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The exercise is as follows. There is a fourth degree polynomial, that when divided by $(x - 3)$ has a $r_1 = 100$, and when divided by $(x + 1)$ a $r_2 = -4$. And the question is what would the rest be when divided by $(x - 3)(x + 1)$.

I have tried using the rest theorem, and guess What the coefficients would be. And with this, I can find more the 1 polynomial that works. I was just wondering if there is a better way to solve this?

I have for example:

$x^4 + x^3 + 2x^2 - 9x + 1$, $r = 22x + 34$

and:

$x^4 + x^3 + x^2 - 5x - 2$, $r = 24x + 28$

when they are divided by

$x^2 - 2x - 3 ((x - 3)(x + 1))$

But the answer to the exercise is

$26x + 22$

Or must I keep guessing?

Ma Ming
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Rptx
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1 Answers1

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Let $f=p(x-3)(x+1)+q$ with $\deg q<2$, say $q=ax+b$. Then $q(3)=100$ and $q(-1)=-4$ by the rest theorem.

It amounts to solve $$ 3a+b=100,\quad -a+b=-4, $$ which have solution $(a,b)=(26,22)$. Whence $f=p(x-3)(x+1)+26x+22$ for arbitary $\deg p=2$.

By the Chinese remainder theorem, the answer is unique. You get two examples, there must be something goes wrong in your argument.

Ma Ming
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