<s>[INST] Remainder of the polynomial A polynomial function $f(x)$  with real coefficients leaves the remainder $15$ when divided by $x-3$, and the remainder $2x+1$ when divided by $(x-1)^2$. Then the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$ is?
What I have thought-The remainder must be of the form $ax^2+bx+c$. Now applying the remainder theorem, I am able to find $2$ equations in $a,b,c$ . eg
Let $f(x)$=$(x-3)$$h(x)$+$15$             ...................................(1)
Also let $f(x)=(x-3)(x-1)^2g(x)+ax^2+bx+c$...........................(2)
Put $x=3$ and using (1) we get $15=9a+3b+c$ 
Similarly I can get another equation using the other information given. But I am only able to get 2 equations in 3 variables. From where do I get the 3rd equation in $a,b,c$ and hence the remainder?
 [/INST] Note that
$$
\frac{1}{4}(x-1)^2-\frac{1}{4}(x+1)(x-3)=1.
$$
You can get this by dividing $(x-1)^2$ by $x-3$. So take
$$
f(x)=15\frac{1}{4}(x-1)^2-(2x+1)\frac{1}{4}(x+1)(x-3).
$$
Then the remainder when $f(x)$ is divided by $x-3$ is 15 and by $(x-1)^2$ is $2x+1$ since $(x+1)(x-3)=[(x-1)+2][(x-1)-2]=(x-1)^2-4$.
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