How to rearrange a quadratic into its factorized form? Like the title says, I'm a bit confused about how the smart people of past centuries figured out that the quadratic:
$$
ax^2+bx+c = a(x-x_1)(x-x_2).
$$
The book I have at hand shows how to do it the other way, going from $a(x-x_1)(x-x_2)$ to $ax^2+bx+c$, and their method is based on the already proven assumption, that the roots (if they exist) of a quadratic equation can be solved using the quadratic formula (acquired by completing the square), that I'm not going to spell out here because doing it in plain text is a pain. Basically they start out with $a(x-x_1)(x-x_2)$, expand it, and substitute the quadratic formula in place of $x_1$ and $x_2$.
So my question is, how does one go from the sum form of a quadratic polynomial $ax^2+bx+c$, to the factorized form $a(x-x_1)(x-x_2)$ (NOT the other way around) simply by manipulating the sentence $ax^2+bx+c$? Is this even possible, and if not, how is it done? Does completing the square have something to do with it?
I'm slightly afraid that the answer is going to be that it isn't possible, and that it is going to entail the fundamental theorem of algebra, since I don't really have any knowledge about how and why it works, but screw it. If somebody could give a detailed answer to my question, that would be great.
 A: Completing square does the job,
$$ax^2+bx+c=a(x^2+\frac{b}{a}x+\frac{c}{a})=a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})=a((x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a^2})=a(x+\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a})(x+\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a})$$
You could also do it with Vietas formulas
$$ax^2+bx+c=a(x^2+\frac{b}{a}x+\frac{c}{a})=a(x^2-(x_1+x_2)x+x_1x_2)=a(x(x-x_1)-x_2(x-x_1))=a(x-x_1)(x-x_2)$$
A: You just complete the square:
\begin{align}
ax^2+bx+c&=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)\\
&=a\left(x^2+2\frac{b}{2a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}\right)\\
&=a\left(\left(x+\frac{b}{2a}\right)^{\!2}-\frac{b^2-4ac}{4a^2}\right)
\end{align}
Now, if $b^2-4ac\ge0$ we can set $\sqrt{b^2-4ac}=d\ge0$ and use $r^2-s^2=(r+s)(r-s)$:
\begin{align}
ax^2+bx+c&=a\left(\left(x+\frac{b}{2a}\right)^{\!2}-\frac{d^2}{4a^2}\right)\\
&=a\left(x+\frac{b}{2a}+\frac{d}{2a}\right)\left(x+\frac{b}{2a}-\frac{d}{2a}\right)
\end{align}
and you're done by setting
$$
x_1=\frac{-b-d}{2a},\qquad
x_2=\frac{-b+d}{2a}.
$$
A: Here is a quite different approach. Given $f(x)=ax^2+bx+c$ and $x_1$ that solves $f(x)=0$, I claim that $(x-x_1)$ divides $f(x)$. From long division of polynomials over the real numbers we know that we can divide $f(x)$ by $(x-x_1)$ and obtain a remainder $r(x)$ of degree strictly less than the divisor $(x-x_1)$:
$$
f(x)=g(x)(x-x_1)+r(x)
$$
Since $r(x)$ has lower degree than $(x-x_1)$ it must be some constant. Now, since $f(x_1)=0=g(x_1)(x_1-x_1)+r(x_1)$ it follows that $r(x_1)=0$ so $r(x)$, which is constant, must be the zero function $r(x)=0$. We thus have
$$
f(x)=g(x)(x-x_1)
$$
Considering degrees, we must have $g(x)$ of degree 1. Thus $g(x)=ax+k$ so we can write
$$
f(x)=(ax+k)(x-x_1)=a(x-x_1)(x+k/a)
$$
So by assigning $x_2:=-k/a$ we have rewritten $f(x)$ as
$$
f(x)=a(x-x_1)(x-x_2)
$$
which is what we wanted to do.
