How do you factor #4n^2 + 68n + 250 #? Algebra Polynomials and Factoring Factorization of Quadratic Expressions 1 Answer Nghi N. Oct 10, 2015 Factor #f(n) = 4n^2 + 68n + 250# Explanation: #f(n) = 2y = 2(2n^2 + 34n + 125).# Factor # y = 2n^2 + 34n + 125 # #D = b^2 - 4ac = 1156 - 4000 = -2844 < 0#. Since its D < 0, this trinomial y can't be factored. Finally, #f(n) = 2(2n^2 + 34n + 125)# 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 1499 views around the world You can reuse this answer Creative Commons License