How do you factor #a^4-18a^2+81#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Alan N. Aug 14, 2016 #(a-3)^2(a+3)^2# Explanation: #a^4-18a^2+81# Factorise: # (a^2-9)(a^2-9)# #=(a^2-3^2)(a^2-3^2)# #=(a+3)(a-3)(a+3)(a-3)# #=(a-3)^2(a+3)^2# 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 2897 views around the world You can reuse this answer Creative Commons License