Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$
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.
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 $b$.
$$g(x)=(x+1)(x+2)$$ $$f(x)=x^{2016}+2x^{2015}-3x+4$$ $$f(-1)=1-2+3+4=6$$ $$f(-2)=-2^{2015}(-2+2)+6+4=10$$ $$f(x)=g(x).q(x)+a.x+b$$ $$f(-1)=b-a=6$$ $$f(-2)=b-2a=10$$ $$\therefore a=-4 ; b=2$$ Remainder is $=-4x+2$ .