Question

which expression shows the complete factored form of $9x^4 - 121y^2$?
a $(9x^2 - 11y)(9x^2 - 11y)$
b $(3x^2 - 11y)(3x^2 - 11y)$
c $(3x^2 - 11y)(3x^2 + 11y)$

Explanation:

Step1: Identify the form

The given expression \(9x^4 - 121y^2\) is a difference of squares. The formula for factoring a difference of squares is \(a^2 - b^2=(a - b)(a + b)\).
Here, \(a^2 = 9x^4\), so \(a=\sqrt{9x^4}=3x^2\), and \(b^2 = 121y^2\), so \(b=\sqrt{121y^2}=11y\).

Step2: Apply the formula

Using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\) with \(a = 3x^2\) and \(b = 11y\), we get:
\(9x^4-121y^2=(3x^2)^2-(11y)^2=(3x^2 - 11y)(3x^2+11y)\)

Answer:

C. \((3x^2 - 11y)(3x^2 + 11y)\)