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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.

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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 $ax+b.$ –  Genomeme Jul 21 at 15:23
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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 at 15:25
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Velcome to our site! –  kjetil b halvorsen Jul 21 at 15:32

2 Answers 2

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

$r(x)=ax+b$

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

$$r(x)=-13x+3$$

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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 $\,f(x).$

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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 at 16:09
    
This is a rather sophisticated way doing it. I like it. –  Genomeme Jul 21 at 16:15

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