Showing that a map is $R$-bilinear Let $I=(2,x)\subset \mathbb{Z}[x]:=R$. 
I want to show that the map $f: I\times I \rightarrow \mathbb{Z}/2\mathbb{Z}$ defined by $f(a_0+a_1x+\cdots + a_nx^n, b_0+b_1x+\cdots +b_mx^m)=\overline{\frac{a_0}{2}b_1}$ is bilinear. 
My partial solution:
First, I get  $f(a_1(x) +a_2(x), b(x))= \overline{\frac{a_{1,0}+ a_{2,0}}{2}b_1}=\overline{\frac{a_{1,0}}{2}b_1} + \overline{\frac{a_{2,0}}{2}b_1} = f(a_1(x), b(x))+f(a_2(x),b(x))$. 
Similarly, we have $f(a(x), b_1(x)+b_2(x))=\overline{\frac{a_0}{2}(b_{1,0}+b_{2,0})}= \overline{\frac{a_0}{2}b_{1,0}}+\overline{\frac{a_0}{2}b_{2,0}}=f(a(x), b_1(x)+b_2(x))$. 
My problem:
Assuming $r(x)\in R$, we get $f(r(x)a(x), b(x)) = \overline{\frac{r_0a_0}{2}b_1}= \ldots$. 
This is where I'm stuck. I want to get $f(r(x)a(x), b(x))= r(x)f(a(x), b(x))$, but I can't see a way to derive this. I'm sure it's pretty easy but I can't see the correct manipulation. I expect I'll need to use the fact that $\mathbb{Z}/2\mathbb{Z} \cong R/I$ and that $a_0, b_0$ are zero $\mod 2$, but I'd appreciate some help. 
Update: To clarify, I want bilinearity over $\mathbb{Z}[x]$. I'm not given an explicit definition of how $\mathbb{Z}/2\mathbb{Z}$ is an $R$-module here, but I'm told that $\mathbb{Z}/2\mathbb{Z}\cong R/I$ is naturally an $R$-module annihilated by both $2$ and $x$.
 A: You should ask yourself how action of $R$ on $\Bbb Z/2\Bbb Z$ is defined. You already wrote yourself that $\Bbb Z/2\Bbb Z\cong R/I$ and there lies the answer: $$(a_0 + a_1x+\ldots + a_nx^n,\overline a)\mapsto \overline{a_0a}\tag{1}$$
How so? Well, this just a standard way to make $R/I$ an $R$ module, action is given by $$(r', r + I) \mapsto r'r + I\tag{2}$$
Observe that $(1)$ really is of the form $(2)$ using the isomorphism.
The exercise should be easy now and I will leave the rest to you, so let me know if you've finished it or need more help.
A: Notice that if $p : R \to R/I$ is the natural projection, then $R/I$ obtains a natural $R$-module structure with the multiplication by scalars given by $r \cdot \widehat s := \widehat {rs}$.
Since the isomorphism $R/I \simeq \Bbb Z / (2)$ is given by $\widehat F \leftrightarrow \overline {F(0)}$ (with the hat denoting classes modulo $I \subset R$ and the bar denoting classes modulo $(2) \subset \Bbb Z$, and $F \in R$), it follows that the multiplication of $F \in R$ and an element $\overline n \in \Bbb Z /(2)$  is given by $F \cdot \overline n : = \overline {F(0)} \overline n = \overline {F(0) n}$.
Since $f(G,H) = \overline {\frac 1 2 G(0) H(0)}$, it follows that
$$f(F \cdot G, H) = \overline {\frac 1 2 (FG)(0) H(0)} = \overline {\frac 1 2 (FGH)(0)} = \overline {F(0)} \overline {\frac 1 2 G(0) H(0)} = \\
\overline {F(0)} f(G,H) = F \cdot f(G,H) ,$$
completing your argument that shows that $f$ is $R$-bilinear.
