Find the sum of $x_1+x_2+x_3$ of intercept points

Question

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

Answer 1

Let $y=ax+b$ the equation of the straight line $L$. Then $x_1,x_2,x_3$ are solutions of the equation

$3x^3-15x^2+(7-a)x-8-b=0$.

Then Vieta says: $x_1+x_2+x_3= - \frac{-15}{3}=5.$

Answer 2

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 $x^{n-1}$.