LCM and GCD Solution1





In this page lcm and gcd solution1 we are going to see solution of some questions with detailed explanation.

(1) Find the LCM of each pair of the following polynomials

(i) x² - 5 x + 6 , x² + 4 x - 12 whose GCD is (x - 2)

Solution:

 L.C.M x G.C.D = f (x) x g (x)

 f (x) = x² - 5 x + 6

 g (x) = x² + 4 x - 12

 G.C.D = (x - 2)

 L.C.M = [ f (x) x g (x) ]/G.C.D

          = [ (x² - 5 x + 6) (x² + 4 x - 12) ]/(x - 2)

To simplify this we have to factorize these quadratic equations.So that we get

          = [ (x - 2) (x - 3) (x + 6) (x - 2) ]/(x - 2)

          = (x - 2) (x - 3) (x + 6)


(ii) x⁴ + 3 x³ + 6 x² + 5 x + 3 , x⁴ + 2 x² + x + 2 whose GCD is x² + x + 1

Solution:

 L.C.M x G.C.D = f (x) x g (x)

 f (x) = x⁴ + 3 x³ + 6 x² + 5 x + 3

 g (x) = x⁴ + 2 x² + x + 2

 G.C.D = x² + x + 1

 L.C.M = [ f (x) x g (x) ]/G.C.D

          = [ (x⁴ + 3 x³ + 6 x² + 5 x + 3) (x⁴ + 2 x² + x + 2) ]/(x² + x + 1)

To simplify this we have to use long division.

          = (x² + 2 x + 3) (x⁴ + 2 x² + x + 2)


(iii) 2 x³ + 15 x² + 2 x - 35 , x⁴ + 8 x² + 4 x - 21 whose GCD is x + 7

Solution:

 L.C.M x G.C.D = f (x) x g (x)

 f (x) = 2 x³ + 15 x² + 2 x - 35

 g (x) = x⁴ + 8 x² + 4 x - 21

 G.C.D =x + 7

 L.C.M = [ f (x) x g (x) ]/G.C.D

          = [ (2 x³ + 15 x² + 2 x - 35) (x⁴ + 8 x² + 4 x - 21) ]/(x + 7)

To simplify this we have to use long division.

          = (2 x² + x - 5) (x⁴ + 8 x² + 4 x - 21)

lcm and gcd solution1 lcm and gcd solution1