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.

  • 4
    $\begingroup$ 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 $ax+b.$ $\endgroup$ Jul 21 '14 at 15:23
  • 1
    $\begingroup$ Hint : Use the horner-scheme. To avoid complications, first use it for 1, then for -1. There is a compact notation, described in wikipedia. $\endgroup$
    – Peter
    Jul 21 '14 at 15:25
  • 1
    $\begingroup$ Velcome to our site! $\endgroup$ 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



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},\ $ hence

$ f(x) =\, \overbrace{(c_0 + c_2\color{#0a0}{ x^2} + c_4\color{#0a0}{ x^4}+\cdots)}^{\large \color{#0a0}{f_0(x)}} \ +\ \overbrace{(c_1\color{#c00} x + c_3\color{#c00}{x^3} + c_5\color{#c00}{x^5} + \cdots)}^{\large \color{#c00}{f_1(x)}}$

$\qquad \equiv \ (c_0 \ +\ c_2\ +\ c_4\ \ + \ \cdots)\,\color{#0a0}1 + (c_1\ +\,\ c_3\ \ +\ \ c_5 \ +\ \cdots)\,\color{#c00}x $

$\qquad \equiv\ f_0(1)\,\color{#0a0}1 + f_1(1)\, \color{#c00}x,\ $ where $\,f_0(x),\ f_1(x)\,$ are the $\rm\color{#0a0}{even}$ and $\rm\color{#c00}{odd}$ parts of $\,f(x).$

e.g. a familiar numerical instance when $\,x=10\,$ in radix $10$ (decimal) arithmetic

$\!\! \bmod 99\!:\ \color{#c00}5\color{#0a0}4\color{#c00}3\color{#0a0}2\color{#c00}1\color{#0a0}0\equiv (\color{#c00}{5\!+\!3\!+\!1}),(\color{#0a0}{4\!+\!2\!+\!0})\equiv \color{#c00}9\color{#0a0}6\equiv \color{#c00}5\color{#0a0}4+\color{#c00}3\color{#0a0}2+\color{#c00}1\color{#0a0}0\ $ by $\,10^2\equiv 1$

  • $\begingroup$ 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. $\endgroup$
    – Lubin
    Jul 21 '14 at 16:09
  • $\begingroup$ This is a rather sophisticated way doing it. I like it. $\endgroup$ Jul 21 '14 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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