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 - 15So v(x) = (x – 5) (x + 2)= x2 – 3x - 10

Q. 54.1( 14 Votes )

The HCF and LCM o

Class 10th Rajasthan Board Mathematics 3. Polynomials

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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