How do you solve #x^2-18x+81=0# by factoring? Algebra Polynomials and Factoring Zero Product Principle 1 Answer Konstantinos Michailidis Oct 1, 2015 It is #x^2-18x+81=0=>x^2-2*9*x+9^2=0=>(x-9)^2=0=>x=9# We used the identity #(a-b)^2=a^2-2ab+b^2# Answer link Related questions What is the Zero Product Principle? How to use the zero product principle to find the value of x? How do you solve the polynomial #10x^3-5x^2=0#? Can you apply the zero product property in the problem #(x+6)+(3x-1)=0#? How do you solve the polynomial #24x^2-4x=0#? How do you use the zero product property to solve #(x-5)(2x+7)(3x-4)=0#? How do you factor and solve #b^2-\frac{5}{3b}=0#? Why does the zero product property work? How do you solve #(x - 12)(5x - 13) = 0#? How do you solve #(2u+7)(3u-1)=0#? See all questions in Zero Product Principle Impact of this question 4976 views around the world You can reuse this answer Creative Commons License