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B
(3x−2y+5)(3x−2y−2)
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C
(3x+2y+5)(3x+2y−2)
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D
(3x−2y−5)(3x−2y−2)
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Solution
The correct option is B(3x−2y+5)(3x−2y−2) ⇒(3x−2y)2+3(3x−2y)−10 Let (3x−2y)=a ⇒a2+3a−10 ⇒a2+5a−2a−10 ⇒a(a+5)−2(a+5) ⇒(a+5)(a−2) Substitutinga=(3x−2y) ⇒(3x−2y+5)(3x−2y−2)