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I encountered this question, and I am unsure how to answer it.

When $P(x)$ is divided by $x - 4$, the remainder is $13$, and when $P(x)$ is divided by $x + 3$, the remainder is $-1$. Find the remainder when $P(x)$ is divided by $x^2 - x - 12$.

How would I proceed? Thank you in advance!

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    $\begingroup$ HINT: $x^2-x-12=(x-4)(x+3)$ $\endgroup$ – Tito Eliatron Oct 12 '20 at 19:41
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    $\begingroup$ I see. Is the answer 2x + 5 then? $\endgroup$ – zotz99 Oct 12 '20 at 19:52
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Use the inverse isomorphism of the isomorphism in the Chinese remainder theorem: as $x^2-x-12=(x+3)(x-4)$, we have an isomorphism \begin{align} K[X]/(X^2-X-12)&\xrightarrow[\quad]\sim K[X]/(X+3)\times K[X]/(X-4) \\ P\bmod(X^2-X-12)&\longmapsto(P\bmod (X+3), P\bmod (X-4)&&(K\text{ is the base field}) \end{align} and given a Bézout's relation $\;U(X)(X+3)+V(X)(X-4)=1$, the inverse isomorphisme is given by $$(S\bmod (X+3), T\bmod(X-4))\longmapsto TU(X+3)+SV(X-4)\bmod(X^2\!-X-12) .$$

Now a Bézout's relation can be found with the extended Euclidean algorithm, but in the present case it is even shorter:$(X+3)-(X-4)=7$, so we simply have $$\frac17(X+3)-\frac17(X-4)=1$$ and given that $\:P\bmod(X+3)=-1$, $P\bmod(X-4)=13$, we obtain readily $$P\bmod(X^2-X-12)=\frac{13}7(X+3)+\frac17(X-4)=2X+5.$$

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$P(x)=(x^2-x-12)Q(x)+ax+b$

$P(4)=4a+b=13$

$P(-3)=-3a+b=-1$

$a=2, b=5$

$P(x)=(x^2-x-12)Q(x)+2x+5$

$R(x)=2x+5$

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