# HCF AND LCM – EXERCISE 3.3.2 – Class 9

HCF AND LCM – EXERCISE 3.3.2 – Class 9

1. Find the HCF of each of the following pairs of polynomials:

(i) 5x – 10 and 5x2 – 20

Solution:

5x – 10 = 5(x – 2)

5x2 – 20 = 5(x2 – 4)

= 5(x – 2) (x + 2)

H.C.F= 5(x – 2)

(ii) 2x + 5 and 5x + 2

Solution:

HCF = 1 since there is no common factor

(iii) a2 – 1 and 3a – 3

Solution:

a2 – 1 = (a + 1) (a – 1)

3a – 3 = 3(a – 1)

HCF = (a – 1)

(iv) x2 – xy and x2 – y2

Solution:

x2 – xy = x (x –y)

x2 – y2 = (x – y) (x + y)

HCF = (x – y)

(v) 2x2 + x and 4x2 – 1

Solution:

2x2 + x = x (2x + 1)

4x2 – 1 = (2x + 1) (2x – 1)

HCF = (2x + 1)

(vi) x3 + 27 and x2 – 9

Solution:

x3 + 27 = (x3 + x3)

= (x + 3) (x3 – 3x + 33)

= (x + 3) (x3 – 3x + 9)

x2 – 9 = x2 – 33 = (x + 3) (x – 3)

HCF = (x + 3)

(vii) x3 + y3 and 3x3 – 3y3

Solution:

x3 + y3 = (x + y) (x2 – xy + y2)

3x3 – 3y3 = 3(x3 – y3) =3(x +y) (x –y)

HCF = (x + y)

(viii) a2 + 125 and a2 – 125

Solution:

a2 + 125 = a3 + 53

= (a + 5) (a2 – 5a + 52)

= (a + 5) (a2 – 5a + 25)

a2 – 125 = a2 – 52 = (a + 5) (a – 5)

HCF = (a + 5)

(ix) 4x2 – 1 and 4x2 + 4x +1

Solution:

4x2 – 1 = (2x)2 – 1 = (2x +1) (2x – 1)

4x2 + 4x +1 = (2x +1)2 [a2 + 2ab + b2 = (a + b)2]

HCF = (2x +1)

(x) 6x2 – 2x and 9x2 – 3x

Solution:

6x2 – 2x = 2x (3x – 1)

9x2 – 3x = 3x (3x –1)

HCF = x(3x –1)

HCF AND LCM – EXERCISE 3.3.2 – Class 9