Question

4y^2-4y+1 factorise using appropriate identities
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Answer 1

$\huge\purple{Required \: Answer :}$

$\huge\sf{Explanation :}$

Given :

$4 {y}^{2} - 4y + 1$

Factorization :

$4 {y}^{2} - 4y + 1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ 4 {y}^{2} - 2y - 2y + 1 = 0 \: \: \: \: \: \: \: \\ (4 {y}^{2} - 2y) - (2y - 1) = 0 \\ 2y(2y - 1) - 1(2y - 1) = 0 \\ (2y - 1)(2y - 1) = 0 \: \: \: \: \: \: \: \: \: \: \\ y = \frac{1}{2}$

Answer 2

Given :-

• 4y² - 4y + 1

Answer :-

The factores of given expression is y = ½

Explaination :-

The identity that can be used to factorize the given expression is:

• (a - b)² = a² - 2ab + b²

Now, using the above identity we get :

→ (2y)² - 2(2y)(1) + (1)² = 0

→ (2 y - 1)² = 0

→ (2y - 1) (2 y - 1) = 0

→ y = ½

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$