Q. 54.1( 14 Votes )

The HCF and LCM o

Answer :

HCF = (x – 5)

LCM = x3 – 19x – 30


= x3 – 19x – 38 + 8


= x3 + 8 – 19x – 38


= x3 + 23 – 19(x + 2)


x3 + 23 = (x + 2) (x2 – 2x + 4) [Using a3 + b3 = (a + b) (a2 – ab + b2)]


= (x + 2) (x2 – 2x + 4) – 19(x + 2)


= (x + 2) (x2 – 2x + 4 – 19)


= (x + 2) (x2 – 2x – 15)


= (x + 2) (x2 – 5x + 3x – 15)


= (x + 2) (x(x– 5) + 3(x – 5))


= (x + 2) (x – 5) (x + 3)


Since HCF = x – 5 it will belong to both polynomials.


So u(x) = (x – 5) (x + 3)


= x2 – 2x - 15


So v(x) = (x – 5) (x + 2)


= x2 – 3x - 10


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