If $p_1,p_2\in \mathbb C[x,y]$ have no common factors then $(p_1)\cap (p_2)=(p_1\cdot p_2).$ For any $q\in \mathbb C[x,y]$ we also have $(p_1p_2,q)\subset (p_1,q)\cap (p_2,q),$ but is it true that $(p_1,q)\cap (p_2,q)\subset (p_1p_2,q)$?
Again $p_1,p_2$ have no common factors and we could assume additionally that $(p_i,q)$ is radical for $i=1,2$, if that helps. On the level of schemes it means that the union of intersections of schemes $(V(p_1)\cap V(q))\cup (V(p_2)\cap V(q))$ is $(V(p_1)\cup V(p_2))\cap V(q)$ so it seems "obvious", but I don't see a solid algebraic argument for it.
What about just taking $p_1=x$, $p_2=y$ and $q=x-y$. Then $(p_i,q)=(x,y)$, so this is also the intersection, but it is strictly larger than $(p_1p_2,q)=(xy,x-y)$.