Every algebraic expression has factors as 1 and itself. If the Monomial is a Common Factor, then the product of two or more numbers or variables, then it will have factors of 1 and itself. Find different problems on the Factorization of algebraic expressions and know the process to solve each problem in an easy way. We have included tips and tricks to solve Factorization Problems that help students to learn the Factorization concept easily.

How to Find Factorization of an Expression when a Monomial is a Common Factor?

Follow the below process to find the Factorization of given problems.

(i) Write an algebraic expression.
(ii) Find the H.C.F. of all the individual terms of the expression.
(iii) Divide each individual term of the expression by the H.C.F.
(iv) Keep the H.C.F. outside the bracket and the quotients obtained within the bracket.

The greatest common factor (GCF) of two or more monomials is the product of the greatest common factor of the coefficients and the greatest common factors of the variables

Solved Examples on Monomial as a Common Factor

1. Factorize each of the following algebraic expressions.

(i) 15a + 10

Solution:
The given expression is 15a + 10
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 15 and 10 is 5.
HCF of literal coefficients:
The lowest power of a is 0.
Therefore, the HCF of literal coefficients is nothing.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 15a + 10 is 5.
Multiply and divide each term of the given expression 15a + 10 with 5a
5((15a/5) + (10/5)) = 5(3a + 2)

The final answer is 5(3a + 2)

(ii) 18mn2 + 9m2n – 12mn

Solution:
The given expression is 18mn2 + 9m2n – 12mn
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 18, 9, and 12 is 3.
HCF of literal coefficients:
The lowest power of m is 1.
The lowest power of n is 1.
Therefore, the HCF of literal coefficients is mn.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 18mn2 + 9m2n – 12mn is 3mn.
Multiply and divide each term of the given expression 15a + 10 with 5a
3mn((18mn2/3mn) + (9m2n/3mn) – (12mn/3mn)) = 3mn (6n + 3m – 4)

The final answer is 3mn (6n + 3m – 4).

(iii) 20x2y – 8xy2 + 32xy

Solution:
The given expression is 20x2y – 8xy2 + 32xy
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 20, 8, and 32 is 4.
HCF of literal coefficients:
The lowest power of x is 1.
The lowest power of y is 1.
Therefore, the HCF of literal coefficients is xy.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 20x2y – 8xy2 + 32xy is 4xy.
Multiply and divide each term of the given expression 20x2y – 8xy2 + 32xy with 4xy
4xy(20x2y/4xy – 8xy2/4xy+ 32xy/4xy) = 4xy (5x – 2y + 8)

The final answer is 4xy (5x – 2y + 8).

(iv) 13a3 + 39a2b

The given expression is 13a3 + 39a2b
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 13 and 39 is 13.
HCF of literal coefficients:
The lowest power of a is 2.
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is a2.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 13a3 + 39a2b is 13a2.
Multiply and divide each term of the given expression 13a3 + 39a2b with 13a2.
13a2(13a3/13a2 + 39a2b/13a2) = 13a2 (a + 3b)

The final answer is 13a2 (a + 3b).

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