Q. 3 D5.0( 1 Vote )

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Answer :

Given 15b2 – 3bx2 – 5b + x2 = (3b×5b – 3b×x2 + ( – 1)× 5b + x2 )


Taking 3b common from 1st and 2nd term and ( – 1) from 3rd and 4th,


15b2 – 3bx2 – 5b + x2= 3b(5b – x2) + ( – 1)(5b – x2)


15b2 – 3bx2 – 5b + x2= (5b – x2) (3b – 1)


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