Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f(x) = x^{10}+5x^9-8x^8+7x^7-x^6-12x^5+4x^4-8x^3+12x^2-5x-5. $

Without using long division (which would be horribly nasty!), find the remainder when $f(x)$ is divided by $x^2-1$.

I'm not sure how to do this, as the only way I know of dividing polynomials other than long division is synthetic division, which only works with linear divisors. I thought about doing $f(x)=g(x)(x+1)(x-1)+r(x)$, but I'm not sure how to continue. Thanks for the help in advance.

share|cite|improve this question
plug in $1$ and $-1$ to get two values of $r(x)$ which is linear. From there you can get what $a,b$ are in $ax+b.$ – ReverseFlow Jul 21 '14 at 15:23
Hint : Use the horner-scheme. To avoid complications, first use it for 1, then for -1. There is a compact notation, described in wikipedia. – Peter Jul 21 '14 at 15:25
Velcome to our site! – kjetil b halvorsen Jul 21 '14 at 15:32

Plug in $1$ and $-1$ to get two values of $r(x)$ which is linear. From there you can get what $a,b$ are in $ax+b.$

Since $$f(x)=g(x)(x+1)(x-1)+r(x)$$

we have

$$ f(1)=g(1)(1+1)(1-1)+r(1)=r(1)=-10$$ $$ f(-1)=g(1)(-1+1)(-1-1)+r(-1)=r(-1)=16$$

We know the remainder is of degree $1$, so


and now we know, $$r(1)=ax+b=a+b=-10$$ $$r(-1)=ax+b=-a+b=16$$

so, solve

$$a+b=-10$$ $$-a+b=16$$

which yields, $a=-13$ $b=3$, so


share|cite|improve this answer

Hint $\ {\rm mod\ }x^{\large 2}\!-1\!:\,\ x^{\large 2}\equiv 1\,\Rightarrow\,\color{#0a0}{x^{\large 2n}\equiv 1}\,\Rightarrow\,\color{#c00}{x^{\large 2n+1}\equiv x}.\ $ Therefore,

$\qquad \ \ \ \ (c_0 + c_2\color{#0a0}{ x^2} + c_4\color{#0a0}{ x^4}+\cdots) + (c_1\color{#c00} x + c_3\color{#c00}{x^3} + c_5\color{#c00}{x^5} + \cdots)$

$\qquad \equiv \ c_0 \ +\ c_2\ +\ c_4\ \ + \ \cdots + \color{#c00}x (c_1\ +\,\ c_3\ \ +\ \ c_5 \ +\ \cdots) $

$\qquad \equiv\ f_0(1)\, +\, f_1(1)\, x,\ $ where $\,f_0(x),\ f_1(x)\,$ are the even and odd parts of $\,f(x).$

share|cite|improve this answer
In other words, add up all the even-degree coeffs and you get the constant term; add up all the odd-deg coeffs and you get the linear term. – Lubin Jul 21 '14 at 16:09
This is a rather sophisticated way doing it. I like it. – ReverseFlow Jul 21 '14 at 16:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.