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 $a+b+c = 4$
 A: 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  $r(x)=-2x^2+2x+4$
A: $\ f = \color{#0a0}{3+x + q\cdot (x\!-\!1)^2}\ $ by hypothesis, and also by hypothesis we have  
$\ f = 4\!+\!2x + \color{#c00}g\cdot x^2.\, $ Put $\,\color{#c00}{g  = a} + \color{#89c}{(x\!-\!1)\,h}\,\ $ ($=$ division of $\,g\,$ by $x\!-\!1)\ $ so 
$\bbox[6px,border:1px solid #c00]{ f = 4\!+\!2x + \color{#c00}a\cdot x^2 + x^2\color{#89c}{(x\!-\!1)\,h}}\, =\, \color{#0a0}{3\!+\!x + q\cdot (x\!-\!1)^2} $
Eval'ed at $\,x=1\:\Rightarrow\: 4+2+\color{#c00}a+\color{#89c}0\: =\, \color{#0a0}{3\!+\!1 +0}\ $ so $\ \bbox[6px,border:1px solid #c00]{\color{#c00}{a = -2}}\ \ $  QED
Remark $ $ If you know (Easy) CRT then it immediately yields the general result as below
$\begin{align}&f\equiv \color{#c00}a\!\!\!\pmod{\!\color{#c00}g}\\ &f\equiv\color{#0a0}b\!\!\!\pmod{\!x\!-\!1}\end{align}\!\!\!\!\iff\!\!  f \equiv a\! +\! g\left[ \dfrac{b\!-\!a}g\bmod x\!-\!1\right]\equiv \color{}a\! +\! \left[ \color{#c00}{\dfrac{\color{#0a0}{b(1)}\!-\!a(1)}{g(1)}}\right] g\ \pmod{(x\!-\!1)g}$
$\begin{align}&f\,\equiv\, \color{#c00}{4+2x}\!\pmod{\!x^2}\\ &f\,\equiv\,\color{#0a0}{3\,+\,x} \pmod{\!x-1}\end{align}\ \ \ \iff\ \ \ \  f\ \equiv\ \color{}{4 + 2x} \ +\ \ \underbrace{\left[\color{#c00}{\dfrac{\color{#0a0}{3\!+\!1}\!-\!(4\!+\!2)}{1^2}  }\right]}_{\Large -2\ \ \ }x^2 \pmod{(x\!-\!1)x^2}$
The computation is so easy because we chose $\,x\!-\!1\,$ (vs.  $x^2)\,$ as the modulus in the formula, which simplifies mod arithmetic since $\,f(x)\bmod x\!-\!1 = \color{#c00}{f(1)}\,$ by the Polynomial Remainder Theorem. Generally CRT computations are simpler when we solve last the congruences with least moduli.
A: I am going to follow your approach. We have
$$f(x)=(x-1)^2p(x)+x+3$$
So, $f(1)=4$.
We also have that 
$$f(x)=x^2q(x)+2x+4$$
And we want to find $a,b,c$ in 
$$f(x)=(x-1)x^2g(x)+ax^2+bx+c$$
Plugin $x=1$ you get $4=a+b+c$.
Plugin $x=0$ doesn't give us enough information since this only gives us the remainder after dividing by $x$(instead of $x^2$).
So, instead of plugin, we look $f$ mod $x^2$. We have
$$f(x)=x^2[(x-1)g(x)+a]+(bx+c)$$
Since we are given that the remainder of dividing $f(x)$ by $x^2$ is $2x+4$, we conclude $bx+c=2x+4$, i.e.  $b=2$, $c=4$. And therefore, $a=-2$.
We conclude that the reminder of $f$ dividing by $ (x-1)x^2$ is $-2x^2+2x+4$.
A: General method, without the formal derivatives. 
Suppose
$$
f(x) = g(x) (x-1)^2 + (x+3)= h(x) x^2 + (2x + 4). 
$$
Then 
$$
(x-1)^2 f(x) = (x-1)^2 x^2 h(x) + (x-1)^2 (2x+4) \tag 1
$$
and
$$
x^2 f(x) = x^2 (x-1)^2 g(x) + x^2(x+3). \tag 2
$$
Now do the Euclidean algorithm to $x^2, (x-1)^2$:
\begin{align*}
x^2 &= (x-1)^2 + (2x-1), \\
(x-1)^2 &= \frac 14 (2x-1)(2x - 3) + \frac 14, 
\end{align*}
therefore
$$
1 = 4(x-1)^2 - (2x - 1) (2x-3) = 4(x-1)^2 - (x^2 - (x-1)^2) (2x-3) = (x-1)^2 (4 + 2x-3) - x^2 (2x - 3) = \color{blue}{(x-1)^2 (2x+1) - x^2 (2x - 3)}. 
$$
Thus $(2x+1) \cdot \mathrm {Eq}(1) - (2x - 3) \cdot \mathrm {Eq} (2)$ yields
$$
f(x) = (x-1)^2 x^2 F(x) + (2x+4) (2x+1)(x-1)^2 - (x+3)(2x-3)x^2.
$$
Since
$$
(2x+4) (2x+1)(x-1)^2 - (x+3)(2x-3)x^2 = (4x^2 + 10x + 4)(x-1)^2 - (2x^2 + 3x -9)x^2 = (4x^3 +6x^2 -6x - 4)(x-1) - ((2x+5)(x-1) -4)x^2 = x^2 (x-1) G(x) + ((-6x -4)(x-1) + 4x^2) = x^2(x-1)G(x) + (\color{red}{-2x^2 +2x +4}), 
$$
we have
$$
f(x) = x(x-1)^2 (G(x)+ xF(x)) + (\color{red}{-2x^2 +2x +4}),
$$
which means the remainder is 
$$
\color{red}{-2x^2 +2x +4}. 
$$
A: Adapting the Extended Euclidean Algorithm, as implemented in this answer, to polynomial division and rotating to go down the page rather than across:
$$
\begin{array}{c|cc|c}
\color{#C00}{x^2}&\color{#C00}{1}&0\\
\color{#090}{x^2-2x+1}&0&\color{#090}{1}\\
2x-1&1&-1&1\\
\color{#C90}{\frac14}&\color{#C00}{-\frac12x+\frac34}&\color{#090}{\frac12x+\frac14}&\frac12x-\frac34\\
0&4x^2-8x+4&-4x^2&8x-4
\end{array}\tag1
$$
which says
$$
\overbrace{(2x+1)}^{4\left(\frac12x+\frac14\right)}(x-1)^2+\overbrace{(-2x+3)}^{4\left(-\frac12x+\frac34\right)}x^2=\overbrace{\ \quad1\ \quad}^{4\cdot\frac14}\tag2
$$
Therefore
$$
(2x+1)(x-1)^2\equiv\left\{\begin{align}1&\pmod{x^2}\\0&\pmod{(x-1)^2}\end{align}\right.\tag3
$$
and
$$
(-2x+3)x^2\equiv\left\{\begin{align}0&\pmod{x^2}\\1&\pmod{(x-1)^2}\end{align}\right.\tag4
$$
Thus, mod $x^2(x-1)^2$, the polynomial is
$$
\begin{align}
&(2x+4)\overbrace{(2x+1)(x-1)^2}^{(3)}+(x+3)\overbrace{(-2x+3)x^2}^{(4)}\\
&=2x^4-x^3-3x^2+2x+4\\
&\equiv3x^3-5x^2+2x+4&&\bmod x^2(x-1)^2\\
&\equiv-2x^2+2x+4&&\bmod x^2(x-1)\tag5
\end{align}
$$
which we can say since $\left.x^2(x-1)\,\middle|\,x^2(x-1)^2\right.$.
A: $f(x)\equiv{x+3}\pmod{(x-1)^2}\wedge f(x)\equiv{2x+4}\pmod{x^2} \overset{CRT}{\implies} f(x)\equiv{3x^3 - 5x^2 + 2x + 4}\pmod{(x-1)^2x^2}$
? 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)

