Does the ( so called) " CAB" factoring method work for every quadratic trinomial? Note : my qustion assumes one is dealing with a trinomial that admits of a factorization.
By "CAB " method I mean he following process:
when confronted with an expression of the form $Ax^2+Bx+C$
(1) compute the product $AC$
(2) find 2 numbers $N_1$ and $N_2$ such that : $N_1\times N_2 = AC $ and $ N_1+N_2 = B$
(3) rewrite the original expression as $Ax^2+(N_1+N_2)x +C$
(4) develop this last expression and factor it by grouping
I roughly see why it works when the factorization to be reached has the form $(ax+b)(x+c)$ but it's much less cleear to me when it has the form $ ( ax+b)(cx+d)$.
Hence my question : does the method work when the factorization to be reached contains two binomials in which the "x-term" has a coefficient different from $1$ ( in each one of these two binomials) ?

Yes, it always works. To see why, suppose that the quadratic $Ax^2+Bx+C$ can be written as $(ax+b)(cx+d)$. If we expand $(ax+b)(cx+d)$, we get $(ac)x^2+(ad+bc)x+bd$. So $A=ac$, $B=ad+bc$, and $C=bd$. There is always a unique pair of numbers $N_1$ and $N_2$ such that $N_1 \cdot N_2=AC$ and $N_1+N_2=B$. Those two numbers are $ad$ and $bc$. Hence,
\begin{align}
Ax^2+Bx+C &= (ac)x^2+(ad+bc)x+bd \\
&= (ac)x^2+(ad)x+(bc)x+bd \\
&= ax(cx+d)+b(cx+d) \\
&= (ax+b)(cx+d) \, .
\end{align}
If we split the middle term $B$ in a different way, then we get
\begin{align}
Ax^2+Bx+C &= (ac)x^2+(bc+ad)x+bd \\
&= (ac)x^2+(bc)x+(ad)x+bd \\
&= cx(ax+b)+d(ax+b) \\
&= (cx+d)(ax+b) \, ,
\end{align}
which is the same thing.
