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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:

$P(x)=(x-1)Q_{1}(x)+1$
$P(x)=(x+1)Q_{2}(x)-1$

$P(x)=\underbrace{(x-1)(x+1)}Q_{3}(x)+A, \; A=?$
$\;\;\;\;\;\;\;\;\;\;\;\;\;=(x^2-1)$

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?

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  • $\begingroup$ The ermainder has the form $Ax+B$. $\endgroup$ – Bernard Jun 24 '15 at 10:04
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HINT : You should have $$P(x)=(x-1)(x+1)Q_3(x)+\color{red}{ax+b}.$$ Now use $$P(1)=1$$ and $$P(-1)=-1.$$

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  • $\begingroup$ 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

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