How do you factor #16n^2-25#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer smendyka Nov 17, 2016 #(4n - 5)(4n + 5)# Explanation: To eliminate the #n# term in a quadratic equation the coefficients for #n# in the factored for must be the same and the constants must be the same. Therefore: #16n^2 - 25 =># #(4n - 5)(4n + 5)# 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 2212 views around the world You can reuse this answer Creative Commons License