How do you solve #16x^2 - 64=0#? Algebra Polynomials and Factoring Factoring Completely 1 Answer Don't Memorise Jun 15, 2015 #color(blue)(x = +-2# Explanation: #16x^2 - 64=0# Taking out the common term #(16)#: #16(x^2 - 4)=0# #x^2 - 4=0# #x^2 =4# # x = sqrt4# #color(blue)(x = +-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 3816 views around the world You can reuse this answer Creative Commons License