L.C.M. of Polynomials by Factorization

Learn how to solve L.C.M. of polynomials by factorization splitting the middle term.

Solved examples on lowest common multiple of polynomials by factorization:

1. Find the L.C.M of m3 – 3m2 + 2m and m3 + m2 – 6m by factorization.

Solution:

First expression = m3 – 3m2 + 2m

                      = m(m2 – 3m + 2), by taking common ‘m’

                      = m(m2 - 2m - m + 2), by splitting the middle term -3m = -2m - m

                      = m[m(m - 2) - 1(m - 2)]                     

                      = m(m - 2) (m - 1)                     

                      = m × (m - 2) × (m - 1)



Second expression = m3 + m2 – 6m

                          = m(m2 + m - 6) by taking common ‘m’

                          = m(m2 + 3m – 2m - 6), by splitting the middle term m = 3m - 2m

                          = m[m(m + 3) - 2(m + 3)]

                          = m(m + 3)(m - 2)

                          = m × (m + 3) × (m - 2)

In both the expressions, the common factors are ‘m’ and ‘(m - 2)’; the extra common factors are (m - 1) in the first expression and (m + 3) in the 2nd expression.

Therefore, the required L.C.M. = m × (m - 2) × (m - 1) × (m + 3)

                                         = m(m - 1) (m - 2) (m + 3)


2. Find the L.C.M of 3a3 - 18a2x + 27ax2, 4a4 + 24a3x + 36a2x2 and 6a4 - 54a2x2 by factorization.

Solution:

First expression = 3a3 -18a2x + 27ax2

                      = 3a(a2 - 6ax + 9x2), by taking common ‘3a’

              = 3a(a2 - 3ax - 3ax + 9x2), by splitting the middle term - 6ax = - 3ax - 3ax

                      = 3a[a(a - 3x) - 3x(a - 3x)]                     

                      = 3a(a - 3x) (a - 3x)                     

                      = 3 × a × (a - 3x) × (a - 3x)

Second expression = 4a4 + 24a3x + 36a2x2

                          = 4a2(a2 + 6ax + 9x2), by taking common ‘4a2

               = 4a2(a2 + 3ax + 3ax + 9x2), by splitting the middle term 6ax = 3ax + 3ax

                          = 4a2[a(a + 3x) + 3x(a + 3x)]

                          = 4a2(a + 3x) (a + 3x)

                          = 2 × 2 × a × a × (a + 3x) × (a + 3x)

Third expression = 6a4 - 54a2x2

                      = 6a2(a2 - 9x2), by taking common ‘6a2

                      = 6a2[(a)2 - (3x)2), by using the formula of a2 – b2

                      = 6a2(a + 3x) (a - 3x), we know a2 – b2 = (a + b) (a – b)

                      = 2 × 3 × a × a × (a + 3x) × (a - 3x)

The common factors of the above three expressions is ‘a’ and other common factors of first and third expressions are ‘3’ and ‘(a - 3x)’.

The common factors of second and third expressions are ‘2’, ‘a’ and ‘(a + 3x)’.

Other than these, the extra common factors in the first expression is ‘(a - 3x)’ and in the second expression are ‘2’ and ‘(a + 3x)’

Therefore, the required L.C.M. = a × 3 × (a - 3x) × 2 × a × (a + 3x) × (a - 3x) × 2 × (a + 3x) = 12a2(a + 3x)2(a - 3x)2


More problems on L.C.M. of polynomials by factorization splitting the middle term:

3. Find the L.C.M. of 4(a2 - 4), 6(a2 - a - 2) and 12(a2 + 3a - 10) by factorization.

Solution:

First expression = 4(a2 - 4)

                      = 4(a2 - 22), by using the formula of a2 – b2

                      = 4(a + 2) (a - 2), we know a2 – b2 = (a + b) (a – b)

                      = 2 × 2 × (a + 2) × (a - 2)

Second expression = 6(a2 - a - 2)

                          = 6(a2 – 2a + a - 2), by splitting the middle term – a = – 2a + a

                          = 6[a(a - 2) + 1(a - 2)]

                          = 6(a - 2) (a + 1)

                          = 2 × 3 × (a - 2) × (a + 1)

Third expression = 12(a2 + 3a - 10)

                      = 12(a2 + 5a – 2a - 10), by splitting the middle term 3a = 5a – 2a

                      = 12[a(a + 5) - 2(a + 5)]

                      = 12(a + 5) (a - 2)

                      = 2 × 2 × 3 × (a + 5) × (a - 2)

In the above three expressions the common factors are 2 and (a - 2).

Only in the second expression and third expression the common factor is 3.

Other than these, the extra common factors are (a + 2) in the first expression, (a + 1) in the second expression and 2, (a + 5) in the third expression.

Therefore, the required L.C.M. = 2 × (a - 2) × 3 × (a + 2) × (a + 1) × 2 × (a + 5)

                                         = 12(a + 1) (a + 2) (a - 2) (a + 5)







8th Grade Math Practice

From L.C.M. of Polynomials by Factorization to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Months of the Year | List of 12 Months of the Year |Jan, Feb, Mar, Apr

    Dec 01, 23 01:16 AM

    Months of the Year
    There are 12 months in a year. The months are January, February, march, April, May, June, July, August, September, October, November and December. The year begins with the January month. December is t…

    Read More

  2. Days of the Week | 7 Days of the Week | What are the Seven Days?

    Nov 30, 23 10:59 PM

    Days of the Weeks
    We know that, seven days of a week are Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. A day has 24 hours. There are 52 weeks in a year. Fill in the missing dates and answer the questi…

    Read More

  3. Types of Lines |Straight Lines|Curved Lines|Horizontal Lines| Vertical

    Nov 30, 23 01:08 PM

    Types of Lines
    What are the different types of lines? There are two different kinds of lines. (i) Straight line and (ii) Curved line. There are three different types of straight lines. (i) Horizontal lines, (ii) Ver…

    Read More