How do you factor #1-(x-y)^2#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Deepak G. Aug 8, 2016 #=(1-x+y)(1+x-y)# Explanation: Since #a^2-b^2# #(a-b)(a+b)# #1-(x-y)^2# #=1^2-(x-y)^2# #=(1-(x-y))(1+(x-y))# #=(1-x+y)(1+x-y)# 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 11632 views around the world You can reuse this answer Creative Commons License