & = & [0],\\ g(x) & = & [1/2]. \end{array} $$

If $\mathbb{Q}(x)/\mathbb{Z}[x]$ were injective, then there would exist a $\mathbb Z[x]$-homomorphism $h: Y \to \mathbb{Q}(x)/\mathbb{Z}[x]$ such that $g = h \circ f$. Then we have: $$ \begin{array}{l} [0] = g(2) = h(f(2)) = h(2) = 2h(1), \\ [1/2] = g(x) = h(f(x)) = h(x) = xh(1).\end{array} $$ So, $h(1) \in \mathbb{Q}(x)/\mathbb{Z}[x]$ has the property that $2h(1)=[0]$ and $xh(1)=[1/2]$. It can be checked that there's no such element in $\mathbb{Q}(x)/\mathbb{Z}[x]$. This is a contradiction, so $\mathbb{Q}(x)/\mathbb{Z}[x]$ is not an injective $\mathbb{Z}[x]$-module.

Please, check this carefully, because I've already made a couple of stupid mistakes today.