Find the sum of $x_1+x_2+x_3$ of intercept points Suppose that the straight line $L$ meets the curve $y=3x^3-15x^2+7x-8$ in three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$. Then $x_1+x_2+x_3=?$
A) 3  $\quad$   B) 4   $\quad$       C) 5   $\quad$    D) 6  $\quad$      E) 7
At the beginning, my main idea is to use Vieta's theorem $x_1+x_2+x_3=-\frac{b}{a}$, and find the answer is $5$, it is correct if $y=0$. But when I use software to draw the grapic of $y=3x^3-15x^2+7x-8$ and a line intercept at three points,I also changed the coefficient of the straight line and use calculator did the summation,still can get $x_1+x_2+x_3=5$, I want to know the general solution of this question  

If polynomial $p(x) = ax^n + bx^{n-1} + ...$ has roots $x_i$, then $x_1+\dots+x_n=-\frac{b}{a}$, as you noted.
Now, though, you don't want $p(x)=0$, you want $p(x)=mx+b$.
So, to use Vieta's theorem, you need to define a new polynomial:
Let $q(x)=p(x)-mx-b$. Then, your $x_i$ are the roots of $q(x)$, Vieta's theorem applies, and (because $p$ is cubic or higher), $p$ and $q$ have the same coefficient of $x^n$ and $x^{n-1}$. 
