How do you factor completely #x^2 - 6x + 8 #? Algebra Polynomials and Factoring Factoring Completely 1 Answer Neh Pandya Apr 29, 2016 For #x^2-6x+8=0# #0=x^2-4x-2x+8# #0=x(x-4)-2(x-4)# #0=(x-4)(x-2)# #(x-4)=0# OR #(x-2)=0# #x=4# OR #x=2# For #x=4 and 2# #x^2-6x+8# is zero Hence #4 and 2# are its factors 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 1904 views around the world You can reuse this answer Creative Commons License