Given remainders from other polynomial divisions. Find the remainder in a polynomial division Let $F(x)$ be a polynomial. 
If $F(x)$ is divided by $(x-1)^2$ the remainder will be $x+1$
and if $F(x)$ is divided by $x^2$ the remainder will be $2x+3$.
What is the remainder if $F(x)$ is divided by $x^2(x-1)$?

My solution : 
 $F(x) = (x-1)^2 P(x) + x+1$, substitute $x=1$ and get $F(1) = 2$.
Differentiate it and get $F'(x) = (x-1)^2 P'(x)+P(x)(2)(x-1)+1$ then $F'(1)=1$.
And do the same to the other equation $( F(0)=3, F'(0) = 2 )$.
Let the remainder of $F(x)$ divided by $x^2(x-1)$ be $R(x)$. $( R(x) = ax^2+bx+c)$
so $F(x) =x^2(x-1)A(x) + R(x)$
$F(0) = R(0) = 3$
$F(1) = R(1) = 2$
$F'(x) = (x)^2(x-1) A'(x)+x^2A(x)+2x(x-1)A(x) + R'(x)$
$F'(0) = R'(0) = 2$ 
from $R(0) = 3$ then $c=3$
from $R'(0) = 2$ then $R'(x) = 2ax+b$ then $R'(0) = b = 2$
$R(x) = ax^2+2x+3$ from $R(1) = 2$, will get $a+5 = 2 , a=-3$
so the remainder will be  $-3x^2+2x+3$
but one said that the answer can be $-3x^2+2x+1$ too, so what's correct answer.

$\!\!\overbrace{\bmod x\!-\!1\!:}^{\large x\ \equiv\ 1}\,\ \ f\equiv \overbrace{1\!+\!x}^{\Large 2} \equiv \overbrace{2x\!+\!3\! +\! a x^2}^{\Large 2\ +\ 3\ +\ a\ \ \ }\!\iff\! a=-3\!\iff\! f\equiv\overbrace{ 2x\!+\!3\!-\!3x^2}^{\large\rm your\ answer\ \checkmark}\!\!\pmod{\!x^2(x\!-\!1)}$
Update You mention $\!\bmod\!$ is unknown, so  we translate the above into more elementary language. $\ f = 2x+3 + ax^2\ $ and $\,1+x\,$ have equal remainders when dividied by $x-1\,$ iff $\,a=-3,\ $ which is easily verified since a polynomial $\, g(x)\, $ has remainder $g(1)$ when divided by $\,x\!-\!1\,$ by the Polynomial Remainder Theorem. Said  more simply and explicitly, with full details, we have:
$ f = \color{#0a0}{\,x\!+\!1 + q\cdot (x\!-\!1)^2}\ $ by hypothesis, and also by hypothesis we have  
$ f = 2x\!+\!3 + \color{#c00}g\cdot x^2.\, $   Substitute $\,\color{#c00}{g  = a + (x\!-\!1)\,h}\ $ (by dividing $\,g\,$ by $x\!-\!1)\,$ to get 
$ f = 2x\!+\!3 + \color{#c00}a\cdot x^2 + x^2\color{#c00}{(x\!-\!1)\,h}\, =\, \color{#0a0}{x\!+\!1 + q\cdot (x\!-\!1)^2} $
Eval at $\,x=1\:\Rightarrow\: 2+3+\color{#c00}{a+0}\: =\: \color{#0a0}{1+1 +0}\ $ so $\ \color{#c00}{a = -3},\, $ which substituted above yields 
$ f = 2x\!+\!3  \color{#c00}{-3}\cdot x^2 + x^2\color{#c00}{(x\!-\!1)\,h},\, $ which agrees with your result.
