HOW TO FIND LCM FROM TWO POLYNOMIALS AND GCD

Let f(x) and g(x) be two polymials. 

We can find the LCM or GCD of the two polynomials using the relationship given below. 

f (x) × g(x)  =  LCM × GCD

Practice Problems

Problem 1 :

Find the LCM of the following polynomials whose GCD is (a - 2).

(a2 + 4a −12)  and  (a2 −5a + 6)

Solution :

Let f(x)  =  a2 + 4a −12, g(x)  =  a2 −5a + 6.

f(x)  =   a2 + 4a −12  

  =  a2 + 6a - 2a −12

  =  a(a + 6) - 2(a + 6)

 f(x)  =  (a + 6)(a - 2)

g(x) = a2 −5a + 6

  =  a2 - 2a - 3a + 6

  =  a(a - 2) - 3(a - 6)

g(x)  =  (a - 3)(a - 2)

GCD is (a -2)

f (x) × g(x) = LCM × GCD

LCM  =  [f(x) × g(x)] / GCD

LCM  =  [(a + 6)(a - 2)  (a - 3)(a - 2)] /  a -2

LCM  =  (a + 6)(a - 2)  (a - 3)

Problem 2 :

Find the LCM of the following polynomials whose GCD is (x - 3a).

(x 4 -27a3x)  and  (x -3a)2

Solution :

Let f(x)  =  x 4 -27a3x, g(x)  =   (x -3a)2

f(x)  =  x(x3 - 27a3)

f(x)  =  x(x3-(3a)3)

  f(x)  =  x(x-3a)(x2-x(3a)+(3a)2)

f(x)  =  x (x- 3a)(x2-3ax+9a2)

g(x)  =   (x -3a)2

GCD is (x -3a)

f (x) × g(x) = LCM × GCD

LCM  =  [f(x) × g(x)] / GCD

LCM  =  [x(x- 3a)(x2-3ax+9a2)(x -3a)2] /  (x -3a)

LCM  =  x(x2-3ax+9a2)(x -3a)2

How to Find GCD from Two Polynomials and LCM

Problem 1 :

Find the GCD of the following polynomials.  

12(x4 -x3)  and  8(x4 −3x3 +2x2)

Given that LCM is 24x3(x -1)(x -2).

Solution :

Let f(x)  =  12(x4 -x3), g(x)  =  8(x4 −3x3 +2x2)

LCM  =  24x3(x -1)(x-2)

f(x)  =  12(x4 -x3)

f(x)  =  12x3(x - 1)

g(x)  =  8(x4 −3x3 +2x2)

g(x)  =  8x2(x2 - 3x + 2)

GCD  =  24x3(x -1)(x -2)

f (x) × g(x) = LCM × GCD

GCD  =  [f(x) × g(x)] / LCM

GCD  =  [12x3(x - 1) 8x2(x2 - 3x + 2)]/ 24x3(x -1)(x -2)

GCD  =  4x2(x-1)

Problem 2 :

Find the GCD of the following polynomials.  

(x3 + y3)  and  (x4 + x2y2 + y4)

Given that LCM is (x3 + y3)(x2 + xy + y2).

Solution :

Let f(x)  =  (x3 + y3), g(x)  =  (x4 + x2y2 + y4)

LCM is (x3 + y3)(x2 + xy + y2)

f(x)  =  (x3 + y3)

g(x)  =  (x4 + x2y2 + y4)

= (x2 + y2)2 - (xy)2

= (x2 + y2)2 - (xy)2

= (x2-xy+ y2 )(x2+ xy+ y2)

LCM  =  (x3 + y3)(x2 + xy + y2)

f (x) × g(x) = LCM × GCD

GCD  =  [f(x) × g(x)] / LCM

GCD  =  [ (x3 + y3)(x2-xy+ y)(x2+ xy+ y2)] / (x3 + y3)(x2 + xy + y2)

GCD  =  (x- xy +  y2)

Problem 3 :

LCM and GCD of the two polynomials p(x) and q(x) and the polynomial p(x) are given below. Find q(x). 

LCM  =  a3 −10a2 +11a + 70

GCD  =  a - 7

p(x)  =  a2 −12a + 35

Solution :

p (x) × q(x)  =  LCM × GCD

Then, 

q(x)  =  (LCM × GCD) / p (x)

=   (a3 −10a2 +11a + 70)( a - 7)/(a2 −12a + 35)

q(x)  =  (a + 2) (a - 7) 

Problem 4 :

LCM and GCD of the two polynomials p(x) and q(x) and the polynomial q(x) are given below. Find p(x). 

LCM = (x2 +y2)(x4 +x2y2+y4)

GCD  =  (x2 -y2)(x4 −y4)

q(x)  =  (x2 +y2 −xy)

Solution :

p (x) × q(x)  =  LCM × GCD

Then, 

p(x)  =  (LCM × GCD)/q(x)

=  (x2 + y2)(x4 + x2y+ y4)(x- y2) / (x− y4)(x+ y2− xy)

=  (x4 - y4)(x4 + x2y+ y4) / (x− y4)(x+ y2− xy)

=  (x4 + x2y+ y4) / (x+ y2− xy)

=  [(x2 + y2)2 - (xy)2] / (x+ y2− xy)

=  (x- xy + y2)(x+ xy + y2(x2  xy + y2)

=  (x2+ xy+ y2)

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 144)

    Apr 14, 25 07:27 PM

    digitalsatmath172.png
    Digital SAT Math Problems and Solutions (Part - 144)

    Read More

  2. Quadratic Equation Problems with Solutions (Part - 1)

    Apr 14, 25 11:33 AM

    Quadratic Equation Problems with Solutions (Part - 1)

    Read More

  3. Quadratic Equation Problems with Solutions (Part - 2)

    Apr 14, 25 11:22 AM

    Quadratic Equation Problems with Solutions (Part - 2)

    Read More