HCF AND LCM – EXERCISE 3.3.2 – Class 9

**Find the HCF of each of the following pairs of polynomials:**

**(i) 5x – 10 and 5x**^{2}**– 20 **

Solution:

5x – 10 = 5(x – 2)

5x^{2} – 20 = 5(x^{2} – 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) a**^{2}**– 1 and 3a – 3 **

Solution:

a^{2} – 1 = (a + 1) (a – 1)

3a – 3 = 3(a – 1)

HCF = (a – 1)

**(iv) x**^{2}**– xy and x**^{2}**– y**^{2 }

Solution:

x^{2} – xy = x (x –y)

x^{2} – y^{2} = (x – y) (x + y)

HCF = (x – y)

**(v) 2x**^{2}**+ x and 4x**^{2}**– 1 **

Solution:

2x^{2} + x = x (2x + 1)

4x^{2} – 1 = (2x + 1) (2x – 1)

**HCF = (2x + 1) **

**(vi) x**^{3}**+ 27 and x**^{2}**– 9 **

Solution:

x^{3} + 27 = (x^{3} + x^{3})

= (x + 3) (x^{3} – 3x + 3^{3})

= (x + 3) (x^{3} – 3x + 9)

x^{2} – 9 = x^{2} – 3^{3} = (x + 3) (x – 3)

**HCF = (x + 3) **

**(vii) x**^{3 }**+ y**^{3}**and 3x**^{3 }**– 3y**^{3}

Solution:

x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

3x^{3} – 3y^{3} = 3(x^{3} – y^{3}) =3(x +y) (x –y)

**HCF = (x + y)**

**(viii) a**^{2}**+ 125 and a**^{2}**– 125 **

Solution:

a^{2} + 125 = a^{3} + 5^{3}

= (a + 5) (a^{2} – 5a + 52)

= (a + 5) (a^{2} – 5a + 25)

a^{2} – 125 = a^{2} – 5^{2} = (a + 5) (a – 5)

**HCF = (a + 5) **

**(ix) 4x**^{2}**– 1 and 4x**^{2}**+ 4x +1 **

Solution:

4x^{2} – 1 = (2x)^{2} – 1 = (2x +1) (2x – 1)

4x^{2} + 4x +1 = (2x +1)^{2} [a^{2} + 2ab + b^{2} = (a + b)^{2}]

**HCF = (2x +1)**

** **

**(x) 6x**^{2}**– 2x and 9x**^{2}**– 3x **

Solution:

6x^{2} – 2x = 2x (3x – 1)

9x^{2} – 3x = 3x (3x –1)

**HCF = x(3x –1)**

HCF AND LCM – EXERCISE 3.3.2 – Class 9

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