Question

select the correct answer.

what is the completely factored form of this polynomial?

(81x^4 - 16y^4)

( ) ((3x^2 + 2y^2)(9x + 4y)(9x - 4y))
( ) ((9x^2 + 4y^2)(9x^2 - 4y^2))
( ) ((9x^2 + 4y^2)(3x - 2y)^2)
( ) ((9x^2 + 4y^2)(3x + 2y)(3x - 2y))

Explanation:

Apply the difference of squares formula to the expression

$$ 81x^4 - 16y^4 = (9x^2)^2 - (4y^2)^2 = (9x^2 + 4y^2)(9x^2 - 4y^2) $$

Factor the remaining difference of squares term

$$ 9x^2 - 4y^2 = (3x)^2 - (2y)^2 = (3x + 2y)(3x - 2y) $$

Combine the factored terms into the final expression

$$ 81x^4 - 16y^4 = (9x^2 + 4y^2)(3x + 2y)(3x - 2y) $$

Answer:

• (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) \((9x^2 + 4y^2)(3x + 2y)(3x - 2y)\) (Correct answer)