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Question

Factorise (2xy)214x+7y18


A

(2xy+9)(2xy2)

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B

(2xy9)(2xy2)

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C

(2xy9)(2x+y+2)

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D

(2x+y+9)(2x+y+2)

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Solution

The correct option is B

(2xy9)(2xy2)


Given (2xy)214x+7y18
Taking 7 common in the middle terms , we get
(2xy)27(2xy)18
Let 2xy=a
a27a18
Using splitting the middle term , we get
a29a+2a18
a(a9)2(a9)
Taking (a-9) as common bracket , we get
(a9)(a2)
Re-substituting for a , we get
(2xy9)(2xy2)


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