How do you factor #2x^2-5x+1#? Algebra Polynomials and Factoring Factorization of Quadratic Expressions 1 Answer salamat Mar 2, 2017 #2 x^2 - 5 x + 1 = 2(x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)# Explanation: #2 x^2 - 5 x + 1 = 2(x^2 - 5/2 x + 1/2)# Consider #x^2 - 5/2 x + 1/2 = 0# and solve by completing a square #x^2 - 5/2 x + 1/2 = 0# #(x -5/4)^2 -(5/4)^2 + 1/2 = 0# #(x -5/4)^2 -25/16 + 1/2 = 0# #(x -5/4)^2 -17/16 = 0# #(x -5/4)^2 = 17/16# #(x -5/4) = +-sqrt (17/16) = +-sqrt 17/4# #x = 5/4 +-sqrt 17/4 = (5+- sqrt 17)/4# #(x^2 - 5/2 x + 1/2) = (x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)# therefore #2 x^2 - 5 x + 1 = 2(x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)# Answer link Related questions How do you factor trinomials? What is factorization of quadratic expressions? How do you factor quadratic equations with a coefficient? What are some examples of factoring quadratic expressions? How do you check that you factored a quadratic correctly? How do you factor #x^2+16x+48#? How do you factor #x^2-9x+20#? Question #3fdac How do you factor #8+z^6#? There is no GCF to be factor out, so is there another method to complete this? How do you factor #2t^2+7t+3#? See all questions in Factorization of Quadratic Expressions Impact of this question 14984 views around the world You can reuse this answer Creative Commons License