<s>[INST] 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.
 [/INST] Your answer is correct.
The second answer cannot be also true, as $(-3x^2 + 2x + 3) - (-3x^2 + 2x + 1) =2 $ is not divisible by $(x-1)x^2$.
Here is another elementary solution:


*

*$(1):F(x) = (x-1)^2p(x) + x+1$

*$(2):F(x) = x^2q(x) + 2x+3$

*Setting $p(x) =x^2f(x)+ax+b$ you get


$$(x-1)^2p(x) + x+1 = x^2q(x) -2bx+ax+b+x+1 $$
$$\stackrel{(2)}{\Rightarrow} b+1 = 3, 1+a-2b = 2\Rightarrow \boxed{b=2, a= 5}$$
$$\Rightarrow p(x) =x^2f(x)+5x+2$$
Plugging this into $(1)$, you need only consider
$$(x-1)^2(5x+2) + x+1 = 5x^3-8x^2+2x+3 $$
$$= 5(x^3 - x^2) \boxed{- 3x^2+2x+3}$$
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