Question

Factorize using factor theorem: x³-23x²+142x-120.

Answer 1

Let p(x) = x³ – 23x² + 142x – 120
The factors of –120 are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
By hit and trial method,

we find that p(1) = 0. So x – 1 is a factor of p(x).
Now, x ³ – 23x ² + 142x – 120 = x ³ – x ² – 22x ² + 22x + 120x – 120
= x ²(x –1) – 22x(x – 1) + 120(x – 1)

= (x – 1) (x ² – 22x + 120) [ On taking (x – 1) common]

Now x ² – 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we get
x ² – 22x + 120 = x ² – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
So, x ³ – 23x ² – 142x – 120 = (x – 1)(x – 10)(x – 12)