Question 1010032
{{{ n=3}}}=> degree {{{3}}}
{{{x[1]=2}}} and {{{x[2]=2i}}} are zeros
if {{{x[2]=2i}}} is zero, then {{{x[3]= -2i}}} is zero too, complex zeros always come in pairs
so, we use zero product theorem to find the polynomial

given also {{{f(-1)=-15}}}, but since all three roots are given, this will not be used

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

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

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

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

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

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

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

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

 {{{f(x) = x^3-2x^2+4x-8}}}=> your answer


now, we can check if {{{f(-1)=-15}}}

{{{f(-1) = (-1)^3-2(-1)^2+4(-1)-8}}}

{{{f(-1) = -1-2(1)-4-8}}}

{{{f(-1) = -1-2-4-8}}}

{{{f(-1) = -15}}}