= Polar bound for singular points of a plane curve
{title2=$s\leq d(d-1)/2$}
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 $2s\leq d(d-1)$.
For an irreducible curve with $d>1$, one can improve this to $s\leq d(d-1)/2-1$. 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 $d_id_j$ intersections per component pair. The sum is $d(d-1)/2$ because $\sum d_i=d$.
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