To factor the polynomial [tex]\( 81x^4 - 16y^4 \)[/tex], we start by recognizing that this polynomial is a difference of squares.
1. Difference of Squares Formula:
The difference of squares formula states that [tex]\( a^2 - b^2 = (a - b)(a + b) \)[/tex].
2. Identifying the Squares:
[tex]\[ 81x^4 = (9x^2)^2 \quad \text{and} \quad 16y^4 = (4y^2)^2 \][/tex]
3. Apply the Difference of Squares:
We apply the difference of squares formula to [tex]\( 81x^4 - 16y^4 \)[/tex]:
[tex]\[
81x^4 - 16y^4 = (9x^2)^2 - (4y^2)^2 = (9x^2 - 4y^2)(9x^2 + 4y^2)
\][/tex]
4. Factoring Further:
The term [tex]\( 9x^2 - 4y^2 \)[/tex] is itself a difference of squares:
[tex]\[
9x^2 - 4y^2 = (3x)^2 - (2y)^2 = (3x - 2y)(3x + 2y)
\][/tex]
5. Combining All Factors:
Therefore, the completely factored form of the polynomial [tex]\( 81x^4 - 16y^4 \)[/tex] is:
[tex]\[
(9x^2 - 4y^2)(9x^2 + 4y^2) = (3x - 2y)(3x + 2y)(9x^2 + 4y^2)
\][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{(9x^2 + 4y^2)(3x + 2y)(3x - 2y)} \][/tex]
So, the answer is option D.
