Question 1155804


function of minimum degree in standard form that has zeros

{{{x[1]=-1}}} 

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

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

then

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

{{{f(x)=(x-(-1))(x-2i)(x-(-2i))}}}

{{{f(x)=(x+1)(x-2i)(x+2i)}}}.....use the rule for difference of squares 

{{{f(x)=(x+1)(x^2-(2i)^2)}}}

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

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

{{{f(x)=(x+1)(x^2+4)}}}

{{{f(x)=x^3+4x+x^2+4}}}

{{{f(x)=x^3+x^2+4x+4}}}