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. 

Since all coefficients are reals, $p$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2+1=x^2-2x+2.$$
We have $$p(x)=x^5-2x^4+ax^3+bx^2-2x+c=$$
$$=x^5-2x^4+2x^3+(a-2)x^3-2(a-2)x^2+2(a-2)x+(b+2a-4)x^2-(2a-2)x+c=$$
$$=(x^2-2x+2)(x^3+(a-2)x)+(b+2a-4)x^2-(2a-2)x+c.$$
We need $$b+2a-4=k,$$ $$2a-2=2k$$ and $$c=2k,$$ which gives
$$b=3-a,$$ $$c=2a-2$$ and
$$p(x)=(x^2-2x+2)((x^3-2x-1+a(x+1)).$$
Now, let $x^3-2x-1+a(x+1)=q(x).$
Thus, $q$ and $q'$ have the same negative integer root.
$$q'(x)=3x^2-2+a,$$ which gives $$a=2-3x^2.$$
Id est, for the same root we obtain the following equation.
$$x^3-2x-1+(2-3x^2)(x+1)=0$$ or
$$2x^3+3x^2-1=0$$ or
$$2x^3+4x^2+2x-x^2-2x-1=0$$ or
$$(x+1)^2(2x-1)=0,$$
which gives $x=-1$ and $a=-1$.
