Question

Factor

`{4y}^(2) - 1 = 0`
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Answer 1

The given expression is factorized as
`4y^2 - 1 = 0  \implies  (2y -1)(2y +1)  = 0`

Explanation:

Here, the given expression to factorize is:
`4y^2 - 1 = 0`

Now, using Algebraic Identity:

`a^2 - b^2 = (a + b) (a-b)`

Now here let a = 2y , b = 1

⇒
`(a)  ^2 =  (2y)^2  = 2y * 2y =  4y^2\\\implies  4y^2 = (2y)^2`

So, The given expression is equivalent to:

`4y^2 - 1 = 0  \implies  (2y)^2 - (1)^2 = 0\\\implies (2y -1)(2y +1)  = 0`

Hence the given expression is factorized as
`4y^2 - 1 = 0  \implies  (2y -1)(2y +1)  = 0`