How do you factor #256n^4-c^4#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Shwetank Mauria Feb 5, 2017 #256n^4-c^4=(16n^2+c^2)(4n+c)(4n-c)# Explanation: As the two monomials #256n^4# and #c^4# are perfect square, we can use the identity #a^2-b^2=(a+b)(a-b)# to factorize it and #256n^4-c^4# = #(16n^2)^2-(c^2)^2# = #(16n^2+c^2)(16n^2-c^2)# = #(16n^2+c^2)((4n)^2-c^2)# = #(16n^2+c^2)(4n+c)(4n-c)# 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 1794 views around the world You can reuse this answer Creative Commons License