How do you factor completely #22y^4-33y^3+11y^2#? Algebra Polynomials and Factoring Factoring Completely 1 Answer Konstantinos Michailidis Apr 30, 2016 It is #22y^4-33y^3+11y^2=22y^4-22y^3-11y^3+11y^2= 22y^3(y-1)-11y^2(y-1)=(22y^3-11y^2)*(y-1)= 11y^2(2y-1)*(y-1)# 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 1343 views around the world You can reuse this answer Creative Commons License