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Let $P(x)$ be a polynomial which when divided by $x-3$ and $x-5$ leaves remainders $10$ and $6$ respectively. If $P(x)$ is divided by $(x-3)(x-5)$ then remainder is:

I assumed $P(x)=Q(x)\cdot (x-3)+10=R(x)\cdot (x-5)+6$ but it hasn't worked.

Any helpful answer will be appreciated by me.

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Let $P(x)=(x-3)(x-5)Q(x)+xA+B$
$P(3)=10=3A+B$
$P(5)=6=5A+B$

This gives $A=-2$,$B=16$

Thus by division by $(x-5)(x-3)$, ans will be $-2x+16$.

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Hint:

Let $P(x)=G(x)(x-3)(x-5)+A(x-3)+B(x-5)$ where $A,B$ are arbitrary constants.

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Using the Chinese remainder theorem, and the formula for solving a system of congruences mod $x-3$ and mod $x-5$:

A Bézout's relation between $x-3$ and $x-5$ is $$\frac12\bigl((x-3)-(x-5)\bigr)=1$$ hence the remainder will be $$R(x)=\frac12\bigl(6(x-3)-10(x-5)\bigr)=-2x+16$$

Note this is but the linear interpolation formula between the points $(3,10)$ and $(5,6)$.

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