HCF AND LCM – EXERCISE 3.3.2 – Class 9

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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)

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HCF AND LCM – EXERCISE 3.3.2 – Class 9

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