Lowest common multiple of Polynomials by Factorization

How to find the lowest common multiple of polynomials by factorization?

Let us follow the following examples to know how to find the lowest common multiple (L.C.M.) of polynomials by factorization.


Solved examples of lowest common multiple of polynomials by factorization:

1. Find out the L.C.M. of a2 + a and a3 – a by factorization.

Solution:

First expression = a2 + a

                      = a(a + 1), by taking common ‘a’


Second expression = a3 - a

                          = a(a2 - 1), by taking common ‘a’

                          = a(a2 – 12), by using the formula of a2 – b2

                          = a(a + 1) (a - 1), we know a2 – b2 = (a + b) (a – b)

The common factors of the two expressions are ‘a’ and (a + 1); (a - 1) is the extra factor in the second expression.

Therefore, the required L.C.M. of a2 + a and a3 – a is a(a + 1) (a - 1)



2. Find out the L.C.M of x2 - 4 and x2+ 2x by factorization.

Solution:

First expression = x2 - 4

                      = x2 - 22, by using the formula of a2 – b2

                      = (x + 2) (x - 2), we know a2 – b2 = (a + b) (a – b)

Second expression = x2 + 2x

                          = x(x + 2), by taking common ‘x’    

The common factor of the two expressions is ‘(x + 2)’.

The extra common factor in the first expression is (x - 2) and in the second expression is x.

Therefore, the required L.C.M = (x + 2) × (x - 2) × x

                                        = x(x + 2) (x - 2)


3. Find out the L.C.M of x3 + 2x2 and x3 + 3x2 + 2x by factorization.

Solution:

First expression = x3 + 2x2

                      = x2(x + 2), by taking common ‘x2

                      = x × x × (x + 2)

Second expression = x3 + 3x2 + 2x

                          = x(x2 + 3x + 2), by taking common ‘x’

                          = x(x2 + 2x + x + 2), by splitting the middle term 3x = 2x + x

                          = x[x(x + 2) + 1(x + 2)]

                          = x(x + 2) (x + 1)

                          = x × (x + 2) × (x + 1)

In both the expressions, the common factors are ‘x’ and ‘(x + 2)’; the extra common factors are ‘x’ in the first expression and ‘(x + 1)’ in the second expression.

Therefore, the required L.C.M. = x × (x + 2) × x × (x + 1)

                                         = x2(x + 1) (x + 2)







8th Grade Math Practice

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