To factor the polynomial [tex]\(81 x^4 - 16 y^4\)[/tex], we need to examine its structure and use algebraic identities.
First, notice that [tex]\(81 x^4 - 16 y^4\)[/tex] can be expressed as a difference of squares:
[tex]\[
81 x^4 - 16 y^4 = (9 x^2)^2 - (4 y^2)^2
\][/tex]
The difference of squares identity states that [tex]\(a^2 - b^2\)[/tex] can be factored as [tex]\((a + b)(a - b)\)[/tex]. Applying this identity, let [tex]\(a = 9 x^2\)[/tex] and [tex]\(b = 4 y^2\)[/tex], we get:
[tex]\[
81 x^4 - 16 y^4 = \left(9 x^2 + 4 y^2\right)\left(9 x^2 - 4 y^2\right)
\][/tex]
Next, we observe that [tex]\(9 x^2 - 4 y^2\)[/tex] is also a difference of squares. Factoring it further, where [tex]\(9 x^2 - 4 y^2 = (3 x)^2 - (2 y)^2\)[/tex], we can apply the difference of squares identity again with [tex]\(a = 3 x\)[/tex] and [tex]\(b = 2 y\)[/tex]:
[tex]\[
9 x^2 - 4 y^2 = \left(3 x + 2 y\right)\left(3 x - 2 y\right)
\][/tex]
Combining these results, the completely factored form of the original polynomial is:
[tex]\[
81 x^4 - 16 y^4 = \left(9 x^2 + 4 y^2\right)\left(3 x + 2 y\right)\left(3 x - 2 y\right)
\][/tex]
Therefore, the correct answer is:
D. [tex]\(\left(9 x^2 + 4 y^2\right)(3 x + 2 y)(3 x - 2 y)\)[/tex]
