4y2 can be written as (2y)24y can be written as 2(1)(2y)1 can be written as 12⇒ 4y2 - 4y + 1 = (2y)2 - 2(1)(2y) + 12Using identity (a - b)2 = a2 - 2ab + b2Here a = 2y and b = 1⇒ 4y2 - 4y + 1 = (2y - 1)2Therefore 4y2 - 4y + 1 = (2y - 1) (2y - 1)

Q. 3 B

Factorise the fol

Class 9th AP- Mathematics 2. Polynomials and Factorisation

Answer :

4y² can be written as (2y)²

4y can be written as 2(1)(2y)

1 can be written as 1²

⇒ 4y² - 4y + 1 = (2y)² - 2(1)(2y) + 1²

Using identity (a - b)² = a² - 2ab + b²

Here a = 2y and b = 1

⇒ 4y² - 4y + 1 = (2y - 1)²

Therefore 4y² - 4y + 1 = (2y - 1) (2y - 1)

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