Factor completely. y^(2)-9x^(6) Answer:

Recognize as difference of squares: Recognize the expression y^2 - 9x^6 as a difference of squares. A difference of squares is a binomial of the form a^2 - b^2, which factors into (a + b)(a - b). Here, y^2 is a perfect square (a^2) and 9x^6 is also a perfect square (b^2) because 9 is a perfect square (3^2) and x^6 is a perfect square (x^3)^2.Write expression as difference of squares: Write the expression as a difference of squares. y^2 - 9x^6 can be written as (y)^2 - (3x^3)^2.Apply difference of squares formula: Apply the difference of squares formula. Using the formula (a^2 - b^2) = (a + b)(a - b), we get: (y)^2 - (3x^3)^2 = (y + 3x^3)(y - 3x^3).Check for additional factoring: Check for any additional factoring. The terms (y + 3x^3) and (y - 3x^3) are not factorable further with real numbers, so the expression is now completely factored.

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Factor completely.

y^(2)-9x^(6)
Answer:

Factor completely. $\newline$ $y^{2}-9 x^{6}$ $\newline$ Answer:

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Algebra 2

Multiply and divide rational expressions

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Q. Factor completely.

\newline

y^{2}-9 x^{6}

\newline

Answer:

Recognize as difference of squares: Recognize the expression $y^2 - 9x^6$ as a difference of squares. $\newline$ A difference of squares is a binomial of the form $a^2 - b^2$ , which factors into $(a + b)(a - b)$ . $\newline$ Here, $y^2$ is a perfect square $(a^2)$ and $9x^6$ is also a perfect square $(b^2)$ because $9$ is a perfect square $(3^2)$ and $x^6$ is a perfect square $a^2 - b^2$ $0$ .
Write expression as difference of squares: Write the expression as a difference of squares. $\newline$ $y^2 - 9x^6$ can be written as $(y)^2 - (3x^3)^2$ .
Apply difference of squares formula: Apply the difference of squares formula. $\newline$ Using the formula $(a^2 - b^2) = (a + b)(a - b)$ , we get: $\newline$ $(y)^2 - (3x^3)^2 = (y + 3x^3)(y - 3x^3)$ .
Check for additional factoring: Check for any additional factoring. $\newline$ The terms $(y + 3x^3)$ and $(y - 3x^3)$ are not factorable further with real numbers, so the expression is now completely factored.