How do you factor #t^2 + t +48#? Algebra Polynomials and Factoring Factorization of Quadratic Expressions 1 Answer Shwetank Mauria May 8, 2018 #t^2+t+48=(t+1/2+sqrt91/2i)(t+1/2-sqrt91/2i)# Explanation: #t^2+t+48# = #t^2+2xxtxx1/2+(1/2)^2-(1/2)^2+48# = #(t+1/2)^2+191/4# = #(t+1/2)^2-(sqrt191/2i)^2#, where #i^2=-1# = #(t+1/2+sqrt91/2i)(t+1/2-sqrt91/2i)# 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 1395 views around the world You can reuse this answer Creative Commons License