How do you factor #x^2-81#? Algebra Polynomials and Factoring Factoring Completely 1 Answer Alan P. May 18, 2016 #x^2-81=color(green)((x-9)(x+9))# Explanation: For the general case: #color(white)("XXX")(x^2-a^2)=(x-a)(x+a)# #(x^2-81 ) = (x^2-9^2)# substituting #9# for #a# in the general case gives: #color(white)("XXX")(x-9)(x+9)# Answer link Related questions What is Factoring Completely? How do you know when you have completely factored a polynomial? Which methods of factoring do you use to factor completely? How do you factor completely #2x^2-8#? Which method do you use to factor #3x(x-1)+4(x-1) #? What are the factors of #12x^3+12x^2+3x#? How do you find the two numbers by using the factoring method, if one number is seven more than... How do you factor #12c^2-75# completely? How do you factor #x^6-26x^3-27#? How do you factor #100x^2+180x+81#? See all questions in Factoring Completely Impact of this question 14177 views around the world You can reuse this answer Creative Commons License