Sum of distinct roots of $g(-x+1)=f(x)$ 
Let $f(x)$ and $g(x)$ be two cubic polynomials such that $g(-x+1)=f(x)$. If the sum of the distinct roots of $f(x)$ is $10$, which of the following cannot be the sum of the distinct roots of $g(x)$?
a) $-10$, b) $-9$, c) $-8$ or d) $-7$.

I tried to approach the problem the following way:
Since $f(x) = g(-x+1)$ and for $f(x)$ we have that $\alpha+\beta=10$, this would imply that $f(\alpha)=0$ and $f(\beta)=0$. So from here $g(-\alpha+1)=0$ and $g(-\beta +1)=0$. And the sum of the roots for $g(x)$ would be $(-\alpha+1)+(-\beta+1)$. So if we let the sum of the roots be equal to some $k$ we would have that $(-\alpha+1)+(-\beta+1) =k$ and from here we can get this expression to the form $\alpha+\beta=-2k$, so $k$ has to be a multiple of $2$. The only ones that aren’t multiples of two from the solutions are $-7$ and $-9$. Turns out that the correct answer for this was $-10$. Could someone enlighten me on where am I going downhill with my reasoning?
Edit: Looking back I seem to have multiple mistakes here…
 A: Let's consider all the cases:

$\bullet$ $$f(x) = (x-a)(x-b)(x-c) $$
Here $a+b+c=10$.
So sum of distinct roots of $g(x)$ is :
$$=(-a+1)+(-b+1)+(-c+1) = -(10)+3 =-7$$

$\bullet$
$$f(x) = (x-a)(x-b)^2 $$
Here $a+b=10$
So sum of distinct roots of $g(x)$ is :
$$=(-a+1)+(-b+1) = -10+2=-8$$

$\bullet$
$$f(x)=(x-a)^3$$
Here $a=10$
So sum of distinct roots of $g(x)$ is :
$$=-a+1=-10+1=-9$$

Here since all cases are considered, $-10$ definitely does not occur as the sum of distinct roots of $g(x)$. 
A: Here are your mistakes:


*

*It seems that you only assume the case that there can be only two distinct roots for $f$. 

*Also from $-\alpha +1 -\beta +1 = k$, $-\alpha - \beta = k-2$. The claim that $\alpha +\beta = -2k$ is not true and I can't follow the reasoning that $k$ must be a multiple of $2$.



Notice that if $\alpha_i$ is a root of $f$, then $-\alpha_i+1$ is the root of $g$. Suppose we have $l$ distinct roots. 
Then the sum of the distinct roots, would be $$\sum_{i=1}^l (-\alpha_i+1)=-\sum_{i=1}^l\alpha_i +l = -10+l \ne -10$$
Note that $l$ takes value from $1$ to $3$ for a cubic equation. 
