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Factoring Trinomials of the form
$\,ax^2 + bx + c\,$
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:
$\frac1c(fx+c)(gx+c)$ |
$= \frac1{p_1p_2}(fx+p_1p_2)(gx+p_1p_2)$ |
$= \frac1{p_1}(fx+p_1p_2)\frac1{p_2}(gx+p_1p_2)$ |
$= (\frac{f}{p_1}x + p_2)(\frac{g}{p_2}x + p_1)$ |
$= ((\text{an integer})x + p_2)((\text{an integer})x + p_1)$ |
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.