How do you factor #12c^2-75# completely? Algebra Polynomials and Factoring Factoring Completely 1 Answer Wataru Oct 28, 2014 #12c^2-75# by factoring out 3, #=3(4c^2-25)# by #a^2-b^2=(a+b)(a-b)#, #=3(2c+5)(2c-5)# I hope that this was helpful. 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 #x^6-26x^3-27#? How do you factor #100x^2+180x+81#? How do you factor #x^5-5x^3-36x# completely? See all questions in Factoring Completely Impact of this question 10875 views around the world You can reuse this answer Creative Commons License