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Let, f (x) = x3 - 23x2 + 142x - 120 The factors of the constant term – 120 are ± ±1, ± ±2, ± ±3, ± ±4, ± ±5, ± ±6, ± ±7, ± ±8, ± ±9, ± ±10, ± ±11, ± ±12, ± ± 15, ± ± 20, ± ± 24, ± ± 30, ± ± 40, ± ± 60, and ± ± 120 Putting x = 1, we have f (1) = (1)3 – 23 (1)2 + 142 (1) – 120 = 1 – 23 + 142 – 120 = 0

So,

(x – 1) is a factor of f (x) Let us now divide f (x) = x3 - 23x2 + 142x - 120 by (x - 1) to get the other factors of f (x) Using long division method, we get x3 - 23x2 + 142x - 120 = (x – 1) (x2 – 22x + 120) x2 – 22x + 120 = x2 – 10x – 12x + 120 = x (x – 10) – 12 (x – 10)

Hence,

x3 - 23x2 + 142x - 120 = (x – 1) (x - 10) (x - 12)