How to solve this nice math Exponential algebra problem | Olympiad Question | b=?

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[walterwen]

Olympiad Question: (b + 1)(b + 3)(b – 2)(b – 4) = 24; b =?
[(b + 1)(b – 2)][(b + 3)(b – 4)] = [(b² – b) – 2][(b² – b) – 12] = 24
(b² – b)² – 14(b² – b) + 24 = 24, (b² – b)² – 14(b² – b) = 0, b(b – 1)(b² – b – 14) = 0
b = 0, b – 1 = 0; b = 1 or b² – b – 14 = 0; b = (1 ± √57)/2
Answer check:
b = 0: (b + 1)(b + 3)(b – 2)(b – 4) = (1)(3)(– 2)(– 4) = 24; Confirmed
b = 1: (1 + 1)(1 + 3)(1 – 2)(1 – 4) = (2)(4)(– 1)(– 3) = 24; Confirmed
b = (1 ± √57)/2: (1/16)(3 ± √57)(7 ± √57)(– 3 ± √57)(– 7 ± √57)
= (1/16)(– 9 + 57)(– 49 + 57) = [(48)(8)]/16 = 24; Confirmed
Final answer:
b = 0; b = 1; b = (1 + √57)/2 or b = (1 – √57)/2

[RealQinnMalloryu]

(b +1) (b +3)(b ➖ 2)(b ➖ 2b+2) =24 ( b ➖ 1b+1) (b ➖ 3b+3) (b ➖ 2b+2) (b ➖ 1b+1) (b ➖ 1b+1) (b ➖ 2b+1) .2^12 2^2^6 1^13^2 1^^1^2 (b ➖ 2b+1).