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Again, for my Equation Theory class, I have the subject question.$p(x)$ has a remainder of 3 when divided by $x-1$ and a remainder of 5 when divided by $x-3$. What is the remainder when $p(x)$ is divided by $(x-1)(x-3)$?

I've been able to list out linear expressions that satisfy each of the first two requirements. The expression I found that satisfies both requirements is $x+2$ and the remainder when divided by $(x-1)(x-3)$ is $\frac{x+2}{(x-1)(x-3)}$.

While I am confident that $p(x)=x+2$ is an acceptable polynomial and the fractional remainder is an acceptable answer, I am also certain that there is another "nicer" answer. I just can't figure out how to find it.

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  • $\begingroup$ The remainder will be of the shape $ax+b$, no denominator, no $(x-1)(x-3)$. $\endgroup$ – André Nicolas Jul 13 '14 at 17:15
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Here is a direct explanation using polynomial division to show how the quotient and remainder come into play.

Using the division algorithm with $(x-1)(x-3)$ and $p(x)$ we know we can express $$p(x)=(x-1)(x-3)q(x)+r(x)=(x-1)(x-3)q(x)+ax+b$$ for some polynomial $q(x)$ which we don't need to compute, and the remainder $r(x)$ with degree less than $2$

Then $p(1)=a+b=3$ and $p(3)=3a+b=5$ whence $a=1, b=2$ as you found and $r(x)=x+2$

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HINT:

Let $\displaystyle p(x)\equiv ax+b\pmod{(x-1)(x-3)}$

$\displaystyle\implies p(x)\equiv ax+b\pmod{(x-1)}\equiv a(x-1)+a+b\equiv a+b$

$\displaystyle\implies a+b=3$

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