Solution
= Solution
Any <unit in a ring> of $R$ must also be a unit of $\mathbb Q[X]$, hence a nonzero constant $q\in\mathbb Q$. Membership of both $q$ and $q^{-1}$ in $R$ requires $q,q^{-1}\in\mathbb Z$, so
$$
\boxed{R^\times=\{1,-1\}.}
$$
= Solution
Any <unit in a ring> of $R$ must also be a unit of $\mathbb Q[X]$, hence a nonzero constant $q\in\mathbb Q$. Membership of both $q$ and $q^{-1}$ in $R$ requires $q,q^{-1}\in\mathbb Z$, so
$$
\boxed{R^\times=\{1,-1\}.}
$$