How do you factor #(4y-2)^2-(3y)^2#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer George C. Jun 27, 2015 This is a difference of squares: #(4y-2)^2-(3y)^2# #= ((4y-2)-3y)((4y-2)+3y)# #=(y-2)(7y-2)# Explanation: For any #a# and #b#, we have #a^2-b^2 = (a-b)(a+b)# Let #a=4y-2# and #b=3y# Then #(4y-2)^2-(3y)^2# #=a^2 - b^2# #=(a-b)(a+b)# #= ((4y-2)-3y)((4y-2)+3y)# #=(y-2)(7y-2)# 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 1418 views around the world You can reuse this answer Creative Commons License