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We have a polynomial $f(x)$ with rational roots that leaves remainders $15, 2x + 1$ when divided by the polynomials $x - 3, (x-1)^2$ respectively. What is the remainder when $f(x)$ is divided by $(x - 3)(x-1)^2$?

From the remainder theorem we know that,

$f(3) = 15$.

I thought quite hard about this and tried various methods, but couldn't come up with anything useful.

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Write $$f(x)=A(x-3)(x-1)^2+B(x-1)^2+C(x-3)+D$$ where $A$ is an integer polynomial or constant

$\implies 15=f(3)=4B+D\implies D=15-4B$

$\implies f(x)=A(x-3)(x-1)^2+B(x-1)^2+C(x-3)+15-4B$

Now, $f(x)\equiv C(x-3)+15-4B\pmod{(x-1)^2}\equiv Cx+15-4B-3C $

But $f(x)\equiv 2x+1\pmod{(x-1)^2}$

$\implies C=2$ and $15-4B-3C=1$ and so on


HINT:

Write $$f(x)=A(x-3)(x-1)^2+Bx^2+Cx+D$$

$\implies 15=f(3)=9B+3C+D\ \ \ \ (1)$

Now, $f(x)\equiv Bx^2+Cx+D\pmod{(x-1)^2}\equiv(2B+C)x+D-B\pmod{(x-1)^2} $

But $f(x)\equiv2x+1\pmod{(x-1)^2}$

$\implies 2B+C=2\ \ \ \ (2)$ and $D-B=1\ \ \ \ (3)$

Can you solve for $B,C,D$ from $(1),(2),(3)$?

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