How do you factor #-9r^2 + 81#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer ProfLayton Apr 22, 2016 #=-9(r-3)(r+3)# Explanation: The difference of squares rule states that: #a^2-b^2=(a-b)(a+b)#, and using this: #-9r^2+81=-9(r^2-9)# #=-9(r^2-3^2)# #=-9(r-3)(r+3)# 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 1670 views around the world You can reuse this answer Creative Commons License