Why do the 3 solutions of $x^3 + ax^2 + bx + c$ add up to $-a$?

Question

Elliptic Tales makes the claim that:

The three solutions of the equation $x^3 + ax^2 + bx +c = 0$ add up to $-a$.

The reason is:

Why? Suppose that the three solutions are $x_1,x_2,x_3$. Then we know that $x^3 + ax^2 + bx + c = (x - x_1)(x - x_2)(x - x_3)$ and multiplying the right-hand side and looking at the coefficient of $x^2$ tells us that $-a = x_1 + x_2 + x_3$.

However, I plugged

$x^3 + ax^2 + bx + c = (x - x_1)(x-x_2)(x-x_3)$, solve for $a$

into Wolfram Alpha & did not get this elegant result. I just got the expected result of:

• subtracting $x^3$, $bx$, and $c$
• dividing by $x^2$

How does the book get this clean result? I've looked at the explanation, but I can't understand it.

Answer 1

From the fact that\begin{multline}(x-x_1)(x-x_2)(x-x_3)=\\=x^3-x_1 x^2-x_2 x^2-x_3 x^2+x_1 x_2 x+x_1 x_3 x+x_2 x_3 x-x_1x_2 x_3=\\=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2x_3)x-x_1x_2x_3.\end{multline}

Answer 2

The identity is obtained by identification, i.e. noting that the two polynomials coincide for all $x$, hence have the same coefficients.

In the same vein but a harder way, differentiate the polynomial twice and set $x=0$. This gives you $2a$, as well as

$$\left.((x-x_1)(x-x_2)+(x-x_0)(x-x_2)+(x-x_0)(x-x_1))'\right|_{x=0} \\=-x_2-x_1-x_0-x_2-x_0-x_1.$$