Question 1100942
If complex number {{{2 + 8i}}} is a zero of a polynomial with real coefficients,
so is the conjugate complex number, {{{2 - 8i}}} .
So, we know four zeros of the function,
and for each {{{zero}}} , there must be a factor of the form {{{x-zero}}} .
As a consequence, the simplest  polynomial function of minimum degree,
with real coefficients, and including those four zeros is
{{{f(x)=(x-7)(x-(-11))(x-(2 + 8i))(x-(2 - 8i))}}} .
Now,  we just have to simplify and multiply.
{{{f(x)=(x-7)(x+11)(x-2- 8i)(x-2+ 8i)}}}
{{{f(x)=(x-7)(x+11)((x-2)- 8i)((x-2)+ 8i)}}}
{{{f(x)=(x-7)(x+11)((x-2)^2- (8i)^2)}}}
{{{f(x)=(x-7)(x+11)(x^2-4x+4-8^2(i)^2)}}}
{{{f(x)=(x-7)(x+11)(x^2-4x+4-64(-1))}}}
{{{f(x)=(x-7)(x+11)(x^2-4x+4+64)}}}
{{{f(x)=(x^2+4x-77)(x^2-4x+68)}}}
{{{f(x)=x^4-25x^2+580x-5230}}}