Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2 (ii) 4y2 - 4y + 1 (iii) x2 - y2/100
Solution:
Using algebraic Identities,
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
(a + b)(a - b) = a2 - b2
(i) 9x2 + 6xy + y2
= (3x)2 + 2(3x)( y) + (y)2
Identity: (a + b)2 = a2 + 2ab + b2
Considering a = 3x and b = y
Hence 9x2 + 6xy + y2 = (3x + y)2
(ii) 4y2 - 4y + 1
= (2y)2 - 2(2y)(1) + (1)2
Identity: (a - b)2 = a2 - 2ab + b2
Considering a = 2y and b = 1
Hence 4y2 - 4y + 1 = (2y - 1)2
(iii) x2 - y2/100
= x2 - (y/10)2
Identity: (a + b)(a - b) = a2 - b2
Considering a = x and b = y/10
Hence, x2 - y2/100 = (x + y/10)(x - y/10)
☛ Check: Class 9 Maths NCERT Solutions Chapter 2
Video Solution:
Factorise the following using appropriate identities: (i) 9x² + 6xy + y² (ii) 4y² - 4y + 1 (iii) x² - y²/100
NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.5 Question 3:
Summary:
The factorized form of the following 9x2 + 6xy + y2, 4y2 - 4y + 1, and x2 - y2/100 using appropriate identities are (3x + y)2, (2y - 1)2, and (x + y/10) (x - y/10) respectively.
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