Question 709832
You seem to be asking for polynomial function with roots: -2,and  1+3i, AND 1-3i.  Complex roots come as conjugate pairs.


-2 is a root:
((-2)-(k))=0
-2-k=0
-k=2
k=-2
The binomial factor is (x-(-2))={{{(x+2)}}}


1+3i is a root:
(1+3i-k)=0
1+3i-k=0
-k=-1-3i
The binomial factor is (x-(-1-3i))
or equal to{{{ (x+(1+3i))}}}


1-3i is a root:
(1-3i-k)=0
-k=-1+3i
k=1-3i
The binomial factor is {{{(x-(1-3i))}}}


A satisfying function can come from {{{(x+2)(x+(1+3i))(x-(1-3i))}}}


Do the multiplications, at least for the factors containing the complex numbers:
... The finished polynomial function may be:
{{{highlight(f(x)=(x+2)(x^2-2x+10))}}} and you could do the rest of the multiplying if needed.