# Factorise the following using appropriate identities:

(i) 9x^{2} + 6xy + y^{2 }(ii) 4y^{2} - 4y + 1 (iii) x^{2} - y^{2}/100

**Solution:**

Using algebraic Identities,

(a + b)^{2} = a^{2} + 2ab + b^{2}

(a - b)^{2} = a^{2} - 2ab + b^{2}

(a + b)(a - b) = a^{2} - b^{2}

**(i)** 9x^{2} + 6xy + y^{2}

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

Identity: (a + b)^{2} = a^{2} + 2ab + b^{2}

Considering a = 3x and b = y

Hence 9x^{2} + 6xy + y^{2} = (3x + y)^{2}

**(ii)** 4y^{2} - 4y + 1

= (2y)^{2} - 2(2y)(1) + (1)^{2}

Identity: (a - b)^{2} = a^{2} - 2ab + b^{2}

Considering a = 2y and b = 1

Hence 4y^{2} - 4y + 1 = (2y - 1)^{2}

**(iii)** x^{2} - y^{2}/100

= x^{2} - (y/10)^{2}

Identity: (a + b)(a - b) = a^{2} - b^{2}

Considering a = x and b = y/10

Hence, x^{2} - y^{2}/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 9x^{2} + 6xy + y^{2}, 4y^{2} - 4y + 1, and x^{2} - y^{2}/100 using appropriate identities are (3x + y)^{2}, (2y - 1)^{2}, and (x + y/10) (x - y/10) respectively.

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