Question 166677




Since {{{4+i}}} and {{{4-i}}} are given zeros this means that:



{{{x=4+i}}} and {{{x=4-i}}}



Get all terms to the left side in each case



{{{x-(4+i)=0}}} and {{{x-(4-i)=0}}}



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



{{{(x-(4+i))(x-(4-i))=0}}}



{{{((x-4)+i)((x-4)-i))=0}}} Regroup the terms.



Now let {{{w=x-4}}}. So the equation is now:



{{{(w+i)(w-i)=0}}}



{{{w^2-i^2=0}}} FOIL



{{{(x-4)^2-(-1)=0}}} Plug in {{{w=x-4}}} and {{{i^2=-1}}}



{{{(x-4)^2+1=0}}} Rewrite {{{(x-4)^2-(-1)}}} as {{{(x-4)^2+1}}}



{{{x^2-8x+16+1=0}}} FOIL



{{{x^2-8x+17=0}}} Combine like terms.




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Answer:


So the polynomial with roots of  {{{4+i}}} and {{{4-i}}} is



{{{x^2-8x+17}}}




Note: you can verify the answer by use of the quadratic formula.