Factorise : $x^{4} + 5 x^{3} + 5 x^{2} - 5 x - 6$

(x^{2} - 1) (x^{2} + 6)

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(x - 1) (x + 2)^{3}

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(x^{2} - 1) (x + 3) (x + 2)

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(x - 1) (x + 2) (x^{3} + 3)

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Solution

The correct option is C $(x^{2} - 1) (x + 3) (x + 2)$
$[A] (x^{2} - 1) (x^{2} + 6) = x^{4} + 6 x^{2} - x^{2} - 6 = x^{4} + 5 x^{2} - 6$

$[B] (x - 1) (x + 2)^{3} = (x - 1) (x^{3} + 2 x^{2} + 4 x + 8) = x^{4} + 2 x^{3} + 4 x^{2} + 8 x - (x^{3} + 2 x^{2} + 4 x + 8) = x^{4} + x^{3} + 2 x^{2} + 4 x - 8$

$[C] (x^{2} - 1) (x + 3) (x + 2) = (x^{3} + 3 x^{2} - x - 3) (x + 2) = x^{4} + 3 x^{3} - x^{2} - 3 x + 2 x^{3} + 6 x^{2} - 2 x - 6 = x^{4} + 5 x^{3} + 5 x^{2} - 5 x - 6$

$[D] (x - 1) (x + 2) (x^{3} + 3) = (x^{2} + x - 2) (x^{3} + 3) = x^{5} + x^{4} - 2 x^{3} + 3 x^{2} + 3 x - 6$

So, here the answer is $[C]$ .

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