Polar bound for singular points of a plane curve ($s\le d(d-1)/2$)

For a complex reduced plane curve of degree $d$, the number $s$ of distinct singular points is at most $d(d-1)/2$. For an irreducible polynomial $F$ of degree $d>1$, $F_x$ is nonzero, since a binary homogeneous polynomial over $\mathbb{C}$ factors into linear factors. It has no common factor with $F$. Each singular point contributes at least two to the intersection with the polar curve $F_x=0$, so Bézout's theorem gives $2s\le d(d-1)$.

For an irreducible curve with $d>1$, one can improve this to $s\le d(d-1)/2-1$. Place a nonsingular point at $p=[1:0:0]$. The Euler theorem for homogeneous functions gives $F_x(p)=0$ and $\nabla F_x(p)=(d-1)\nabla F(p)$, so the polar curve is nonsingular and tangent to the curve at $p$. This consumes at least two further units of intersection multiplicity. For a reducible curve with component degrees $d_i$, add the irreducible bounds and at most $d_i{d}_j$ intersections per component pair. The sum is $d(d-1)/2$ because $\sum d_i=d$.