Polynomial ring is not a UFD

Let $$K$$ be a field, and consider the ring $$R=\{f\in K[x]\mid f'(1)=f''(1)=0\}$$. Show that $$R$$ is not a UFD (Unique Factorization Domain).

My thoughts: I can show that elements such as $$(x-1)^3$$ and $$(x-1)^4$$ are irreducible in $$R$$. Can this be used to show $$R$$ is not a UFD? I am not sure the best route to take. Should we exhibit an element with a non-unique factorization into irreducibles, or should we find two elements who do not have a GCD? Another thing we may be able to do, is consider a quotient of $$R$$ by an irreducible polynomial and show it has zero-divisors (hence it is not a domain, so the polynomial we choose isn't prime, but every irreducible polynomial in a UFD must be prime).

• An element with non-unique factorisation is easiest, methinks. May 11, 2020 at 15:27
• IF you knew $z^3$ and $z^4$ were irreducible, then you would have no trouble coming up with inequivalent factorizations of $z^{12}$. May 11, 2020 at 15:29

Silly me. Following the advice of @rschwieb, we have the non-unique factorizations $$(x-1)^{12}=(x-1)^3(x-1)^3(x-1)^3(x-1)^3$$ and $$(x-1)^{12}=(x-1)^4(x-1)^4(x-1)^4$$ into irreducibles. Therefore $$R$$ is not a UFD. Just to add this bit of detail: $$(x-1)^3$$ and $$(x-1)^4$$ are irreducible in $$R$$ since if they weren't, they'd either have a linear or quadratic factor. But any linear or quadratic factor will either have a non-zero constant first or second derivative, violating the definition of $$R$$.