Factorization when Monomial is Common

In factorization when monomial is common factor we know, that an algebraic expression is the sum or the difference of monomials.

In order to factorize follow the following steps:

Step 1: Write the algebraic expression.

Step 2: Find the HCF of all the terms of the given algebraic expression.

Step 3:  Express each terms of the algebraic expression as the product of H.C.F and the quotient when it is divided by the H.C.F.

i.e. divide each term of the given expression by the HCF.

Step 4:  Now use distributive property of multiplication over addition or subtraction to express the algebraic expression as the product of H.C.F and the quotient of the expression divided by the H.C.F.

i.e. write the given expression as the product of this HCF and the quotient obtained in step 2. 

Step 5:  Keep the H.C.F. outside the bracket and the quotients obtained within the bracket.


Solved examples of factorization when monomial is common:

1. Factorize each of the following: 

(i) 5x + 20

Solution: 

5x + 20

= 5(x + 4)


(ii) 2n2 + 3n

Solution:

2n2 + 3n

= n(2n + 3)


(iii) 3x2y - 6xy2

Solution:

3x2y - 6xy2

= 3xy(x - 2y)

(iv) 6ab - 9bc

Solution:


6ab - 9bc

= 3b(2a - 3c)


2. Factorize 6a2b2c + 27abc.

Solution:

The H.C.F. of 6a2b2c and 27abc = (H.C.F. of 6 and 27) × (H.C.F. of a2b2c and abc)

The H.C.F. of 6 and 27 = 3

The H.C.F. of a2b2c and abc = abc

Therefore, the H.C.F. of 6a2b2c and 27abc is 3abc.

Now, 6a2b2c + 27abc = \(3abc(\frac{6a^{2}b^{2}c}{3abc} - \frac{27abc}{3abc})\)

                              = 3abc(2ab + 9)

Therefore, the factor of 6a2b2c + 27abc are 3abc and (2ab + 9).



3. Factorize the expression:

18a3 - 27a2b

Solution:

18a3 - 27a2b

HCF of 18a3 and 27a2b is 9a2.

Therefore, 18a3 - 27a2b = 9a2(2a - 3b).





8th Grade Math Practice

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