How do you find all the real and complex roots of #x(x^2 - 4)(x^2 + 16) = 0#? Precalculus Complex Zeros Complex Conjugate Zeros 1 Answer P dilip_k Mar 5, 2016 #x=0,+-2,+-4i# Explanation: #x(x^2-4)(x^2+16)=0# #x(x+2)(x-2)(x+4i)(x-4i)=0# #:.##x=0,+-2,+-4i# Answer link Related questions What is a complex conjugate? How do I find a complex conjugate? What is the conjugate zeros theorem? How do I use the conjugate zeros theorem? What is the conjugate pair theorem? How do I find the complex conjugate of #10+6i#? How do I find the complex conjugate of #14+12i#? What is the complex conjugate for the number #7-3i#? What is the complex conjugate of #3i+4#? What is the complex conjugate of #a-bi#? See all questions in Complex Conjugate Zeros Impact of this question 1331 views around the world You can reuse this answer Creative Commons License