Cubic and Conic curves has an linear factor in common?

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Let $V(f)$ be a cubic projective curve and a $V(g)$ be a conic projective curve, with $f,g \in \mathbb{C}[X,Y,Z]$. If $V(f)\cap V(g)$ has more than six points, then $f$ and $g$ has a linear factor in common?

A friend of mine said that it's false, but I couldn't find any conterexample for this. Just need one conterexample, can you help me? Thanks.

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Your friend is absolutely correct - stated in that generality, the claim is false, and trivially so.

Take any irreducible conic $C$ and any line $L$, and let the cubic be $C \cup L$. Obviously the conic and the cubic have all of $C$ in common (infinitely many points!) but they don't have a line in common, since $C$ was chosen irreducible.

Bezout's theorem has some conditions under which it holds; obviously, my counter-example does not satisfy those conditions. For the claim to be true, it has to explicitly exclude extreme cases, such as my counter-example.

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Hint: what does Bézout's theorem say, and what are its assumptions?