How do you factor #n^4 - 1#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Tom Dec 17, 2015 #(n-1)(n+1)(n^2+1)# Explanation: By using this formula #a^2-b^2 = (a-b)(a+b)# here we have #a = n^2# and #b = 1# #(n^2)^2 - 1# (remember #(a^n)^m = a^(nm)#) #(n^2-1)(n^2+1)# You do it again with #(n^2-1)(n^2+1)# and you have your answer 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 4960 views around the world You can reuse this answer Creative Commons License