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Question

Factorise 32a48a4

A
8a2(2a1)(2a+1)
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B
8a(2a1)(2a+1)
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C
4a2(2a1)(2a+1)
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D
32a2(2a+1)
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Solution

The correct option is A 8a2(2a1)(2a+1)
32a48a4
Taking 8 a2 common we have
=8a2(4a21
= 8a2((2a)21) {a2b2=(a+b)(ab)}
= 8a2(2a1)(2a+1)

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