How do you factor #16x^4 - 4y^2#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Meave60 Jun 26, 2015 #(16x^4-y^2)=4(2x^2-y)(2x^2+y)# Explanation: #16x^4-y^2# Factor out the GCF #4#. #4(4x^4-y^2)# #(4x^4-y^2)# is in the form of the difference of squares: #(a^2-b^2)=(a-b)(a+b)#. #a=2x^2 and b=y# #4((2x^2)^2-y)# = #4(2x^2-y)(2x^2+y)# Answer link Related questions How do you factor special products of polynomials? How do you identify special products when factoring? How do you factor #x^3 -8#? What are the factors of #x^3y^6 – 64#? How do you know if #x^2 + 10x + 25# is a perfect square? How do you write #16x^2 – 48x + 36# as a perfect square trinomial? What is the difference of two squares method of factoring? How do you factor #16x^2-36# using the difference of squares? How do you factor #2x^4y^2-32#? How do you factor #x^2 - 27#? See all questions in Factor Polynomials Using Special Products Impact of this question 2815 views around the world You can reuse this answer Creative Commons License