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Question states as following:

A polynomial $f(x)$ in the $4^{th}$ degree is divisible by $x^2-1$. Find the remainder if $f(x)$ is divided by $(x+1)(x-1)(x-2)$.

Did a bit of googling, in which I found this post, but that didn't really help much since the question states that it is divisible, in which I don't think that I'll be able to solve for $a$ and $b$.

I've also tried doing $$f(x)=Q(x+1)(x-1)+R$$ And that didn't really helped since all I've gotten is that $f(1)=0$ and $f(-1)=0$.

Also tried equating $f(x)$ so that it becomes $$Q_1(x+1)(x-1)=Q_2(x+1)(x-1)(x-2)+R_2$$, solving for $R_2$ which is $$R_2=(x+1)(x-1)[Q_1-Q_2(x-2)]$$

Apparently the answer to the question is $$\frac 13f(2)[x^2-1]$$ but I can't seem to make $Q_1-Q_2(x-2)$ equal to $\frac 13f(2)$. This question has been bugging me for a while so thanks in advance for helping.

Jyrki Lahtonen
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Christopher Yap
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1 Answers1

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Write $f(x)=(x^2-1)q(x)$, where $q(x)$ is a quadratic polynomial. Note that $$ \frac{f(x)}{(x^2-1)(x-2)}=\frac{q(x)}{x-2}.$$ Further, by Euclidean division, we can write $q(x)=p(x)(x-2)+q(2)$ for some linear polynomial $p(x)$. Thus $$\frac{f(x)}{(x^2-1)(x-2)}=\frac{(x^2-1)(p(x)(x-2)+q(2))}{(x^2-1)(x-2)},$$ and hence the remainder is $q(2)(x^2-1)$. Finally note that $f(2)=(2^2-1)q(2)$, so we can write the remainder as $\tfrac13f(2)(x^2-1)$.

cqfd
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    What is $p(x)$ suppose to resemble? And why can we write $Q(x)=P(x)(x-2)+Q(2)$? – Christopher Yap Dec 25 '23 at 06:23
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    @ChristopherYap $p(x)$ is the quotient when you divide $q(x)$ by $(x-2)$. And $q(x)=p(x)(x-2)+q(2)$ follows by remainder theorem. – D S Dec 25 '23 at 06:31