Finding real $a$, $b$, $c$ such that $x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$ has $1+i$ as a zero, and one negative integer as a zero with multiplicity $2$ 
Find $a, b, c \in \mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where: 
  $$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$$

I know that one zero of $p$ is $1-i$, so I can reduce it one degree down after dividing with $x-2$, but after that I'm stuck. 
I tried using Vieta's formulas, but with no result. 

If $ 1 + i $ is the root of the polynomial 
$$
p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
$$
and  $a,b,c\in\mathbb{R}$ then $1-i$ is root of $p(x)$. Soon $q(x)=(x-1+i)\cdot(x-1-i)=(x^2-2x+2)$ divide $p(x)$ and 
\begin{align}
x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
=&
(x^2-2x+2)\big( x^3 -(2r+s)x^2+(r^2+2rs)x-r^2s\big)
\\
=&x^5
\\
&\hspace{0.5cm} -2x^4-(2r+s)x^4
\\
&\hspace{1.0cm}  +2x^3+2(2r+s)x^3+(r^2+2rs)x^3
\\
&\hspace{1.5cm}  -2(2r+s)x^2-2(r^2+2rs)x^2-r^2sx^2
\\
&\hspace{2.0cm}  +2(r^2+2rs)x+2r^2sx
\\
&\hspace{2.5cm}  +2(r^2+2rs)x+2r^2sx
\\
&\hspace{3.0cm}  -2r^2s
\end{align}
Note that 
\begin{align}
-2-(2r+s)=&-2\quad \implies \quad 2r+s=0\\
+2+2(2r+s)+(r^2+2rs)=&a\\
-2(2r+s)-2(r^2+2rs)-r^2s=&b\\
+2(r^2+2rs)+2r^2s=&-2 \quad \implies \quad 2r^2+4rs+2r^2s=-2\\
-2r^2s=&c\\
\end{align}
