The LCM and HCF of two polynomials are respectively (2a−5)2(a+1)and(2a−5) If one of the polynomials is 4a2−20a+25 the other one is
A
4a2+20a+5
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B
4a2−25
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C
2a2+3a−5
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D
2a2−3a−5
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Solution
The correct option is D2a2−3a−5 LCM=(2a−5)(2a−5)(a+1) HCF=(2a−5) One polynomial=4a2−20a+25 ⇒4a2−10a−10a+25 ⇒2a(2a−5)−5(2a−5) ⇒(2a−5)(2a−5) LCM×HCF=Productoftwonumber Let another number is x (2a−5)(2a−5)(a+1)(2a−5)=(2a−5)(2a−5)×x x=(2a−5)(2a−5)(a+1)(2a−5)(2a−5)(2a−5) x=(a+1)(2a−5)⇒2a2−3a−5