The logic of the AC Method of factoring quadratic polynomials This could be primary school stuff. But I want to ask it.
In factoring $x^2+bx+c$ (i.e. $a = 1$ in $ax^2+bx+c$), we find $m$ and $n$ such that $m+n = b$ and $mn=c$. We can reason this well as follows:

In the expansion of $(x+m)(x+n)$, $mx$ and $nx$ should add up to $bx$ and $mn$ should result in $c$.

However I am not able to come up with the logic for following:
For factoring $qx^2+bx+c$ (i.e. $a \neq 1$ in $ax^2+bx+c$), we find $m$ and $n$ such that $m+n=b$ and $mn = ac$. 


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*I am not able to realize the logic behind why it is $mn = ac$ and 

*how doing this result in correct factors


If $m,n $ are the roots of $ax^2 + bx  +c$ then we may write 
$$ax^2 + bx  +c = \color{red} d\ (x- m)(x-n) = dx^2 -d(m+ n)x + dmn$$
Using the equality of polynomials we have taht 
$$d = a ; d(m+n) = -b ; mnd = c $$
that is 
$$a = d ; m+n = \color {red}{-\frac{b}{a}} ; mn = \color{red}{\frac{c}{a}}$$
