Scriem astfel:
(x + 3)³ - 4x - 12 = (x + a)(x + b)(x + c)
(x + 3)³ - 4(x + 3) = (x + a)(x + b)(x + c)
(x + 3) [(x + 3)² - 2²] = (x + a)(x + b)(x + c)
(x + 3)(x + 3 - 2)(x + 3 + 2) = (x + a)(x + b)(x + c)
(x + 3)(x + 1)(x + 5) = (x + a)(x + b)(x + c)
[tex]\implies \bf a = 3, b = 1, c = 5[/tex]
