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Question

Factorize the below equation:
5−(3a2−2a)(6−3a2+2a)

A
(a)(a+1)(a1)(a+1)
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B
(a)(a+1)(a1)(3)
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C
(a5)(a1)(a1)(a+1)
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D
(3a5)(a+1)(a1)(3a+1)
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Solution

The correct option is C (3a5)(a+1)(a1)(3a+1)
5(3a22a)(63a2+2a)
=56(3a22a)+(3a22a)2
=[(3a22a)25(3a22a)][(3a22a)5]
=(3a22a)(3a22a5)1(3a22a5)
=(3a22a5)(3a22a1)
=(3a25a+3a5)(3a23a+a1)
=[3a(a+1)5(a+1)][3a(a1)+(a1)]
=(a+1)(3a5)(a1)(3a+1)

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