How do you solve #x^3-2x^2-x+2=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Binayaka C. May 2, 2016 #x=2 ; x=1; x=-1# Explanation: #x^3-2x^2-x+2=0 or x^3-x^2-x^2+x-2x+2=0 or x^2(x-1)-x(x-1)-2(x-1)=0# or #(x^2-x-2)(x-1)=0 or (x^2+x-2x-2)(x-1)=0 or (x(x+1)-2(x+1))(x-1)=0# or #(x-2)(x+1)(x-1)=0 :. x=2 ; x=1; x=-1#[Ans] Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 3965 views around the world You can reuse this answer Creative Commons License