$\mathbb Z[X]/(2,X) \cong \mathbb{Z}_2$ Let $\mathbb Z[X]$ be the polynomial ring with integer coefficients. How do we know that $\mathbb Z[X]/(2,X) \cong \mathbb{Z}_2$? 
 A: The more general fact is that $R[x]/(I,a_1,\ldots,a_n)\cong (R/I)[x]/(\overline{a_1},\ldots,\overline{a_n})$ where $\overline{a_i}$ is the image of $a_i$ in $(R/I)[x]$. 
Try the obvious map $R[x]\to (R/I)[x]/(\overline{a_1},\ldots,\overline{a_n})$ defined something like 
$$\displaystyle \sum_{i=1}^{n}a_i x^i\mapsto \sum_{i=1}^{n}(a_i+I)x^i+(\overline{a_1},\ldots,\overline{a_n})$$
Check surjectivity, and then apply the FIT.
In your case, the above fact tells us what? It tells us that 
$$\mathbb{Z}[x]/(2,x)\cong (\mathbb{Z}/2\mathbb{Z})[x]/(x)$$
Why is this helpful?
A: Hints:
Try to justify each of the following steps (with $\,\Bbb Z_p=\Bbb Z/p\Bbb Z\,$):
$$\Bbb Z[x]/\langle2,x\rangle\cong\left(\;\Bbb Z[x]/\langle 2\rangle\;\right)\left(\;\langle 2,x\rangle/\langle 2\rangle\;\right)\cong\Bbb Z_2[x]/\langle x\rangle\cong\Bbb Z_2$$
For example, you can define
$$\rho:\langle 2,x\rangle\to\langle x\rangle\;,\;\;\rho(2f(x)+xh(x)):=xh(x)\;,\;\;\ker\rho=\ldots ?$$
$$\phi:\Bbb Z[x]\to\Bbb Z_2[x]\;,\;\;\phi(f(x)):=f(x)\pmod 2\;,\;\;\ker\phi=\ldots ?$$
$$\psi:\Bbb Z_2[x]\to\Bbb Z_2\;,\;\;\psi(f(x)):=f(0)\;,\;\;\ker\psi=\ldots ?$$
Of course, the very last isomorphism also almost follows from the fact that $\,\langle x\rangle\le\Bbb Z_2[x]\,$ is a maximal ideal
A: Hint $\rm\ \Bbb Z[x]/(2,x)\:$ has $\,2\,$ elements, since $\rm\ mod\ (x,2)\!:\ f(x) \equiv f(0)\ mod\ 2\equiv 1\ \ or\ \ 0,\:$ and $\rm\: 1\not\equiv 0 \:$ (else $\rm\:1\in (2,x)\:\Rightarrow\:1 = 2 f + x g\stackrel{x\,=\,0}{\Rightarrow} 1 = 2\,f(0),\:$ contra $\rm\:f(0)\in\Bbb Z).$
