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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