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
 A: Notice: $x^2+3x+2=(x+1)(x+2)$ and let the remainder be $ax+b$ (must be at most linear since we divide by quadratic polynomial).
Write $$x^{2016}+2x^{2015}-3x+4=k(x)(x+2)(x+1)+ax+b$$
Put $x=-1$ we get $$6=-a+b$$ and put $x=-2 $ we get $$10 = -2a+b$$ Now solve this system and you get an answer.
A: The remainder must have degree $1$, so you're looking for $a$ and $b$ in
$$f(x) = q(x)g(x) + (ax+b).$$
Note that $g(-1) = g(-2) = 0$ and that it's easy to evaluate $f(-1)$ and $f(-2)$.  So plug $-1$ and $-2$ into the above and get two equations in $a$ and $b$.
A: $$g(x)=(x+1)(x+2)$$
$$f(x)=x^{2016}+2x^{2015}-3x+4$$
$$f(-1)=1-2+3+4=6$$
$$f(-2)=-2^{2015}(-2+2)+6+4=10$$
$$f(x)=g(x).q(x)+a.x+b$$
$$f(-1)=b-a=6$$
$$f(-2)=b-2a=10$$
$$\therefore a=-4 ; b=2$$
Remainder is $=-4x+2$ .
A: 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 $f(x)\equiv \color{blue}{10}(x+1)\color{red}{(-1)}+\color{blue}6(x+2)\color{red}{(1)}\bmod x^2+3x+2$.
A: Hint:
First, observe that the roots of $g(x)$ can be found by the Rational roots theorem. We find that $\;g(x)=(x+1)(x+2)$.
Next , rewrite the dividend as
$$f(x)=x^{2015}(x+2)-3x+4=(x+2)(x^{2015}-3)+10.$$
Last, the division of this quotient by $x+1$ yields
$$x^{2015}-3=(x+1)q(x)-4\qquad\text{(why?)}$$
Can you deduce the remainder now?
