What is the remainder when $x^{200}+2x^{20}-x^{2}+x-2$ is divided by $x^{2}+x$ Question
What is the remainder when $x^{200}+2x^{20}-x^{2}+x-2$ is divided by $x^{2}+x$
my ideals & thought process
I was really stuck on this problem so i tried long division and plugged in $0x^{?}$ into the missing gaps where the question mark is a place holder for the variable which looked like this (assume the square root is the division sign)
$x^{2}+x \sqrt{x^{200}+0x^{199}+0x^{198}..... 2x^{20}+0x^{19}....-x^{2}+x-2}$
Answer
the answer is supposed to be -x-2
Other
A lot of helpers are suggesting to use Q(x)(x2+x)+ax+b in my process. Please explain how to do it
 A: Using the standard equation for polinomial division to see that:
$$x^{200}+2x^{20}-x^{2}+x-2=Q(x)(x^2+x)+ax+b$$
So for $x=0$ we get $-2=b$ and for $x=-1$ we get $-1=-a+b$ and then $a=-1$. The remainder is $-x-2$
A: HINT:
Let  $x^{200}+2x^{20}-x^{2}+x-2=f(x) x(x+1)+A(x+1)+Bx$
$-2=f(0)=A(0+1)$
$-1= f(-1)=B(-1)$
A: ${\rm mod}\,\ x^2+x\!:\,\ x(\color{#c00}{-x})\equiv x.\,$ But $\,xf\equiv x\equiv xg\,\Rightarrow\, x\color{#c00}{fg}\equiv x,\ $ so $\ x\color{#c00}{(-x)^n}\equiv x.\,$ 
Thus $\,x^{\large\color{#0a0}{ 2N}}\equiv -x(-x)^{2N-1}\equiv -x\,\ \ $ so 
$\quad  \begin{align}&\ \  
x^{\large\color{#0a0}{ 200}}\, +\, 2\ x^{\large\color{#0a0}{20}}\ -\ x^{\large\color{#0a0}2}\, +\ x-2\\ 
\equiv\, &-x\  +\  2(-x)-(-x)+x-2\\ 
\equiv\, & -x-2  \end{align}$
A: If you're unsure, just do the polyomial long division -- you will see a pattern very quickly that should allow you to fast-forward from $x^{195}$ish to $x^{21}$ in one step -- and then afterwards from $x^{19}$ to $x^3$.

A slightly faster approach in this particular case is to note that $x^3\equiv x^1$ modulo $x^2+x$, and therefore $x^{n+2k}\equiv x^n$ as long as $n\ge 1$.
Therefore,
$$ x^{200}+2x^{20}-x^{2}+x-2 \equiv x^2+2x^2-x^2+x-2 = 2x^2+x-2 $$
and from there the division is very easy.
A: At high-school level: $ $ write it as $\ (x^{\large\color{#c00}{200}}\!+x) + 2(x^{\large\color{#c00}{20}}+x) - (x^{\large\color{#c00}2}+x)\,\color{#0a0}{ -x-2}.\ $ Notice that  $\,x(x\!+\!1)\mid f(x) = x^{\large\color{#c00}{2N}}\!+x\,$ by $\,f(0)=0=f(-1)\,$ so the remainder is $\color{#0a0}{-x-2}$

$ \begin{align} {\rm Alternatively}\quad f(x) &= -2 \,+\, x\overbrace{(1-x+2x^{19}+x^{199})}^{\Large g(x)\ \ {\rm so}\ \ \color{#c00}{g(-1)\, =\, -2}}\\
&= -2\, +\, x(g(-1)+(x+1)\,h(x)\\
&= -2\, +\, x\, \underbrace{\color{#c00}{g(-1)}}_{\large\color{#c00}{ -2}} + (x^2+x)h(x)
\end{align}
$
