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) ?
 A: 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.
A: Finding $N_1$ and $N_2$ is the same as finding the roots of the polynomial.
Indeed, the roots $r_1$ and $r_2$ satisfy
$$
r_1+r_2=-\frac{B}{A},\qquad r_1r_2=\frac{C}{A}
$$
Set $N_1=-Ar_1$ and $N_2=-Ar_2$. Then
$$
N_1+N_2=-Ar_1-Ar_2=-A(r_1+r_2)=A\frac{B}{A}=B
$$
and
$$
N_1N_2=(-Ar_1)(-Ar_2)=A^2r_1r_2=A^2\frac{C}{A}=AC
$$
Conversely, given $N_1$ and $N_2$ with the properties you mention, it's easy to see that $r_1=-N_1/A,r_2=-N_2/A$ are the roots.
Note that the factorization can always be written as
$$
A(x-r_1)(x-r_2)
$$
and I don't find a real convenience in the “CAB method”.
A: This becomes more intuitive if you get rid of $A$ by multiplying instead of by dividing:
$$\begin{align}A\times(Ax^2 + Bx + C)&=A^2x^2 + ABx + AC\\
&=(Ax)^2 + B(Ax) + AC
\end{align}$$
So factoring a quadratic with coefficients ${A,B,C}$ is equivalent to factoring one with coefficients ${1,B,AC}$.
