Factorize the below equation:
${(2 a + b)}^{2} - 6 a - 3 b - 4$

(2 a + b - 4) (2 a + b + 1)

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(2 a + b) (b + 1)

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(2 a + b - 7) (a + b + 1)

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

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Solution

The correct option is B $(2 a + b - 4) (2 a + b + 1)$
${(2 a + b)}^{2} - 6 a - 3 b - 4$
$= {(2 a + b)}^{2} - 3 (2 a + b) - 4$
$= {(2 a + b)}^{2} - 4 (2 a + b) + (2 a + b) - 4$
$= (2 a + b) ((2 a + b) - 4) + ((2 a + b) - 4)$
$= ((2 a + b) - 4) ((2 a + b) + 1)$
$= (2 a + b - 4) (2 a + b - 1)$

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(2 a + b)^{2} - 6 a - 3 b - 4

The factors of $8 a^{3} + b^{3} - 6 a b + 1$ are:

Factorise : 8a³ + b³ + 12a²b + 6ab²

(2 a + b^{2} - 4 (2 a + b))

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