Question 1162040

given:

 degree {{{3 }}}

zeros:
{{{x[1]= 5}}} 

{{{x[2]=2i}}}->complex zeros always come in pairs, so you also have  

{{{x[3]=-2i}}}


{{{f(x)=(x-x[1])(x-x[2])(x-x[3])}}}

{{{f(x)=(x-5)(x-2i)(x-(-2i))}}}

{{{f(x)=(x-5)(x-2i)(x+2i)}}}

{{{f(x)=(x-5)(x^2-(2i)^2)}}}

{{{f(x)=(x-5)(x^2-4i^2)}}}

{{{f(x)=(x-5)(x^2-4(-1))}}}

{{{f(x)=(x-5)(x^2+4)}}}

{{{f(x)=x^3 - 5x^2 + 4x - 20}}}-> your answer