How do you write #16x^2 – 48x + 36# as a perfect square trinomial? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Wataru Oct 29, 2014 Since #{(sqrt{16x^2}=4x),(-48" is negative"),(sqrt{36}=6):}#, we have #16x^2-48x+36=(4x-6)^2# I hope that this was helpful. 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? 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#? How do you factor #x^3+64#? See all questions in Factor Polynomials Using Special Products Impact of this question 9402 views around the world You can reuse this answer Creative Commons License