This page is intended for advanced students and teachers. Other people interested in the Factor by Grouping Method should visit: Factoring Trinomials of the form $\,ax^2 + bx + c\,$
Let $\,a\,$, $\,b\,$, and $\,c\,$ be integers, with $\,a\ne 0\,$ and $\,c\ne 0\,$.
The trinomial $\,ax^2 + bx + c\,$ is factorable over the integers
if and only if
there exist integers $\,f\,$ and $\,g\,$ such that
$\,fg = ac\,$ and $\,f+g = b\,$.
Proof
Suppose there exist integers $\,f\,$ and $\,g\,$ such that $\,fg = ac\,$ and $\,f + g = b\,$.
$$ \begin{align} ax^2 + bx + c &= \frac{fg}{c} x^2 + (f+g)x + c\\ &= \frac{fg}{c} x^2 + fx + gx + c\\ &= \frac1c(fg x^2 + cfx + cgx + c^2)\\ &= \frac1c(fx + c)(gx + c) \end{align} $$Since $\frac{fg}{c} = a$ and $\,a\,$ is an integer, one of three things must happen:
- $\,c\,$ divides $\,f\,$ evenly
- $\,c\,$ divides $\,g\,$ evenly
- $\,c = p_1p_2\,$ where $\,p_1\,$ divides $\,f\,$ and $\,p_2\,$ divides $\,g\,$
In all cases, a factorization with integer coefficients results.
Conversely, suppose $\,ax^2 + bx + c = (dx + e)(fx + g)\,$.
$$ \begin{align} (dx + e)(fx + g) &= (df)x^2 + (dg + ef)x + (eg) \end{align} $$Thus, $\,a = df\,$, $\,b = dg + ef\,$, and $\,c = eg\,$.
Let $\,F = dg\,$ and $\,G = ef\,$.
$F + G = b$ and $FG = ac$
Q.E.D.