How do you find all the zeros of #f(x)= x^4+x^3+2x^2+4x-8# with its multiplicities?

1 Answer
Aug 3, 2016

Answer:

#f(x)# has zeros #1#, #-2# and #+-2i#

Explanation:

#f(x) = x^4+x^3+2x^2+4x-8#

First note that the sum of the coefficients of #f(x)# is zero. That is:

#1+1+2+4-8 = 0#

Hence #f(1) = 0#, #x=1# is a zero of #f(x)# and #(x-1)# is a factor:

#x^4+x^3+2x^2+4x-8 = (x-1)(x^3+2x^2+4x+8)#

The remaining cubic is very regular and can be factored in a number of different ways. Let's factor it by grouping:

#x^3+2x^2+4x+8#

#=(x^3+2x^2)+(4x+8)#

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

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

So #x=-2# is a zero.

The remaining quadratic has no Real zeros since #x^2+4 >= 4 > 0# for any Real value of #x#, but it does have Complex zeros:

#x = +-sqrt(-4) = +-2i#