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Iosif Pinelis
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If all zeros of $Q$ in $\mathbb C^2$ are zeros of $P$, then, by Hilbert's Nullstellensatz, $Q$ divides $P^r$ for some natural $r$.

If $Q$ is also irreducible, then it follows that $Q$ divides $P$.

If $Q$ is not irreducible, then $Q$ does not have to divide $P$, as noted in my comment.

Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229