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:

First expression = m

= m(m

= m(m

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

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

= m × (m - 2) × (m - 1)

Second expression = m

= m(m

= m(m

= 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)

First expression = 3a

= 3a(a

= 3a(a

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

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

= 3 × a × (a - 3x) × (a - 3x)

= 4a

= 4a

= 4a

= 4a

= 2 × 2 × a × a × (a + 3x) × (a + 3x)

Third expression = 6a

= 6a

= 6a

= 6a

= 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) = 12aMore
problems on L.C.M. of polynomials by factorization **splitting the middle term:**

First expression = 4(a

= 4(a

= 4(a + 2) (a - 2), we know a

= 2 × 2 × (a + 2) × (a - 2)

Second expression = 6(a

= 6(a

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

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

= 2 × 3 × (a - 2) × (a + 1)

Third expression = 12(a= 12(a

= 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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