Question #30ee9

1 Answer
David B.
Feb 24, 2016

The three zeros are #x=1+i, 1-i#, and #-4#

Explanation:

We start with some knowledge of the nature of cubic polynomials, in that there are only 3 possible cases for its roots:

  1. the equation has three distinct real roots
  2. the equation has a multiple root and all its roots are real
  3. the equation has one real root and two non-real complex conjugate roots

The question has determined for us that situation 3 applies since the first root is complex: #1+i#. Therefore, we also know another root which is the complex conjugate of the first: #1-i#. Now there is only one root left to find. Let's find the answer by assuming the form of the solution where our unknown root is #a#:

#0=(x-a)(x-1-i)(x-1+i)#

lets multiply it out and set it equal to our original polynomial

#x^3-(a+2)x^2+2(a+1)x-2a = x^3+2x^2−6x+8#

By setting the coefficient of each power of #x# on either side of the equation equal to the other, we get

#-(a+2)=2#
#2(a+1)=-6#
#-2a=8#

therefore #a=-4#

(in all three cases which means it is consistent)

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