How do you factor # 36m^2-132mn+121n^2#? Algebra Polynomials and Factoring Factoring Completely 1 Answer GiĆ³ · Jacobi J. Jun 29, 2018 I tried this: Explanation: This is the classic: #(a-b)^2=a^2-2ab+b^2# where in your case we have: #a=6m# and #b=11n# This can be factored as: #(6m-11n)^2# 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 1974 views around the world You can reuse this answer Creative Commons License