How do you factor #81x^4 -256#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer 1s2s2p Apr 18, 2018 #(9x^2+16)(3x+4)(3x-4)# Explanation: Using the difference of two squares (#a^2-b^2=(a+b)(a-b)#) we can get: #(9x^2+16)(9x^2-16)# #9x^2-16=(3x+4)(3x-4)# #(9x^2+16)(3x+4)(3x-4)# Answer link Related questions How do you factor special products of polynomials? How do you identify special products when factoring? How do you factor #x^3 -8#? What are the factors of #x^3y^6 – 64#? How do you know if #x^2 + 10x + 25# is a perfect square? How do you write #16x^2 – 48x + 36# as a perfect square trinomial? What is the difference of two squares method of factoring? How do you factor #16x^2-36# using the difference of squares? How do you factor #2x^4y^2-32#? How do you factor #x^2 - 27#? See all questions in Factor Polynomials Using Special Products Impact of this question 5336 views around the world You can reuse this answer Creative Commons License