Question 1005610

{{{P(x)=0}}} has the given roots: 
{{{x[1]=-1}}} and {{{x[2]=-2i}}}

if we have {{{x[2]=-2i}}} then we also have {{{x[3]=2i}}} because complex roots always come in pairs

use zero product formula

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

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

{{{P(x)=(x+1)(x+2i)(x-2i)}}}

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

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

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

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

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

{{{highlight(P(x)=x^3+x^2+4x+4)}}}

{{{ graph( 600, 600, -10, 10, -10, 10, x^3+x^2+4x+4) }}}