unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.

Here's the question (just the concept as I can't remember precisely). 

An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the number) and when this polynomial is divided by $x^2 $, it leaves $2x + 4$ (again, not sure about the number). From the given conditions, if this polynomial is divided by $(x-1)x^2$, what would be the remainder?

The solution as far as I figured out is this: 

first, from the division of $(x-1)^2$, I got that $f(1) = 3$
in the same way from division of $x^2$, I got $f(0) = 4.$

I can write the polynomial as follows: 

$f(x) = (x-1)(x)(x) g(x) + ax^2 +bx +c$

$ax^2 + bx + c$ is the remainder.  And to find $a,b,c$, I can use the conditions above, so I got $c = 4$ by substituting $x = 0,$ and I got $a+b+4 = 3$ by substituting $x = 1.$

This leaves $a + b = -1,$ and I can't figure out how to continue; please help.

Edit : I made a mistake $f(1)$ should be equal to $4$ and $a+b+c = 4$
We have 

$$f(x)=(x-1)^2q_1(x)+x+3$$

$$f'(x)=2(x-1)q_1(x)+(x-1)^2q_1'(x)+1$$

Where for $x=1$ we have $f(1)=4$ and $f'(1)=1$

Then from   

$$f(x)=x^2q_2(x)+2x+4$$
$$f'(x)=2xq_2(x)+x^2q_2'(x)+2$$

Where for $x=0$ we have $f(0)=4$ and $f'(0)=2$

Now from $$f(x)=x^2(x-1)q_3(x)+ax^2+bx+c$$ and 

$$f'(x)=x^2q_3(x)+2x(x-1)q_3(x)+(x-1)x^2q_3'(x)+2ax+b$$

When we substitute the values $x=0$ and $x=1$ in $f$ and $f'$ we get 

$f(0)=c=4$ and $f(1)=a+b+4=4$ $$a+b=0$$

$f'(0)=b=2$ from this we have $a=-2$.
 Thus the remainder is  $r(x)=-2x^2+2x+4$
$f(x)\equiv{x+3}\pmod{(x-1)^2}\wedge f(x)\equiv{2x+4}\pmod{x^2} \overset{CRT}{\implies} f(x)\equiv{3x^3 - 5x^2 + 2x + 4}\pmod{(x-1)^2x^2}$

? chinese(Mod(x+3,(x-1)^2),Mod(2*x+4,x^2))
%1 = Mod(3*x^3 - 5*x^2 + 2*x + 4, x^4 - 2*x^3 + x^2)


Then $f(x)\equiv{-2x^2 + 2x + 4}\pmod{(x-1)x^2}$

? Mod(3*x^3 - 5*x^2 + 2*x + 4,(x-1)*x^2)
%2 = Mod(-2*x^2 + 2*x + 4, x^3 - x^2)