Chinese Remainder Theorem on polynomials 
Let $f(x) = x^5 + 3x^2 + 4$. Find all solution to the congruence. $f(x)\equiv 0 \pmod{12}$.  

I don't understand how to apply CRT on polynomials.
 A: Here, you don't need to do any sort of polynomial applications. Find the values of $x\bmod 3$ for which $f(x)\equiv 0\bmod 3$, find the values of $x\bmod 4$ for which $f(x)\equiv 0\bmod 4$, and use CRT on integers (with moduli $3$ and $4$) to combine them into all solutions for $x\bmod 12$.

To illustrate this technique, I'll work through a simpler example. Say we want to find all solutions to the congruence
$$x^2+x\equiv 0\bmod 6.$$
We see that $6=2\cdot 3$, and first find all solutions to $x^2+x\equiv 0\bmod 2$. Checking each residue class, we see that they each work, so any integer equivalent to a member of the set $\{0,1\}\bmod 2$ will satisfy $x^2+x\equiv 0\bmod 2$.
For the $\bmod 3$ part, we see that $x^2+x\equiv 0\bmod 3$ if and only if $x$ is equivalent to a member of $\{0,2\}\bmod 3$.
Combining these two results, we have $x^2+x\equiv 0\bmod 6$ if and only if
$$x\in\{0,1\}\bmod 2,\ x\in\{0,2\}\bmod 3.$$
Using CRT, this is equivalent to $x\equiv 0,2,3,5\bmod 6$, and we thus have our answer.
