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Q.

Factorise: $x^{3} - 3 x^{2} - 9 x - 5$

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a

(x - 5) {(x - 1)}^{2}

b

{(x - 5)}^{2} {(x - 1)}^{2}

c

(x + 5) {(x + 1)}^{2}

d

(x - 5) {(x + 1)}^{2}

answer is D.

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Detailed Solution

Concept- The hit-and-trial approach is used to identify the factors or roots of any polynomial with a degree greater than two. Generally speaking, the goal of this strategy is to vary the values of the integers until the polynomial is satisfied by zero.
Given:

x^{3} - 3 x^{2} - 9 x - 5

Put,

x = 0, 0^{3} - 3 (0^{2}) - 9 (0) - 5 = 0

= - 5

Hence, x is not a factor,
Now,

put x = 1,

1^{3} - 3 (1^{2}) - 9 (1) - 5 = - 16

Hence, x-1 is not a factor,
Now, put

x = - 1

(- 1^{3}) - 3 (- 1^{2}) - 9 (- 1) - 5 = 0

So

, x + 1

is the factor

x^{3} - 3 x^{2} - 9 x - 5 = (x + 1) (x^{2} - 4 x - 5)

= (x + 1) (x^{2} - 5 x + x - 5)

= (x + 1) (x - 5) (x + 1)

x^{3} - 3 x^{2} - 9 x - 5 = (x - 5) {(x + 1)}^{2}

Hence, the correct option is 4.

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Factorise: x3-3x2-9x-5