HCF AND LCM – EXERCISE 3.3.2 – Class 9
- 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
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