Question

select the correct answer.
what is the completely factored form of this polynomial?
81x⁴ - 16y⁴
a. (3x² + 2y²)(9x + 4y)(9x - 4y)
b. (9x² + 4y²)(9x² - 4y²)
c. (9x² + 4y²)(3x - 2y)²
d. (9x² + 4y²)(3x + 2y)(3x - 2y)

Explanation:

Step1: Recognize as difference of squares

The polynomial \(81x^4 - 16y^4\) can be written as \((9x^2)^2-(4y^2)^2\). Using the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\), we get:
\((9x^2 + 4y^2)(9x^2 - 4y^2)\)

Step2: Factor \(9x^2 - 4y^2\) further

Notice that \(9x^2 - 4y^2\) is also a difference of squares, where \(a = 3x\) and \(b = 2y\). Applying the difference of squares formula again:
\(9x^2 - 4y^2=(3x + 2y)(3x - 2y)\)

Step3: Combine the factors

Substituting the factored form of \(9x^2 - 4y^2\) back into the previous expression, we have:
\(81x^4 - 16y^4=(9x^2 + 4y^2)(3x + 2y)(3x - 2y)\)

Answer:

D. \((9x^2 + 4y^2)(3x + 2y)(3x - 2y)\)