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Question

Factorise 2a7128a


A

2a(a+2)(a2)(a22a+4)(a2+2a+4)

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B

(a+2)(a2)(a22a+4)(a2+2a+4)

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C

2a(a+2)(a2)(a22a+4)

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D

2a(a+2)(a22a+4)(a2+2a+4)

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Solution

The correct option is A

2a(a+2)(a2)(a22a+4)(a2+2a+4)


2a7128a = 2a(a664)

= 2a[(a3)2(8)2]

= 2a(a3+8)(a38)

= 2a(a3+23)(a323)

= 2a(a+2)(a22a+4)(a2)(a2+2a+4)
= 2a(a+2)(a2)(a22a+4)(a2+2a+4)


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