Question 1003853

given zeros:
{{{x[1]=-1}}},real zero
{{{x[2]=-2+2i}}}, complex zero

recall, complex zeros always coming in pairs; so, this function also have {{{x[3]=-2-2i}}} as a zero

now, use zero product theorem to find a function that has these three zeros

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

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

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

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

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

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

{{{f(x)=(x+1)(x^2+4x+8))}}}

{{{f(x)=x^3+4x^2+8x+x^2+4x+8}}}

{{{f(x)=x^3+5x^2+12x+8}}}