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) ?
Best Answer
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”.