How do you find all the real and complex roots of #x^3 + 4x^2 + x + 4=0#?

1 Answer
George C.
Feb 3, 2016

Factor by grouping to find roots:

#x=-4#, #x=i#, #x=-i#

Explanation:

Factor by grouping:

#x^3+4x^2+x+4#

#= (x^3+4x^2)+(x+4)#

#= x^2(x+4)+1(x+4)#

#= (x^2+1)(x+4)#

#(x+4)# is zero when #x=-4#

#(x^2+1)# is zero when #x=+-i#

So the roots of #x^3+4x^2+x+4 = 0# are #x=-4# and #x=+-i#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
1552 views around the world
You can reuse this answer
Creative Commons License