Question

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

Explanation:

Step1: Recognize as difference of squares

The polynomial \(81x^4 - 16y^4\) is a difference of squares since \(81x^4=(9x^2)^2\) and \(16y^4=(4y^2)^2\). The formula for difference of squares is \(a^2 - b^2=(a + b)(a - b)\). So, \(81x^4 - 16y^4=(9x^2 + 4y^2)(9x^2 - 4y^2)\).

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 get \(81x^4 - 16y^4=(9x^2 + 4y^2)(3x + 2y)(3x - 2y)\).

Answer:

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