I think that if $A$ is any commutative ring with unity and $q\in A$, $p\in A[x]$ then we have $A[x]/(q,p)\cong A/(q)[x]/(\bar{p})$ where $\bar{p}$ denotes the class of $p$ in $A/(q)[x]$. Is this true?
Also is there some isomorphism for $A[x]/(q,p)$ with $q\in A[x]$?
On
To complement Martin's answer in accordance with the wishes expressed in the comments of the OP, why is the kernel of the canonical map $A[x]/(q,p)\to (A/q)[x]/(\bar{p})$ zero? Suppose $f=a_nx^n+\cdots+a_1x+a_0\in A[x]$ is in the kernel. This means that if we reduce the coefficients modulo $(q)$, to get $\bar{f}=\bar{a}_nx^n+\cdots+\bar{a}_1x+\bar{a}_0$, then in $(A/q)[x]$, $\bar{p}$ divides $\bar{f}$. That is, $\bar{f}=\bar{g}\bar{p}$ for some $\bar{g}\in A[x]$ (and we can assume, as the notation suggests, that $\bar{g}$ is the image in $(A/q)[x]$ of a polynomial $g\in A[x]$). Then the image of $f-qp\in A[x]$ in $(A/q)[x]$ is zero. That means that every coefficient of the polynomial $f-gp$ lies in $qA$, i.e., is divisible by $q$. The set of polynomials in $A[x]$ all of whose coefficients are divisible by $q$ is exactly the ideal $qA[x]$, the ideal of $A[x]$ generated by $q$ (verify this if it's not clear to you). So we have $f-gp\in qA[x]$. Thus $f=gp+qh$ for some $h\in A[x]$, and $f\in(q,p)$, the ideal of $A[x]$ generated by $q$ and $p$. This proves the desired injectivity.
As for the second question, regarding the case when $q,p\in A[x]$ but $q$ is not necessarily in $A$, there is nothing that can be said in general. It's just $A[x]/(q,p)$. What would you expect this ring to be isomorphic to other than itself?
You can easily construct an isomorphism using the universal properties of quotient and polynomial rings.
Start as follows: We have $A \to A/q \to A/q [x] \to A/q[x]/(p)$ and the element $x$ in $A/q[x]/(p)$. Hence, we get $A[x] \to A/q[x]/(p)$. Clearly $q$ and $p$ lie in the kernel, so that we get $A[x]/(p,q) \to A/q[x]/(p)$.
Now construct the inverse morphism.
(As always, there is also a one-line proof using representable functors.)