Please the box method for factoring trinomial of the form $ax^2+bx=c$ I was given this method for factoring trinomials of the form $ax^2 + bx + c$
This is the method:

Find numbers $p$ and $q$ such as $a\cdot c=p\cdot q$ and $b=p+q$
With $p$ and $q$ $(GCF(a,q)x + GCF(c,p))(GCF(a,p)x+GCF(c,q))= ax^2+bx+c$

I have no idea how to prove this method or why it works.
Please help me.
 A: The greatest common factor is also called the greatest common divisor.  I will call it the gcd for greatest common divisor.  
There is an implicit assumption here that $\gcd(a, b, c) = 1$. For instance, 
\begin{align*}
8x^2 + 24x + 18 & = 8x^2 + 12x + 12x + 18\\
                & = 4x(2x + 3) + 6(2x + 3)\\
                & = (4x + 6)(2x + 3)\\
                & = 2(2x + 3)^2
\end{align*}
so $p = q = 12$, $\gcd(8, 12) = 4$, $\gcd(18, 12) = 6$, $ac = 8 \cdot 18 = 144 = 12 \cdot 12 = pq$ and $b = 24 = 12 + 12 = p + q$.  However, $a = 8 \neq \gcd(8, 12)\gcd(8, 12) = \gcd(a, p)\gcd(a, q)$.  
There also seems to be an implicit assumption that $c \neq 0$.  Consider the factorization
$$6x^2 + 5x = x(6x + 5) = (x + 0)(6x + 5)$$ 
of $6x^2 + 5x = 6x^2 + 5x + 0$.  Observe that $ac = 6 \cdot 0 = 0$ and that $b = 5$.  If $pq = ac = 0$ and $p + q = b = 5$, then $p = 5$ and $q = 0$ or $p = 0$ and $q = 5$.  Since $\gcd(0, 0)$ is undefined, the formula does not make sense if $c = 0$.
Assume $a, b, c, s, t, u, v$ are integers such that
$$ax^2 + bx + c = (sx + t)(ux + v)$$ 
with $a, c\neq 0$ and $\gcd(a, b, c) = 1$.  
We will show that $a = su$, $b = sv + tu$, and $c = tv$.  Since the equation $$ax^2 + bx + c = (sx + t)(ux + v)$$ 
is an algebraic identity, it must hold for each real number $x$.  In particular, it holds when $x = 0, 1, -1$.  Since it holds when $t = 0$,
$$c = tv$$
Since the identity holds when $t = 1$, 
$$a + b + c = (s + t)(u + v) = su + sv + tu + tv$$
Since $c = tv$, we obtain 
$$a + b = su + sv + tu$$
Since the identity holds when $t = -1$, 
$$a - b + c = (-s + t)(-u + v) = su - sv - tu + tv$$
Since $c = tv$, we obtain
$$a - b = su - sv - tu$$
Adding the equations for $a + b = su + sv + tu$ and $a - b = su - sv - tu$ yields
$$2a = 2su$$
Hence, 
$$a = su$$  
Substituting $su$ for $a$ in the equation $a + b = su + sv + tu$ shows that 
$$b = sv + tu$$
Take $p = sv$ and $q = tu$.  Then 
$$pq = svtu = (su)(tv) = ac$$
and 
$$p + q = sv + tu = b$$
Note that if $\gcd(s, t) = d > 1$.  Then $s = ds_0$, $t = dt_0$ for some integers $s_0, t_0$, so 
\begin{align*}
a & = su = ds_0u\\
b & = sv + tu = ds_0v + dt_0u = d(s_0v + t_0u)\\
c & = tv = dt_0v
\end{align*}
which implies that $\gcd(a, b, c) \geq d > 1$, contrary to our assumption that $\gcd(a, b, c) = 1$. Hence, $\gcd(s, t) = 1$. A similar argument shows that $\gcd(u, v) = 1$.  
Therefore, 
\begin{align*}
\gcd(a, p) & = \gcd(su, sv) = s\\
\gcd(a, q) & = \gcd(su, tu) = u\\
\gcd(c, p) & = \gcd(tv, sv) = v\\
\gcd(c, q) & = \gcd(tv, tu) = t\\
\end{align*}
Thus,
\begin{align*}
ax^2 + bx + c & = (sx + t)(ux + v)\\
              & = [\gcd(a, p)x + \gcd(c, q)][\gcd(a, q)x + \gcd(c, p)]
\end{align*} 
