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Question

Factorize the equation (x2+4y29z2)216x2y2

A
(x+y+3z)(x+y+z)(x2yz)
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B
(x2y+z)(x+2y3z)(x2y3z)
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C
(x+y+z)(x+y3z)(x2yz)
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D
(x+2y+3z)(x+2y3z)(x2y3z)
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Solution

The correct option is D (x+2y+3z)(x+2y3z)(x2y3z)
(x2+4y29z2)216x2y2
=(x2+4y29z2)2(4xy)2
Using,
a2b2=(a+b)(ab)
=(x2+4y29z2+4xy)(x2+4y29z24xy)
Using,
(a+b)2=a2+2ab+b2 and (ab)2=a22ab+b2
=[(x+2y)29z2][(x2y)29z2]
=(x+2y+3z)(x+2y3z)(x2y+3z)(x2y3z)

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