Question 957068
{{{f(x)}}} with real coefficients having the given degree and zeros:
Degree; 
zeros: 
{{{x[1]=-4-2i}}};complex roots coming always in pairs, so you also have  {{{x[2]=-4+2i}}}
{{{x[3]=1}}} multiplicity 2-double root

use zero product rule to find {{{f(x)}}}

{{{f(x)=(x-(-4-2i))(x-(-4+2i))(x-1)}}} since {{{(x-1)}}} has multiplicity 2 we can square it

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

{{{f(x)=(x^2+4x+cross(2x*i)+4x+16-cross(8i)-cross(2x*i)+cross(8i)-4i^2)(x^2-2x+1)}}} 

{{{f(x)=(x^2+8x+16-4(-1))(x^2-2x+1)}}} 

{{{f(x)=(x^2+8x+16+4)(x^2-2x+1)}}} 

{{{f(x)=(x^2+8x+20)(x^2-2x+1)}}} 

{{{f(x)=x^4+8x^3+20x^2-2x^3-16x^2-40x+x^2+8x+20}}} 

{{{f(x)=x^4+6x^3+5x^2-32x+20}}}