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Question

The factorisation of the 1−2a−3a2 is equal to.

A
(1+a)(1- 3a)
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B
(1+a)(1+ 3a)
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C
(1-a)(1- 3a)
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D
(1-a)(1+ 3a)
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Solution

The correct option is A (1+a)(1- 3a)
The given expression is 12a3a2
The product of the first and last term is 3a2
The middle term has to be split in such a way that the difference is -2a and product is 3a2
The terms will be -3a and a as (-3a) (a) = 3a2 and -3a + a = -2a
Spliting the middle term we get,
1+a3a3a2
= 1(1+a) - 3a(1+a) [Taking '3a' common]
= (1+a)(1- 3a)

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