How do you factor completely 18x^(16n+2) - 21x^(8n+1) - 9 ? #18x^(16n+2)-21x^(8n+1)-9# Algebra Polynomials and Factoring Factoring Completely 1 Answer Shwetank Mauria Feb 23, 2017 #18x^(16n+2)-21x^(8n+1)-9=3(3x^((8n+1))+1)(2x^((8n+1))-3)# Explanation: #18x^(16n+2)-21x^(8n+1)-9# = #3(6x^(2xx(8n+1))-7x^(8n+1)-3)# - as #3# is common = #3(6x^(2xx(8n+1))-9x^(8n+1)+2x^(8n+1)-3)# = #3[3x^((8n+1))(2x^((8n+1))-3)+1(2x^(8n+1)-3)]# = #3(3x^((8n+1))+1)(2x^((8n+1))-3)# 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 1394 views around the world You can reuse this answer Creative Commons License