Factorise : $4 b^{2} - 1 - 2 a - a^{2}$

(2 b + 1 + a) (2 b + 1 - a)

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(2 b + 1 + a) (2 b - 1 - a)

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(2 b - 1 + a) (2 b - 1 - a)

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(b + 1 + a) (2 b - 1 - a)

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Solution

The correct option is C $(2 b + 1 + a) (2 b - 1 - a)$
$4 b^{2} - 1 - 2 a - a^{2} = 4 b^{2} - (1 + 2 a + a^{2})$
$= {(2 b)}^{2} - {(1 + a)}^{2}$
$= (2 b + 1 + a) (2 b - 1 - a)$

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