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. |