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

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