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If $d=1$, $C$ is a line and has no singular points. Suppose $d>1$ and take $P$ to be an irreducible polynomial. First, $P_x$ is not the zero polynomial: otherwise $P\in \mathbb{C}[y,z]$, and a homogeneous polynomial in two variables of degree greater than one factors into linear factors over $\mathbb{C}$, contrary to irreducibility.

Let $D=V(P_x)$ be the polar curve in this coordinate direction. Since $\mathrm{deg}{P}_x=d-1<d$, the irreducible polynomial $P$ cannot divide $P_x$. Thus $C,D$ have no common irreducible component, and Bézout's theorem applies. One may count repeated factors of $P_x$ with their multiplicities; alternatively, passing to its reduced equation only decreases $\mathrm{deg}D$.

Every singular point $p$ of $C$ satisfies $P_x(p)=0$ and has $I_p(C,D)\ge 2$ by part (a). Consequently, if $s$ is the number of distinct singular points,This argument also proves that the singular locus is finite.

There is a nonsingular point of an algebraic curve on $C$, since part (b) makes its singular locus finite whereas a projective curve has infinitely many complex points. Apply a projective transformation taking one such point to $p=[1:0:0]$, and use $P$ for the transformed homogeneous polynomial. As in part (b), $P_x$ is nonzero and $D=V(P_x)$ has no common irreducible component with $C$.

The Euler theorem for homogeneous functions and its differentiated identities giveandSince $p$ is nonsingular on $C$, at least one of $P_y(p),P_z(p)$ is nonzero. Because $d>1$, the polar curve $D$ is also nonsingular at $p$, and these identities show that $T_{p}D=T_{p}C$. Hence $I_p(C,D)\ge 2$.

This additional intersection is separate from all the singular points of $C$. Thus Bézout's theorem givesThe improvement comes from making the polar curve tangent at a chosen nonsingular point, thereby reserving at least two of the available intersections.

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 $\sum d_i=d$.