Showing a divisor is effective

Question

On $\mathbb{P}^n$, consider the prime divisor $C=Z(x_0)$. Let $f \in k[D(x_0)]$ be degree $d$. I want to show that the divisor $(f)+\nu_C(f)C$ is effective. Now, $f=g/x_0^d$ with $g \in k[x_1,\ldots,x_n]$ having degree $d$. Hence, $\nu_C(f)=\nu_C(g)-d\nu_C(x_0)=-d$ because $g$ does not vanish along $C$. Then the divisor of $f$ should be $(f)=(g)-d(x_0)=(g)-dC$, so $(f)+\nu_C(f)C=(g)-2dC$. So how can this be effective? The divisor $(g)$ is effective, but it has no zeros along $C$, so it would seem that this is not the case. Any help would be much appreciated.