Question

HCF of 18a^2^b^2 c and 24abc^2 with explanation

Answer 1

Answer:

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(6a2b2c3abc−27abc3abc)3abc(6a2b2c3abc−27abc3abc)

= 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