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Question

Factorize the below equation:
3a212a

A
(2a1)(3a1)
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B
(a1)(3a+1)
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C
(2a1)(a+1)
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D
(a+1)(2a+1)
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Solution

The correct option is C (a1)(3a+1)
3a212a
=3a22a1
=3a23a+a1
=3a(a1)+1(a1)
=(a1)(3a+1)

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