Solve $x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod {(4\cdot3)}$ Solve $x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod {(4\cdot3)}$

Work:
Let $P(x) = x^{10} +4x^3+3x+4 \equiv 0 \pmod 2$ and $12=4\cdot3=2^2\cdot3$
We have $P(x) \equiv0 \pmod {2^2}$ and $P(x)\equiv0\pmod3$
For $\pmod {2^2}$,
$x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod 2$
Try $[0],[1]$, 
$[0]$ works since $P(0) =4 \equiv0\pmod2$, say $a_1 =0$
$[0]$ lifts to a unique solution to a class that $[a_2] \in \mathbb{Z}_{2^2}$ so that $P(a_2) \equiv 0 \pmod {2^2}$
Solving $P'(a_1)\cdot x =3\cdot{x}\equiv1\pmod2$, $\overline{P'(a_1)}=x=1$
any representative in class of 
$a_2 = a_1 -\overline{P'(a_1)}\cdot {P'(a_1)}=3-1\cdot3=0\equiv{0}\pmod{2^2}$
$0\pmod{2^2}$ the only solution to $P(x)\equiv{0}\pmod{2^2}$
For $\pmod3$,
$x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod 3$
Try $[0],[1],[2]$, $[1]$ works. Say $a_2=1$
$P(1)$ is a solution to $\pmod3$
and how do I combine 2 cases??

If you want $n \equiv 1 \pmod 3$ and $n \equiv 0 \pmod 4$ you can just try the multiples of $4$ until you find one that works.  In this case it is $4$.  So $P(4)\equiv 0 \pmod {12}$.  Here is a check from Alpha
