Let $f = a_0 + a_1 t + \dotsc + a_n t^n$ be a polynomial over some nontrivial, possibly noncommutative ring $R$. When is $f$ invertible in $R[t]$?
When $R$ is commutative, the answer is well-known: $a_0$ is a unit and $a_1,\dotsc,a_n$ are nilpotent. There are direct element proofs for this, but there is also very nice proof which reduces the claim to integral domains $R$ by using that the radical of $R$ is the intersection of all prime ideals of $R$.
What is known about the noncommutative case? Clearly $a_0$ has to be a unit (apply the homomorphism $t \mapsto 0$). Since $a_0^{-1} f$ is invertible iff $f$ is invertible, we may therefore assume that $a_0 = 1$. If $f = 1 + a_i t^i$ for some $i$, it is still true that $f$ is a unit iff $a_i$ is nilpotent.