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Question

Factorize: 6a2a15

A
(3a+2)(2a5)
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B
(2a3)(3a5)
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C
(2a+3)(5a3)
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D
(2a+3)(3a5)
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Solution

The correct option is D (2a+3)(3a5)
6a2a15

Split (1) into two parts such that
their product is (6)(-15) = -90 and their sum is -1

6a2+9a10a15

3a(2a+3)5(2a+3)

(2a+3)(3a5)

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