How do you factor #4x^4 - y^2#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Don't Memorise Jul 3, 2015 #4x^4 - y^2 =color(blue)((2x^2+y)(2x^2-y)# Explanation: #4x^4 - y^2# is of the form #color(blue)(a^2 - b^2# (where #a=2x^2 , b=y)# , and as per the identity: #color(blue)(a^2 - b^2= (a+b)(a-b)# So, #4x^4 - y^2 =color(blue)((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 2294 views around the world You can reuse this answer Creative Commons License