Simply factoring a quadratic equation

Question

On pp 255 - 256 (footnote 7) of "Love & Math", Edward Frenkel states that we can factor a quadratic in terms of its solutions $x_1$ and $x_2$ as:

$ax^2 + bx + c = a(x - x_1)(x - x_2)$

Where does this come from? I just cannot derive that.

I can get his next result $-\frac{b}{a} = (x_1 + x_2)$ just by simultaneously solving $a{x_1}^2 + bx_1 + c = 0$ and $a{x_2}^2 + bx_2 + c = 0$:

$a({x_1}^2 - {x_2}^2) + b(x_1 - x_2) = 0$

$a(x_1 + x_2) + b = 0$

$-\frac{b}{a} = (x_1 + x_2)$

I appreciate that this is probably quite a simple result but it has passed me by.

Answer 1

If $p(x)$ is a polynomial of the set $P^n(X)$, and $x_1$ satisfies $p(x_1) = 0$ then we have

$p(x) = q(x).(x-x_1)$ where $q(x)$ is a polynomial of the set $P^{n-1}(X)$.

Thus we can write $p(x) = h(x).(x-x_1).(x-x_2)$ where $h(x)$ is an element of $P^0(X)$, i.e. a constant.

Expanding both equations:

$ax^2 + bx + c = h.( x^2 - (x_1+x_2).x + x_1.x_2 ) \implies h = a$

Answer 2

Let $p(x) = ax^2 + bx + c$. By solutions they mean $p(x_1) = p(x_2) = 0$. Using the quadratic formula then the $x_i$ are of the form;

$$ \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Without loss let;

$$x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a} \ : \ x_2 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$

Now compute;

$$(x-x_1)(x-x_2) = \left(x + \frac{b- \sqrt{b^2-4ac}}{2a} \right)\left(x+ \frac{b+ \sqrt{b^2-4ac}}{2a} \right)$$

If you expand the above then you get;

$$x^2 + \frac{b}{a} + \frac{c}{a}$$

Now multiplying by $\textbf{a}$ we have;

$$ax^2+b x +c$$