How do you factor #(x+a)^2-b^2 = 0#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer F. Javier B. May 26, 2018 Use identity #(a+b)(a-b)=a^2-b^2# #(x+a)^2-b^2=(x+a+b)(x+a-b)=0# Then #x+a+b=0# #x+a-b=0# #x=-(a+b)# #x=b-a# 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 3644 views around the world You can reuse this answer Creative Commons License