Question

Factor = 0" alt=" {4y}^{2} - 1 = 0" align="absmiddle" class="latex-formula">
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Answer 1

Answer:

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

Step-by-step 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

⇒ $(a) ^2 = (2y)^2 = 2y \times 2y = 4y^2\\\implies 4y^2 = (2y)^2$

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