I was working on this problem, but I don't see how I can solve it. I was given a hint, but I don't know how to use it. Can anyone help me? Thanks in advance!
Let $f \in \Bbb C[x_0, x_1, x_2]$ be an irreducible homogeneous polynomial of degree 4. Show that $V (f) \in \mathbb{CP}^2$ can have a maximum of 3 multiple points. Hint: One possible strategy is to use a proof by contradiction to provide a counterexample to the theorem of Bézout.
If three of the multiple points are collinear, then the line through these three multiple points has intersection multiplicity at least 6 with the quartic, contradicting Bezout. If no three of these multiple points are collinear, then we can take a conic through the four points with a specified tangent direction at one of them - making that tangent direction the same as one of the tangent directions of the quartic at the point, we have a total intersection multiplicity of the conic and quartic of at least $3+2+2+2=9$, so again we have a contradiction to Bezout.
Alternatively, the genus of the normalization of the curve is non-negative and equal to $$\frac{1}{2}(4-1)(4-2)-\sum \frac12 \mu_P(\mu_P-1)=3-\sum \frac12 \mu_P(\mu_P-1)$$ where the sum is taken over all multiple points (including infinitely near points) and $\mu_P$ is the multiplicity of that point. If the multiplicity is greater than 1, then $\frac12 \mu_P(\mu_P-1) > 0$, and therefore there may be at most three multiple points.