How do you factor completely #w^2 - w^2 y^4#? Algebra Polynomials and Factoring Factoring Completely 1 Answer JL · Stefan V. Apr 4, 2018 #w^2(1+y)(1-y)(1+y^2)# Explanation: This is just a difference of squares. When you have #a^2 - b^2#, it's is equal to #(a+b)(a-b)#. Just simply expand it to see why. In this case, it would be #(w)^2 - (w*y^2)^2# #(w- wy^2)(w + wy^2)# #w^2(1-y^2)(1+y^2)# #w^2(1+y)(1-y)(1+y^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 1378 views around the world You can reuse this answer Creative Commons License