How do you solve #a^4 + 2a^3 + a^2=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer sente Apr 7, 2016 #a = 0# or #a = -1# Explanation: By factoring: #a^4+2a^3+a^2 = 0# #=> a^2(a^2+2a+1) = 0# #=> a^2(a+1)^2 = 0# #=> a^2 = 0# or #(a+1)^2 = 0# #=> a = 0# or #a+1 = 0# #:. a = 0# or #a = -1# 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 1423 views around the world You can reuse this answer Creative Commons License