Use the variation-of-squares formula: a^2 - b^2 = (a + b)(a - b) substitute 4x^2 (sq. root of 16x^4) for a and 9 (sq. root of 80 one) for b: 16x^4 - 80 one = (4x^2 + 9)(4x^2 - 9) observe the comparable rule to the (4x^2 - 9): 4x^2 - 9 = (2x + 3)(2x - 3) That makes the comprehensive factoring: (4x^2 + 9)(2x + 3)(2x - 3)

always remember : a^2-b^2=(a+b)(a-b)

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

So your answer should read (2x-3y)

(4x^2)^2-(9y^8)^2