FACTORING BY GROUPING METHOD:   PROOF

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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,

 $ax^2 + bx + c$ $\displaystyle= \frac{fg}c x^2 + (f+g)x + c$ $\displaystyle= \frac{fg}c x^2 + fx + gx + c$ $\displaystyle= \frac1c(fg x^2 + cfx + cgx + c^2)$ $\displaystyle= \frac1c(fx + c)(gx + c)$

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, $\frac1c(fx+c) = (\frac{f}{c}x + 1) = ((\text{an integer})x + 1)$
• $\,c\,$ goes into $\,g\,$ evenly:
in this case, $\frac1c(gx+c) = (\frac{g}{c}x + 1) = ((\text{an integer})x + 1)$
• $\,c = p_1p_2\,$, where $\,p_1\,$ and $\,p_2\,$ are integers, $\,p_1\,$ divides $\,f\,$, and $\,p_2\,$ divides $\,g\,$:
in this case,  $\frac1c(fx+c)(gx+c)$ $= \frac1{p_1p_2}(fx+p_1p_2)(gx+p_1p_2)$ $= \frac1{p_1}(fx+p_1p_2)\frac1{p_2}(gx+p_1p_2)$ $= (\frac{f}{p_1}x + p_2)(\frac{g}{p_2}x + p_1)$ $= ((\text{an integer})x + p_2)((\text{an integer})x + p_1)$
In all three cases, one obtains the desired factorization with integer coefficients.

Conversely, suppose there exist integers $\,d\,$, $\,e\,$, $\,f\,$, and $\,g\,$
such that $\,ax^2 + bx + c = (dx + e)(fx + g) = (df)x^2 + (dg + ef)x + (eg)\,$.
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.