Please help me for solving this problem of class 9th ncert book of Polynomials chapter of exercise 2.5 of math of question no.3(2) give me the best and simple way for solving this question Factorize the following using appropriate identities:(ii) 4y2−4y+1
Share
4y2−4y+1 = (2y)2–(2×2y×1)+1
Using identity, x2 – 2xy + y2 = (x – y)2
Here, x = 2y
y = 1
4y2−4y+1 = (2y)2–(2×2y×1)+12
= (2y–1)2
= (2y–1)(2y–1)