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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

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  • $\begingroup$ Do you know the Chinese Remainder Theorem (for polynomials)? You have to compute that polynomial $\pmod{x(x+1)}$. $\endgroup$ – Crostul Nov 24 '16 at 16:06
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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 $-x-2$

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  • $\begingroup$ My question is how did you get $Q(x)(x2+x)+ax+b$ ? That part is where I am stuck on now $\endgroup$ – John Rawls Nov 24 '16 at 16:16
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    $\begingroup$ When we divide two polynomials we get a quotient $Q(x)$ and a remainder $r(x)$ wich has a degree smaller than the degree of $x^2+x$. So $r(x)$ has the form $ax+b$. That is a standard concept about polynomials. $\endgroup$ – Arnaldo Nov 24 '16 at 16:20
  • $\begingroup$ @JohnRawls This answer is not correct - see my answer for the correct result. $\endgroup$ – Bill Dubuque Nov 24 '16 at 16:35
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HINT:

Let $x^{200}+2x^{20}-x^{2}+x-2=f(x) x(x+1)+A(x+1)+Bx$

$-2=f(0)=A(0+1)$

$-1= f(-1)=B(-1)$

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  • $\begingroup$ How did you get $f(x)x(x+1)+A(x+1)+Bx$ $\endgroup$ – John Rawls Nov 24 '16 at 16:12
  • $\begingroup$ @JohnRawls, If you divide a polynomial by a quadratic polynomial, the result is mono-nomial $\endgroup$ – lab bhattacharjee Nov 24 '16 at 16:14
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${\rm mod}\,\ x^2+x\!:\,\ x(\color{#c00}{-x})\equiv x.\,$ 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 $\quad \begin{align}&\ \ x^{\large\color{#0a0}{ 200}}\, +\, 2\ x^{\large\color{#0a0}{20}}\ -\ x^{\large\color{#0a0}2}\, +\ x-2\\ \equiv\, &-x\ +\ 2(-x)-(-x)+x-2\\ \equiv\, & -x-2 \end{align}$

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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.

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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} $

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