Blockage

$f(x)=(x-1)^2Q_1(x)+(2x-1)\quad \Rightarrow \quad \color{#D61F06}{f(1)=1}$

$f'(x)=[2(x-1)Q_1(x)+(x-1)^2Q'_1(x)]+2=(x-1)[2Q_1(x)+(x-1)Q'_1(x)]+2 \quad \Rightarrow \quad \color{#D61F06}{ f'(1)=2 }$

$f(x)=(x-3)Q_2(x)+15 \quad \Rightarrow \quad \color{#D61F06}{ f(3)=15}$

$f(x)=(x-1)^2(x-3)Q_3(x)+(ax^2+bx+c)\quad \Rightarrow \quad \color{#D61F06}{ f(1)=a+b+c=1} \wedge \color{#D61F06}{f(3)=9a+3b+c=15}$

$f'(x)=(x-1)[2(x-3)Q_3(x)+(x-1)[(x-3)Q_3(x)]']+2ax+b \quad \Rightarrow \quad \color{#D61F06}{f'(1)=2a+b=2}$

$\begin{vmatrix} a+b+c=1 \\ 9a+3b+c=15 \\ 2a+b=2 \end{vmatrix} \quad \Rightarrow \quad \begin{vmatrix} a=\frac{5}{2} \\ b=-3 \\ c=\frac{3}{2} \end{vmatrix}$

$a+2b=\frac{5}{2}-6=-\frac{7}{2}=-3.5$

It is given that $f(x) = (x-1)^2(x-3)g(x) + ax^2+bx+c$ .

And that $\dfrac{f(x)}{(x-1)^2}$ leaves a remainder of $2x-1$ .

$\begin{aligned} \Rightarrow \dfrac{ax^2+bx+c}{(x-1)^2} & = a(x-1)^2 + \dfrac{(b+2a)x+c-a}{(x-1)^2} \end{aligned}$

$\Rightarrow \begin{cases} b+2a = 2 & \Rightarrow \color{#3D99F6} {b = 2 -2a} \\ c -a = -1 & \Rightarrow \color{#D61F06} {c=a-1 } \end{cases}$

And that $\dfrac{f(x)}{x-3}$ leaves a remainder of $15$ .

$\begin{aligned} \Rightarrow 3^2a + 3\color{#3D99F6}{b} + \color{#D61F06}{c} & = 15 \\ 9a + 3\color{#3D99F6}{(2-2a)} + \color{#D61F06}{a-1} & = 15 \\ 4a & = 10 \\ a & = 2.5 \\ \Rightarrow b & = 2 - 2(2.5) = -3 \\ \Rightarrow a + 2b & = 2.5 + 2(-3) = \boxed{-3.5} \end{aligned}$