Question

Factorise the following using appropriate identities: 

(i) 9x ² + 6xy + y 2 

(ii) 4y ² – 4y + 1 

(iii) x ² – \(\frac{y^2 }{ 100}\)

Answer 1

Factorisation Using Identities

1. Identity Recall: 

• Square of a binomial: \( (a+b)^2 = a^2 + 2ab + b^2 \)
• Difference of squares: \( a^2 - b^2 = (a-b)(a+b) \)

(i) \( 9x^2 + 6xy + y^2 \)

Step 1: Observe that \( 9x^2 = (3x)^2 \), \( y^2 = (y)^2 \), and \( 6xy = 2(3x)(y) \) Step 2: Apply the square of a binomial formula: \[ 9x^2 + 6xy + y^2 = (3x + y)^2 \]

(ii) \( 4y^2 - 4y + 1 \)

Step 1: Observe that \( 4y^2 = (2y)^2 \) and \( 1 = (1)^2 \), middle term = -4y = 2*(2y)*(-1) Step 2: Apply the square of a binomial formula: \[ 4y^2 - 4y + 1 = (2y - 1)^2 \]

(iii) \( x^2 - \frac{y^2}{100} \)

Step 1: Express as a difference of squares: \[ x^2 - \left(\frac{y}{10}\right)^2 \] Step 2: Apply difference of squares formula: \[ x^2 - \frac{y^2}{100} = \left(x - \frac{y}{10}\right)\left(x + \frac{y}{10}\right) \]

Answer:

• (i) \( 9x^2 + 6xy + y^2 = (3x + y)^2 \)
• (ii) \( 4y^2 - 4y + 1 = (2y - 1)^2 \)
• (iii) \( x^2 - \frac{y^2}{100} = \left(x - \frac{y}{10}\right)\left(x + \frac{y}{10}\right) \)