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Question

Factorise 4a2b−12a2−16b+48


A

4xz

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B

4xy

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C

4xz

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D

4xyz

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Solution

The correct option is B

4xy


Given 4a2b12a216b+48

Taking common factor as 4a2 from the first two terms and 16b as from the last two terms we get

4a2b12a216b+48 = 4a2(b3)16(b3)

= (4a216)(b3)

= (2a)2(4)2)(b3) ( converting in the form a2b2)

= (2a4)(2a+4)(b3))


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