How do you find all zeros of the function #P(x)=2x^3-5x^2+6x-2# given the zero 1+i?

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Answer 1 · 2016-07-16T22:35:25

Zeros: #1+i#, #1-i#, #1/2#

#P(x) = 2x^3-5x^2+6x-2#

We are told that #x=1+i# is a zero.

Since the coefficients of #P(x)# are all Real, the Complex conjugate #1-i# must be a zero too. So #P(x)# must be divisible by:

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

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

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

#= x^2-2x+2#

We find:

#2x^3-5x^2+6x-2 = (x^2-2x+2)(2x-1)#

So the remaining zero is #x=1/2#