Dividing polynomials A polynomial $f(x)$ 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 reminder when $f(x)$ is divided by $(x-3)(x-1)^2 $ is 
$a)$ $2x^2 -2x+3$
$b)$ $6x-3$
$c)$ $x^2 + 2x$ 
$d)$ $3x+6$   
I have tried using the rule $ f(x) = p(x)q(x) + r(x) $ and then later trying to substitute the zeros of the roots in the same expression. 
That is if we substitute x=3 in the equation written just above we get the result that $ f(3) = 15 $ since the first term tends to zero as 3 is the zero for $p(x)$.
Two , from the following logic I tried checking the options to see whether they satisfy the conditions $f(3) = 15$ and $f(1) = 3$.
Where I am getting stuck is that options 1 , 2 and 3 all of them satisfy the conditions mentioned above. 
The correct answer mentioned is $ (a) $.

Suppose $f=p.(x-3)(x-1)^2+r$ and $r=q_1(x-3)+r_1$ and $r=q_2(x-1)^2+r_2$ 
Now $$f=p(x-3)(x-1)^2+r=(x-3)(p(x-1)^2+q_1)+r_1$$ and $$f=p(x-3)(x-1)^2+r=(x-1)^2(p(x-3)+q_2)+r_2$$  
Thus we should have $$r_1=15 \quad *$$ and $$r_2=2x+1 \quad **$$
Because the reminder of $f$ by $x-3$ and $(x-1)^2$ respectively should be $15$ and $2x+1$ .  
The first three options satisfied first equation, but we see
$2x^2-2x+3=2(x-1)^2+2x+1$
$6x-3=0.(x-1)^2+6x-3$
$x^2+2x=(x-1)^2+4x-1$  
So the answer is a !
