Let us prove it by double inclusion.
First of all, 3x∈IJ,2x∈IJ so 3x−2x=x∈IJ. Also, 3⋅2=6∈IJ, so we have <x,6>⊆IJ.
Now, let us prove the other inclusion. An element of I is of a form x⋅P(x)+2a0 and an element of J is of a form x⋅Q(x)+3b0. Therefore, an element of IJ is of a form x⋅(xP(x)Q(x)+2a0Q(x)+3b0P(x))+6a0b0. We see, that it is an element of <x,6>, so we conclude that
IJ=<x,6>
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