Using the "scientific approach", the author proves directly the original Beal conjecture (and not the equivalent conjecture) that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. One will let r, s, and t be prime factors of A, B and C, respectively, where D, E, and F are positive integers, such that A = Dr, B = Es, C = Ft. Then, the equation A^x + B^y = C^z becomes D^xr^x + E^ys^y = F^zt^z. Seven numerical Beal equations were factored. Based on the consistent pattern of the structure of the relationships between the prime factors on the left sides of the equations and the prime factors on the right sides of the equations in the factorizations, the author conjectured the equalities, r^x = t^x and s^y = t^y, which would imply that r = s = t, and establish that the Beal conjecture is true. The proof would be complete after showing that r^x = t^x, s^y = t^y and r = s = t, The proof in this paper is an expansion of a previous paper (viXra:2001.0694. by the author. The proof of the above equalities will be complete after showing that the ratios, (r^x)/(t^x) = 1 and (s^y)/(t^y) =1. To accomplish these relationships, one will factor out r^x on the left side of the equation, D^xr^x + E^ys^y = F^zt^z, followed by factoring out s^y of the same equation. The main principle for obtaining relationships between the prime factors on the left side of the equation and the prime factor on the right side of the equation is that the greatest common power of each prime factor on the left side of the equation equals the same power of the prime factor on the right side of the equation. High school students can learn and prove this conjecture as a bonus question on a final class exam.