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Question

Factorise 4x2−12ax−y2−z2−2yz+9a2

A
(2x23a+y+z)(2x3a2yz)
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B
(2x3a+y+z)(2x3ayz)
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C
(2x3a+y+z)(2x23ayz3)
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D
(2x3a+y+z)(2x3a3y+z)
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Solution

The correct option is B (2x3a+y+z)(2x3ayz)
4x212axy2z22yz+9a2
=4x212ax+9a2y2z22yz
=[(2x)212ax+(3a)2][y2+z2+2yz]
{ a22ab+b2=(ab)2}
=[2x3a]2[y+z]2
{ a2b2=(a+b)(ab)}
=[(2x3a)+(y+z)][(2x3a)(y+z)]
=(2x3a+y+z)(2x3ayz)


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