I've seen this exercise in the book "Lectures on curves, surfaces and projective varieties" by Beltrametti and others:
Show that three polynomials $\phi_2(x_1,x_2,x_3), \phi_3(x_1,x_2,x_3), \phi_4(x_1,x_2,x_3)$ homogeneous of degree $deg(\phi_i)=i$ can be chosen in a way such that $\phi_2 \phi_4 - \phi_3^2$ is the product of 6 linear forms.
Under the statement the outhors put also their solution, but there is a passage that I can't really get. They say: in the projective complex plane, take an irreducible conic of equation $\phi_2=0$ and a cubic of equation $\phi_3=0$ that instersects the conic in exactly six distinct points $P_1,..., P_6$. Consider the lines $l_i$ tangent to the conic at $P_i$. Then, from the linear system of curves of equation $\phi_3^2- \lambda l_1...l_6$ take the curve $C$ that intersect the conic in an other point $Q$ distinct from all the $P_i$. Then, $C$ has the conic as irreducible component, so the thesis follows.
My question is: why does $C$ split into the conic and an other curve of degree four? I thought I could use Bézout theorem, but it seems to me that the conic and $C$ have only 7 points in common, and I don't really see how one can compute the intersection multiplicity of these two curves in the points in which tey intersect.
Any help would be appreciated.