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Question

Factorize the following equation by the grouping method
a2+1a223a+3a

A
(a2a)(a1a1)
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B
(a1a)(a13a1)
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C
(a2a)(a1a3)
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D
(a1a)(a1a3)
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Solution

The correct option is D (a1a)(a1a3)
a2+1a223a+3a
=a2+1a22×a×1a3a+1a
=(a1a)23(a1a)
=(a1a)(a1a3)

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