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Question

Factorise:4y212y+9

A

(7y5)(7y5)

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B

(5y3)(5y3)

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C

(2y3)(2y3)

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D

(2y5)(2y5)

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Solution

The correct option is C

(2y3)(2y3)


The given expression can be rewritten as,
(2y)22(2y)(3)+32.
Comparing with the identity
(ab)2=a22ab+b2
we get,
(2y)22(2y)(3)+32
=(2y3)2
=(2y3)(2y3)


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