How do you factor completely #6y^2 - 54#? Algebra Polynomials and Factoring Factoring Completely 1 Answer Tarik Apr 22, 2016 #6(y-3)(y+3)# Explanation: #6y^2-54=6y^2-6*9=6(y^2-9)=6[(y)^2-(3)^2]=6(y-3)(y+3)# 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 1787 views around the world You can reuse this answer Creative Commons License