An Alternate AC method

The goal is to solve

        ax² + bx + c

First find ac and then find two numbers that multiply to ac and add to b.  Call them u and v.  Then reduce the fractions u/a and v/a.  Call these reduced fractions s₁/s₂ and t₁/t₂.    Then there is a constant k with

        ax² + bx + c  =  k(s₂x + s₁)(t₂x + t₁)

Proof:

        ax² + bx + c  =  ax² + ux + vx + c

        =  ax(x + u/a) + v(x + c/v)        Since ac = uv,  c = uv/a

        =  ax(x + u/a) + v[x + (uv/a)/v]

        =  ax(x + u/a) + v(x + u/a)

        =  (ax + v)(x + u/a)

        =  a(x + v/a)(x + u/a)

        =  a(x +  t₁/t₂)(x + s₁/s₂)

        =  a(x + s₁/s₂)(x +  t₁/t₂)  Pull s₂ out of the first factor and t₂ out of the second

        =  a/(s₂t₂)(s₂x + s₁)(t₂x + t₁)

        =  k(s₂x + s₁)(t₂x + t₁)

Note that if s₂t₂  =  a, then k = 1.  Also note that

        k(s₂x + s₁)(t₂x + t₁)  =  k[s₂t₂x² + (s₂t₁ + s₁t₂)x + s₁t₁]

so that if GCF(a,b,c) = 1, then k = 1.

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