Factorise $x^{3} - 23 x^{2} + 142 x - 120$ , knowing that $(x - 1)$ is a factor of it.

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(x - 1) (x - 12) (x - 10)

(x - 4) (x - 12) (x - 10)

(x - 1) (x - 2) (x - 10)

(x - 1) (x - 12) (x - 1)

answer is A.

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Detailed Solution

Given that,

(x - 1)

is a factor of

x^{3} - 23 x^{2} + 142 x - 120

.
The factorisation is

x^{3} - 23 x^{2} + 142 x - 120 = x^{3} - 22 x^{2} + 120 x - x^{2} + 22 x - 120

\Rightarrow x^{3} - 23 x^{2} + 142 x - 120 = {x (x}^{2} - 22 x + 120) - 1 (x^{2} - 22 x + 120)

\Rightarrow (x - 1) (x^{2} - 22 x + 120)

\Rightarrow (x - 1) (x^{2} - 12 x - 10 x + 120)

\Rightarrow (x - 1) (x (x - 12) - 10 (x - 12))

\Rightarrow (x - 1) (x - 10) (x - 12)

So, the factors are

(x - 1) (x - 12) (x - 10)

.
Therefore, option (1) is correct.

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