= Solution
<Coprimality> gives <polynomials> $a,b\in\mathbb Q[x]$ with $ag+bh=1$, so the <ideals> $(g)$ and $(h)$ are comaximal. Also
$$
(g)\cap(h)=(gh).
$$
Indeed, if $u=gr=hs$, multiplying $ag+bh=1$ by $u$ writes $u=aghs+bhgr$, a multiple of $gh$; the reverse inclusion is immediate. Part (b), with $f=gh$, gives
$$
\boxed{\mathbb Q[x]/(f)\cong\mathbb Q[x]/(g)\times\mathbb Q[x]/(h).}
$$
The explicit <isomorphism> takes the class of a <polynomial> $r$ to its classes modulo $g$ and $h$.
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