How do you factor #81x^2 - 36y^2? Algebra Polynomials and Factoring Factoring Completely 1 Answer Meave60 · George C. Jun 21, 2015 The answer is #9(3x+2y)(3x-2y)#. Explanation: Factor #81x^2-36y^2#. Factor out #9#. #9(9x^2-4y^2)# #(9x^2-4y^2)# is in the form the difference of squares: #a^2-b^2=(a+b)(a-b)# #a=3x;##b=2y# #9((3x)^2-(2y)^2)# = #9(3x+2y)(3x-2y)# Answer link Related questions What is Factoring Completely? How do you know when you have completely factored a polynomial? Which methods of factoring do you use to factor completely? How do you factor completely #2x^2-8#? Which method do you use to factor #3x(x-1)+4(x-1) #? What are the factors of #12x^3+12x^2+3x#? How do you find the two numbers by using the factoring method, if one number is seven more than ... How do you factor #12c^2-75# completely? How do you factor #x^6-26x^3-27#? How do you factor #100x^2+180x+81#? See all questions in Factoring Completely Impact of this question 474 views around the world You can reuse this answer Creative Commons License