# How to utilize the remainder theorem when the quotient is unknown?

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!

• HINT: $x^2-x-12=(x-4)(x+3)$ – Tito Eliatron Oct 12 '20 at 19:41
• I see. Is the answer 2x + 5 then? – zotz99 Oct 12 '20 at 19:52

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 $$P\bmod(X^2-X-12)=\frac{13}7(X+3)+\frac17(X-4)=2X+5.$$

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

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

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

$$a=2, b=5$$

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

$$R(x)=2x+5$$