Q. 225.0( 3 Votes )

The LCM and HCF o

Answer :

Given:

HCF = x – 1


Let the two polynomials be u(x) and v(x)


So (x – 1) is common to both u(x) and v(x)


LCM = x3 – 7x + 6


= x3 – 7x + 6


= x3 – (1 + 6) x + 6


= x3 – x - 6x + 6


= x(x2 – 1) - 6(x - 1)


(x2 – 1) = (x + 1) (x – 1)


a2 – b2 = (a + b) (a – b)


Here a = x and b = 1


= x(x + 1) (x – 1) - 6(x - 1)


= (x-1) [x(x + 1) - 6]


Now solving the inner quadratic equation,


x2 + x - 6 = 0


x2 + 3x - 2x - 6 = 0


x(x + 3) - 2(x + 3) = 0


(x + 3) (x – 2) = 0


LCM = (x-1) (x + 3) (x – 2)


Since HCF = (x-1), which implies that both u(x) and v(x) contains (x – 1)


Therefore u(x) = (x – 1) (x + 3)


= x2 + 2x - 3


v (x) = (x – 1) (x – 2)


= x2 – 3x + 2


Therefore the polynomial are x2 + 2x – 3 and x2 – 3x + 2.


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