Question 167738
Since {{{i}}}, {{{i}}}, and {{{2}}} are given zeros this means that:



{{{x=i}}}, {{{x=i}}}, and {{{x=2}}}



Get all terms to the left side in each case



{{{x-i=0}}}, {{{x-i=0}}}, and {{{x-2=0}}}



Now use the zero product property in reverse to join the factors.



{{{(x-i)(x-i)(x-2)=0}}}



FOIL {{{(x-i)(x-i)}}} to get {{{x^2-2xi-1}}}



{{{(x^2-2xi-1)(x-2)=0}}}



{{{(x-2)(x^2-2xi-1)=0}}} Rearrange the terms.




{{{x(x^2-2xi-1)-2(x^2-2xi-1)=0}}} Expand



{{{x^3-2x^2i-x-2x^2+4xi+2=0}}} Distribute



{{{x^3+(-2x^2i-2x^2)+(-x+4xi)+2=0}}} Group like terms.



{{{x^3-x^2(2i+2)-x(1-4i)+2=0}}} Factor out the GCF from each group



{{{x^3-(2i+2)x^2-(1-4i)x+2=0}}} Rearrange the terms.



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Answer:


So the polynomial with roots of  {{{i}}}, {{{i}}}, and {{{2}}} is


{{{y=x^3-(2i+2)x^2-(1-4i)x+2}}}