an isomorphism to product ring $\mathbb{C}\times \mathbb{C}$ Examine whether there is a polynomial $f(X)\in \mathbb{R}[X]$ such that $\frac{\mathbb{R}[X]}{(f(X))}$ is isomorphic to the product ring $\mathbb{C}\times \mathbb{C}$.

I have found a particular example if the polynomial ring would be of two variable, e.g. $\frac{\mathbb{R}[X,Y]}{(X^2+1,Y^2+1)}$ which is isomorphic to $\mathbb{C}\times \mathbb{C}$, or if we allow the ring over $\mathbb{C}$, e.g. $\frac{\mathbb{C}[X]}{X^2+1}$. I think it is not possible to find such $f$ over $\mathbb{R}[X]$ but I am unable to provide a justification there. All I have figured out that $\mathbb{C}\times \mathbb{C}$ is neither an integral domain nor a field, hence $(f(X))$ is neither prime nor maximal ideal. So $f(X)$ has to be a reducible polynomial over $\mathbb{R}[X]$, if it exists. Can we conclude something from here?
Hint: something similar $\mathbb{Z}/6\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$ can happen over polynomial rings.

Note that $(x^2+1)^2$ won't work, similar to $\mathbb{Z}/4\mathbb{Z}\not\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$. However, unlike with $\mathbb{Z}$, it is possible to quotient $\mathbb{R}[x]$ by different polynomials and get the same quotient ring $\mathbb{C}$.

(If you're familiar with the general form of the Chinese Remainder Theorem for rings, good.)
[My thinking :] As we know that irreducible polynomials in R[x] have degree 1 or 2. If we find any polynomial of degree polynomial of degree 4 that can be reducible into lower degree. So there is no polynomial f(x) in R[x].