How to derive this factorization $ax^2 + bx + c = a(x − x_1)(x − x_2) $? I faced this factorization formula $$ax^2 + bx + c = a(x − x_1)(x − x_2)$$ where $x_1$ is the first solution and $x_2$ is the second one. But I don't understand how the formula gets derived... Could anybody explain it to me? Thanks!
I know I can solve a quadratic equation and substitute the solutions into this formula $a(x − x_1)(x − x_2)$, but I want to understand how it works.
 A: By the Factor theorem for a polynomial of degree $n$ we have $$p_n(x_1)=0 \iff p_n(x)=(x-x_1)q_{n-1}(x)$$
therefore for $p_2(x)=ax^2+bx+c$, given the roots $x_1$ and $x_2$, we have
$$p_2(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$$
A: Since $$x_1+x_2 = -{b\over a} \implies b= -a(x_1+x_2)$$ and $$x_1x_2 = {c\over a} \implies c= ax_1x_2$$
so \begin{eqnarray} ax^2+bx+c &=& ax^2-a(x_1+x_2)x+ax_1x_2\\ &=& a(x^2-x_1x-x_2x+x_1x_2)\\ &=& a(x-x_1)(x-x_2)\end{eqnarray}
A: You can also complete the square. Assuming $a \neq 0$ (necessary to be a quadratic):
\begin{align*}
ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) \\
&= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} + \frac{c}{a} - \frac{b^2}{4a^2}\right) \\
&= a\left(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right).
\end{align*}
Then, using a difference of two squares factorisation,
$$ax^2 + bx + c = a\left(x + \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}\right)\left(x + \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}\right),$$
as requested.
A: $a\neq 0$ so you can reduce to the case 
$x^2+b’x+c’=0$ where $b’=\frac{b}{a}$ and $c’=\frac{c}{a}$
$x^2+b’x+c’=(x^2+2\frac{b’}{2}x+(\frac{b’}{2})^2)-((\frac{b’}{2})^2-c’)=$
$(x+\frac{b’}{2})^2-\frac{1}{4}(b’^2-4c’)=$
$
(x+\frac{b’}{2}-\frac{1}{2}\sqrt{b’^2-4c’})(x+\frac{b’}{2}+\frac{1}{2}\sqrt{b’^2-4c’})=0$
and so 
$x_{1,2}=\frac{-b’ +/- \sqrt{b’^2-4c’}}{2}$
and 
$x_{1,2}=\frac{-b +/- \sqrt{b^2-4ac}}{2a}$
Now if you calculate $ a(x-x_1)(x-x_2)$
with that solutions you get the initial equation 
$ax^2+bx+c$
There is a more elegant method with which you can prove the equality, the  polinomial division. 
A: These are two common ways of describing a polynomial: a polynomial written the first way is relatively easy to compute the value of at a given $x$, the second makes it easy to find the roots, being $x_0$ and $x_1$.
They are, of course, otherwise equivalent, as can be seen by expanding the roots form and comparing the coefficients of the various powers of $x$: we get $a=a$, $b=a(x_0+x_1)$, and $c=ax_0x_1$.
The fact that polynomials can be factored in this way is called The Fundamental Theorem Of Algebra: a polynomial of degree $n$ (that is, its highest exponent with a non-zero coefficient is $n$) has exactly $n$ complex, possibly duplicate roots.  Unfortunately a full justification requires way more than just algebra to complete, but notice how:


*

*Both $ax^2+bx+c$ and $a(x-x_0)(x-x_1)$ have, when expanded, $x^2$,$x$, and constant terms, and

*Both have three free constants.


These suggest at least that they should be able to cover the same ground.
