Factor by Grouping Technique: Proof

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 $\,ax^2 + bx + c\,$

THEOREM the Factor by Grouping Method

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 $\,f + g = b\,.$ Then,

$$ \begin{align} &ax^2 + bx + c\cr\cr &\qquad = \frac{fg}c x^2 + (f+g)x + c\cr\cr &\qquad = \frac{fg}c x^2 + fx + gx + c\cr\cr &\qquad = \frac1c(fg x^2 + cfx + cgx + c^2)\cr\cr &\qquad = \frac1c(fx + c)(gx + c) \end{align} $$

Since $\displaystyle\frac{fg}{c} = a\,$ and since $\,a\,$ is an integer, one of three things must happen:

$\,c\,$ goes into $\,f\,$ evenly: in this case, $$\begin{align} &\frac1c(fx+c)\cr &\qquad = (\frac{f}{c}x + 1)\cr &\qquad = ((\text{an integer})x + 1) \end{align} $$
$\,c\,$ goes into $\,g\,$ evenly: in this case, $$ \begin{align} &\frac1c(gx+c)\cr &\qquad = (\frac{g}{c}x + 1)\cr &\qquad = ((\text{an integer})x + 1) \end{align} $$
$\,c = p_1p_2\,,$ where $\,p_1\,$ and $\,p_2\,$ are integers, $\,p_1\,$ divides $\,f\,,$ and $\,p_2\,$ divides $\,g\,$: in this case, $$ \begin{align} &\frac1c(fx+c)(gx+c)\cr &\qquad = \frac1{p_1p_2}(fx+p_1p_2)(gx+p_1p_2)\cr\cr &\qquad = \frac1{p_1}(fx+p_1p_2)\frac1{p_2}(gx+p_1p_2)\cr\cr &\qquad = (\frac{f}{p_1}x + p_2)(\frac{g}{p_2}x + p_1)\cr\cr &\qquad = ((\text{an integer})x + p_2)((\text{an integer})x + p_1) \end{align} $$

In all three cases, one obtains the desired factorization with integer coefficients.

Conversely, suppose there exist integers $\,d\,,$ $\,e\,,$ $\,f\,,$ and $\,g\,$ such that: $$ \begin{align} &ax^2 + bx + c\cr\cr &\qquad = (dx + e)(fx + g)\cr\cr &\qquad = (df)x^2 + (dg + ef)x + (eg) \end{align} $$ Thus, $\,a = df\,,$ $\,b = dg + ef\,,$ and $\,c = eg\,.$

Define $\,F := dg\,$ and $\,G := ef\,.$ Then:

$\,F\,$ and $\,G\,$ are both integers,

$\,F + G = dg + ef = b\,,$ and

$\,FG = (dg)(ef) = (df)(eg) = ac\,.$

Q.E.D.