How do you find all the zeros of # X^3-3x^2+4x-2#?

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Answer 1 · 2016-05-02T22:23:30

This cubic polynomial has zeros: #x=1#, #x=1-i#, #x=1+i#

First note that the sum of the coefficients is zero. That is:

#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 only has Complex zeros, which we can find by completing the square and using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=(x-1)# and #b=i# as follows:

#x^2-2x+2#

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

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

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

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

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

Hence zeros #x=1+-i#