How do you factor #(x^8)-16#? Algebra Polynomials and Factoring Factoring Completely 1 Answer Konstantinos Michailidis Nov 4, 2015 It is #x^8-16=(x^4)^2-4^2=(x^4-4)*(x^4+4)=(x^2-2)*(x^2+2)*[(x^2+2)^2-4x^2]=(x^2-2)*(x^2+2)*(x^2+2-2x)*(x^2+2+2x)# 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 4821 views around the world You can reuse this answer Creative Commons License