How do you find all the roots of #f(x) = x^4 + 2x^3 + x^2 - 2x - 2#?

1 Answer
May 23, 2016

Answer:

This quartic polynomial has zeros #+-1# and #-1+-i#.

Explanation:

First note that the sum of the coefficients is #0#. That is, #1+2+1-2-2 = 0#. Hence we can deduce that #x=1# is a zero and #(x-1)# a factor:

#x^4+2x^3+x^2-2x-2 = (x-1)(x^3+3x^2+4x+2)#

If you reverse the signs of the coefficients of the terms of odd degree in the remaining cubic #(x^3+3x^2+4x+2)# you get: #-1+3-4+2 = 0#. Hence #x=-1# is a zero and #(x+1)# a factor:

#x^3+3x^2+4x+2 = (x+1)(x^2+2x+2)#

The remaining quadratic factor has negative discriminant, but you can factor it by completing the square with Complex coefficients:

#x^2+2x+2 = x^2+2x+1+1 = (x+1)^2-i^2 = (x+1-i)(x+1+i)#

Hence zeros #x=-1+-i#