Isomorphism from $B[y]/IB[y]$ onto $(B/I)[y]$ For some reason I can't crack the following problem:

Let $B$ be a ring, $I$ an ideal, and $A := B[y]$ the polynomial ring.
  Construct an isomorphism from $A/IA$ onto $(B/I)[y]$.

How to interpret the (polynomial) quotient ring $A/IA$? And how do I construct the isomorphism knowing this?
Hope you can help!
 A: Consider the natural surjective homomorphism $B \to B/I,$ where $b \in B \mapsto b + I \in B/I.$ Show that that induces a surjective homomorphism $h: B[y] \to (B/I)[y],$ where a polynomial $f(y) = b_ny^n +b_{n-1}y^{n-1}+ \cdots+b_0 \in B[y]$  gets mapped to $(b_n + I)y^n +(b_{n-1} + I)y^{n-1}+ \cdots+(b_0+I).$
Now to establish the isomorphism using the first isomorphism theorem, we need to prove that the kernel of $B[y] \to (B/I)[y]$ is precisely the ideal $IB[y]$ (i.e. the ideal $I,$ now generated in $B[y].$) Look closely at the definition of an ideal, where now 
$$IB[y] = \{a_my^m+a_{m-1}y^{m-1}+\cdots+a_0| \; a_i \in I, \; \forall 1 \leq i \leq m  \}.$$
Clearly any element of $IB[y]$ lies in the kernel of $h$ as 
$$a_my^m+a_{m-1}y^{m-1}+\cdots+a_0 \mapsto (a_m+I)y^m+(a_{m-1}+I)y^{m-1}+\cdots+(a_0+I)$$
via $h,$ where $a_i+I = I$ for all $1 \leq i \leq m$ because $a_i \in I$ (by definition). Therefore $(a_m+I)y^m+(a_{m-1}+I)y^{m-1}+\cdots+(a_0+I) = 0$ in $(B/I)[y].$ Hence the $IB[y]$ is contained $ \subseteq$ in the kernel of $h$. Now we need to show that the kernel is contained in $IB[y]$ in order to prove the equality. Take an element of the kernel $g(y) = c_ky^k+\cdots+c_0.$ As it is in the kernel it means that all $c_i \in I$ otherwise it won't be an element of the kernel.
P.S. I tried to explain as extensive as possible and I hope it will be helpful.
