FIND LCM WHEN POLYNOMIALS AND GCD ARE GIVEN

Relationship between LCM and GCD :

f(x) ⋅ g(x)  =  LCM (f(x), g(x)) ⋅ GCD (f(x), g(x))

LCM(f(x), g(x))  =  [f(x) ⋅ g(x)]/GCD (f(x), g(x))

Find the LCM of each pair of the following polynomials

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

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

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

(iv)  2x3-3x2-9x+5, 2x4-x3-10x2-11x+8 whose GCD is 2x-1

(i)  Answer :

LCM ⋅ GCD  =  f(x) ⋅ g(x)

LCM  =  [f(x) ⋅ g(x)]/GCD

f(x)  =  x2-5x+6

g(x)  =  x2+4x-12

GCD  =  (x-2)

x2-5x+6  =  (x-2)(x-3)

x2+4x-12  =  (x+6)(x-2)

LCM  =  (x-2)(x-3)(x+6)(x-2)/(x-2)

By canceling common factors, we get

LCM  =  (x-2) (x-3) (x+6)

So, the required LCM is (x-2) (x-3) (x+6).

(ii)  Answer :

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

LCM  =  [f(x) ⋅ g(x)]/GCD

f(x)  =  x4+3x3+6x2+5x+3

g(x)  =  x4+2x2+x+2

GCD  =  x2+x+1

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

To simplify this, we have to use long division.

LCM  =  (x2+2x+3) (x4+2x2+x+2)

So, the required LCM is (x2+2x+3) (x4+2x2+x+2).

(iii)  Answer :

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

LCM  =  [f(x) ⋅ g(x)]/GCD

f(x)  =  2x3+15x2+2x-35

g(x)  =  x4+8x2+4x-21

GCD  = x+7

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

To simplify this we have to use long division.

LCM  =  (2x2+x-5) (x4+8x2+4x-21)

So, the required LCM is  (2x2+x-5) (x4+8x2+4x-21).

(iv)  Answer :

2x3-3x2-9x+5, 2x4-x3-10x2-11x+8 whose GCD is 2x-1

LCM  =  [f(x) ⋅ g(x)]/GCD

f(x)  =  2x3-3x2-9x+5

g(x)  =  2x4-x3-10x2-11x+8

GCD  =  2x-1

LCM  =  [(2x3-3x2-9x+5) (2x4-x3-10x2-11x+8)]/(2x-1)

To simplify this we have to use long division.

LCM  =  (x3-5x-8) (2x3-3x2-9x+5)

So, the LCM is (x3-5x-8) (2x3-3x2-9x+5).

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