Expand using identity : $(2 a - 4 b + 7) (2 a + 4 b + 7)$

4 a^{2} + 24 a + 49 - 16 b^{2}

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a^{2} + 8 a + 49 - 16 b^{2}

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a^{2} + 14 a + 49 - 16 b^{2}

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4 a^{2} + 28 a + 49 - 16 b^{2}

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Solution

The correct option is D $4 a^{2} + 28 a + 49 - 16 b^{2}$
$(2 a - 4 b + 7) (2 a + 4 b + 7)$
$= (2 a + 7 + 4 b) (2 a + 7 - 4 b)$
$= (2 a + 7)^{2} - (4 b)^{2}$ ....[Using $a^{2} - b^{2} = (a + b) (a - b)$ ]
$= ((2 a)^{2} + 2 (2 a) (7) + (7)^{2}) - 16 b^{2}$ ...[Using $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ ]
$= 4 a^{2} + 28 a + 49 - 16 b^{2}$

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Similar questions

a^{2}

b^{2}

16 b^{2} + 9 a^{2} = a^{2} b^{2}

36 b^{2} + 4 a^{2} = a^{2} b^{2}

a^{2} - 16 b^{2} - 2 a - 8 b

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

Factorise Completely : $a^{2} - 16 b^{2} - 2 a - 8 b$

3 a - 4 b = 7

a b = 10

9 a^{2} + 16 b^{2}

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