How do you factor #256z^2-4-192z^2+3#? Algebra Polynomials and Factoring Factoring Completely 1 Answer Faisal Mar 25, 2018 Simplify & use identities Explanation: =#256z^2-192z^2-1# = #64z^2-1# Use #a^2-b^2=(a+b)(a-b)# = #(8z)^2-(1)^2# #(8z+1)(8z-1=0# #z=1/8 , -1/8# 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 1720 views around the world You can reuse this answer Creative Commons License