Question 140325
{{{x=1+2i}}} or {{{x=1-2i}}} Start with the given zeros



{{{x-1=2i}}} or {{{x-1=-2i}}} Subtract 1 from both sides



{{{(x-1)^2=4i^2}}} or {{{(x-1)^2=4i^2}}} Square both sides



{{{(x-1)^2=4(-1)}}} or {{{(x-1)^2=4(-1)}}} Replace {{{i^2}}} with -1



{{{(x-1)^2=-4}}} or {{{(x-1)^2=-4}}} Multiply



{{{(x-1)^2+4=0}}} or {{{(x-1)^2+4=0}}} Add 4 to both sides




{{{x^2-2x+5=0}}} or {{{x^2-2x+5=0}}} Foil and simplify




Now since {{{x^2-2x+5}}} is a factor of {{{x^4-4x^3+6x^2-4x-15}}}, we can use long division to find another factor






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So another factor of {{{x^4-4x^3+6x^2-4x-15}}} is {{{x^2-2x-3}}}





{{{x^2-2x-3=0}}} Set the factor equal to zero


{{{(x-3)(x+1)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x-3=0}}} or  {{{x+1=0}}} 


{{{x=3}}} or  {{{x=-1}}}    Now solve for x in each case





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Answer:




So the other zeros are 


 {{{x=3}}} or  {{{x=-1}}}