How do you factor # 9y^2 - 1 #? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Candy Mar 19, 2018 #9y^2-1=(3y+1)(3y-1)# Explanation: #9y^2-1# #=(3y)^2-(1)^2# Using the identity #a^2-b^2=(a+b)(a-b),# You can further simplify #((3y)^2-(1)^2)# as #(3y)^2-(1)^2=(3y+1)(3y-1)# 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 5929 views around the world You can reuse this answer Creative Commons License