<s>[INST] Solve the non-linear congruence $x^3+2x^2+5x+4\equiv 0\pmod {60}$. Solve the non-linear congruence $x^3+2x^2+5x+4\equiv 0\pmod {60}$. 
For this question, I think I might need to use the Chinese remainder Theorem to simplify the problem to several smaller moduli (e.g. possibly $\pmod 4,\pmod 5,\pmod 3$ and simplify the $x^3+2x^2+5x$ terms). 
 [/INST] Modulo $4$ we have $x^3+2x^2+x=x(x+1)^2\equiv0$, 
which means $x\equiv0$ or $x+1\equiv2$ or $4$.
Modulo $5$ we have $x^3+2x^2+5x+4=(x+1)(x^2+x+4)\equiv(x+1)(x-2)^2\equiv0$, 
which means $x\equiv 4$ or $2$.
Modulo $3$ we have $x^3+2x^2+5x+4=(x+1)(x^2+x+4)\equiv(x+1)(x-1)^2\equiv0$, 
which means $x\equiv 2$ or $1$.  
Can you take it from here with the Chinese remainder theorem 
to find $12$ solutions modulo $60=4\times5\times3$?
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