<s>[INST] Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$ Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$ 
I try to factor them, but I failed. I know that it's impossible to divide it
 [/INST] First note $x^2+3x+2=(x+1)(x+2)$.
Modulo $x+2$, $x\equiv-2$, so $f(x) = x^{2016}+2x^{2015}-3x+4\equiv2^{2016}-2^{2016}+6+4=\color{blue}{10}.$
Modulo $x+1$, $x\equiv-1$, so $f(x) = x^{2016}+2x^{2015}-3x+4\equiv1-2+3+4=\color{blue}{6}.$
$\color{red}1(x+2)+\color{red}{-1}(x+1)=1$, 
so by the Chinese remainder theorem $f(x)\equiv \color{blue}{10}(x+1)\color{red}{(-1)}+\color{blue}6(x+2)\color{red}{(1)}\bmod x^2+3x+2$.
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