Factor completely. 36y^(4)-1 Answer:

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Factor completely.  36y^(4)-1 Answer:

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Recognize Difference of Squares: Step Title: Recognize the Difference of Squares Concise Step Description: Identify that the expression is a difference of squares, which can be factored into the product of a sum and difference. Step Calculation: Recognize that $ 36y^4 $ is a perfect square, as is 1. The expression can be written as $ (6y^2)^2 - 1^2 $. Step Output: Expression as a difference of squares: $ (6y^2)^2 - 1^2 $Apply Squares Formula: Step Title: Apply the Difference of Squares Formula Concise Step Description: Use the difference of squares formula, which states that $ a^2 - b^2 = (a + b)(a - b) $, to factor the expression. Step Calculation: Apply the formula with $ a = 6y^2 $ and $ b = 1 $ to get $ (6y^2 + 1)(6y^2 - 1) $. Step Output: Factored form using the difference of squares: $ (6y^2 + 1)(6y^2 - 1) $Factor Further if Possible: Step Title: Factor Further if Possible Concise Step Description: Check if the resulting terms can be factored further. Step Calculation: Notice that $ 6y^2 - 1 $ is also a difference of squares, as it can be written as $ (2\sqrt{3}y)^2 - 1^2 $. Step Output: Recognition of further difference of squares: $ 6y^2 - 1 = (2\sqrt{3}y)^2 - 1^2 $Factor Remaining Squares: Step Title: Factor the Remaining Difference of Squares Concise Step Description: Apply the difference of squares formula to the term $ 6y^2 - 1 $. Step Calculation: Factor $ 6y^2 - 1 $ using the formula with $ a = 2\sqrt{3}y $ and $ b = 1 $ to get $ (2\sqrt{3}y + 1)(2\sqrt{3}y - 1) $. Step Output: Fully factored form: $ (6y^2 + 1)(2\sqrt{3}y + 1)(2\sqrt{3}y - 1) $

Factor completely. $\newline$ $36 y^{4}-1$ $\newline$ Answer:

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Factor completely. $\newline$ $36 y^{4}-1$ $\newline$ Answer: