If $g \vert f,$ where $g=(X-1)^2,\ f=aX^{n+2}+bX^n+2,\ a,b\in\mathbb{R}$, determine $q(1)$, where $q\in\mathbb{R}[X]$ is the quotient of $f:g$
This seems like an easy problem, but I'm not quite sure how to find $q(1)$ after determining $a$ and $b$. My logic was: $$g\vert f \Rightarrow \begin{cases}f(1) = 0 \iff a+b+2 = 0 \\ f'(1) = 0 \iff (n+2)a+n\cdot b=0\end{cases} \Rightarrow a=n, b =-(n+2)$$ Here, however, is where I'm quite stuck. My attempt was to now calculate $$f''(1)=n(n+2)(n+1) -(n+2)n(n-1) = 2n^2+4n$$ which turns out to be very close to the real answer my workbook gives, which is $q(1)=n^2+2n$, but I was never taught if this is the correct approach. It feels more like it just so happened for my approach to be close to the answer, so I don't think I'm correct at all. Any hints towards the rigorous solution are much appreciated!
Your computations are correct - but if $f = (x-1)^2 q$, then $f' = 2(x-1)q + (x-1)^2 q'$ and $f'' = 2q+4(x-1)q'+(x-1)^2q''$, so $f''(1) = 2q(1)$.