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When the polynomial $P(x)$ is divided by $x^2 -1$, the remainder is $3x-1$. What is the remainder when $P(x)$ is divided by $x-1$?

I'm not sure what to do. I know that $P(1)=P(-1)=3x-1$, but I don't think that's how you're supposed to do it.

Any help would be appreciated, thanks.

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    $\begingroup$ Hint: $x^2-1 = (x-1)(x+1)$ $\endgroup$ – Tpofofn Sep 15 '13 at 2:14
  • $\begingroup$ $P(1)=3(1)-1=2$ $\endgroup$ – oldrinb Sep 15 '13 at 2:16
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Note you've found that $$P(x)=(x^2-1)Q_1(x)+(3x-1)$$Since $x^2-1=(x+1)(x-1)$ it follows that:$$P(x)=(x-1)(x+1)Q_1(x)+(3x-1)$$Since clearly if we let $x=1$ we have $x-1=0$, we can focus on just our remainder polynomial $3x-1$.$$P(1)=(1-1)[(1+1)Q_1(x)]+(3-1)=3-1=2$$Now note that if we instead merely divided by $x-1$ we would expect $$P(x)=(x-1)Q_2(x)+R(x)\\P(1)=(1-1)Q(x)_2+R(1)\\P(1)=R(1)$$hence since $P(1)=2$ it follows $R(1)=2$.

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HINT:

$$\begin{align*} P(x)&=(x^2-1)Q(x)+(3x-1)\\ &=(x-1)(x+1)Q(x)+3x-1\\ &=(x-1)\Big((x+1)Q(x)+3\Big)+\;? \end{align*}$$

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