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Question


Factorise:4y212y+9


Options:
A .   (7y5)(7y5)
B .   (5y3)(5y3)
C .   (2y3)(2y3)
D .   (2y5)(2y5)
Answer: Option C
:
C

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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