Question

question
factor completely.
\\(121x^4 - 9y^4\\)

Explanation:

Step1: Recognize as difference of squares

The expression \(121x^4 - 9y^4\) can be written as \((11x^2)^2-(3y^2)^2\). Using the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\), we get:
\((11x^2 + 3y^2)(11x^2 - 3y^2)\)

Step2: Check for further factoring

Now, \(11x^2 - 3y^2\) is also a difference of squares (since \(11x^2=(\sqrt{11}x)^2\) and \(3y^2 = (\sqrt{3}y)^2\)). Applying the difference of squares formula again:
\((11x^2 + 3y^2)(\sqrt{11}x+\sqrt{3}y)(\sqrt{11}x - \sqrt{3}y)\)
But if we consider integer coefficients only, we stop at \((11x^2 + 3y^2)(11x^2 - 3y^2)\). However, for complete factoring over real numbers, we use the square roots.

Answer:

\((11x^2 + 3y^2)(\sqrt{11}x+\sqrt{3}y)(\sqrt{11}x - \sqrt{3}y)\) (or \((11x^2 + 3y^2)(11x^2 - 3y^2)\) if restricting to integer coefficients)