Factors of $63 a^{2} - 112 b^{2}$ are

none of these

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7 (3 a + 4 b) (3 a - 4 b)

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

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7 (3 a + 2 b) (3 a - 2 b)

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Solution

The correct option is B $7 (3 a + 4 b) (3 a - 4 b)$
Factors of $63 a^{2} - 112 b^{2}$ are
$= 7 (9 a^{2} - 16 b^{2} = 7 [(3 a)^{2} - (4 b)^{2}] = 7 (3 a + 4 b) (3 a - 4 b)$

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