Remainder of dividing a polynomial $P(x),$ $ \left (\deg{P(x)\geqslant2} \right ) $ with $(x-1)$ is $1$ while remainder of dividing the same polyinomial with $(x+1)$ is $-1$. Find the remainder of dividing $P(x)$ with $(x^{2}-1)$.

In short:


$P(x)=\underbrace{(x-1)(x+1)}Q_{3}(x)+A, \; A=?$

I've written like four pages of manipulation with what's given and either came to where I had begun, or had got nothing useful. I also tried putting roots of binomials instead of x but then I get $P(1)=1=A \wedge P(-1)=-1=A$ which confuses me even more.

Hints on what to do?

  • $\begingroup$ The ermainder has the form $Ax+B$. $\endgroup$ – Bernard Jun 24 '15 at 10:04

HINT : You should have $$P(x)=(x-1)(x+1)Q_3(x)+\color{red}{ax+b}.$$ Now use $$P(1)=1$$ and $$P(-1)=-1.$$

  • $\begingroup$ Could you please explain why (ax+b) ? $\endgroup$ – tyr Jun 24 '15 at 10:05
  • $\begingroup$ Because the degree of $x^2-1$ is $2$. $\endgroup$ – mathlove Jun 24 '15 at 10:06
  • $\begingroup$ So, if I had a polynomial of the degree 3 I would put $ax^2+bx+c$ ? $\endgroup$ – tyr Jun 24 '15 at 10:08
  • $\begingroup$ @lemniscate: Yes, exactly. $\endgroup$ – mathlove Jun 24 '15 at 10:09

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.