Let $R$ be an Artinian regular ring with unity. We say that $a\in R$ is regular if $a=axa$ for some some $x\in R$. $R$ is regular if every element of $R$ is regular. Let $R[x]$ be the ring of polynomials in an indeterminate $x$ and consider the ideal $K[x]$ of $R[x]$.
Question: Is the quotient ring $R[x]/K[x]$
(1) a Noetherian ring?
(2) an Artinian ring?
Well let's take a simple case of what (I guess) you are asking, and suppose $K\lhd R$. (You have already required $K\subseteq R$ and I don't know what else you could mean by $K[x]$ other than the polynomials with coefficients in $K$.)
Then $R[x]/K[x]\cong (R/K)[x]$, which is never Artinian when $R/K$ is nonzero. This should be obvious: in a polynomial ring over one variable, $(x)\supsetneq (x^2)\supseteq (x^3)\supseteq\cdots$.
Moreover, if we take $R$ to be a von Neumann regular ring such that $R/K$ is non-Noetherian, it should be obvious $(R/K)[x]$ is not Noetherian: any strictly ascending chain $A_1\subseteq A_2\subseteq A_3\subseteq \cdots$ immediately gives a chain $A_1[x]\subseteq A_2[x]\subseteq A_3[x]\subseteq\cdots$ in $(R/K)[x]$.