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Find the remainder when $P(x)=5x^6 + x^5 - 2x^3 - x^2 + 1$ is divided by $x^2+1$.

$$ P(x)=(x^2+1)\cdot Q(x)+R(x) $$

When $x^2=-1$,

$$ P(x)=-5+x+2x+1+1 = 3x-3 $$

Exactly, why the method above, does not work for the following question?

Find the remainder when $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x$ is divided by $x^3-x$.

$$ P(x)=(x^3-x)\cdot Q(x)+R(x)=x(x+1)(x-1)\cdot Q(x) + R(x) $$

$P(1)=5$, but $P(-1)=-5$, and $P(0)=0$

However, if I factor $P(x)$ by $x$,

$$ P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1) $$

and consider $P(1)$ and $P(-1)$ without the factored $x$, then I get $5x$ in both cases. (The correct answer.)

Question: I can solve the question by slightly different methods where there is no confusion or ambiguity involved. I just like to know the above method does not work with this problem. (And by “exactly why” I mean to ask, how am I supposed to foresee that the above method should not be applied directly.)

blackened
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    You got $R(1)=P(1)=5$, $R(-1)=P(-1)=-5$, $R(0)=P(0)=0$. Aren't these all compatible with $R(x)=5x$?? In other words, I think the problem was in the step when you tried to extract $R(x)$ from those three data points. The Chinese remainder theorem is usually called upon at that point. – Jyrki Lahtonen Dec 30 '18 at 10:00

2 Answers2

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I think the problem is, with P(1), P(-1) and P(0), you are actually getting the value of r(1), r(-1) and r(0), instead of r(x).

r(x) is linear in this case, so you are able to get the correct answer by factoring P(x) by x

Mengqi Chen
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The first method does work:

When $x^3=x$, we have $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x= x+x+x+x+x = 5x$.

When $x^3=x$, we have $P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)= x(x^2+x^2+x^2+x^2+1) = x(4x^2+1)=4x^3+x=5x$.

lhf
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