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 $2$

Question

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.

Answer 1

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 $a=-1$.

Answer 2

$$x_1=1+i$$ $$x_2=1-i$$ $$x_3=x_4 = m<0$$ and $$x_5=k$$ then we have $$x_1+x_2+x_3+x_4+x_5=2$$

$$2+2m+k=2\implies \boxed{k=-2m}$$ and $$x_1x_2x_3x_4x_5({1\over x_1}+{1\over x_2}+{1\over x_3}+{1\over x_4}+{1\over x_5}) =-2$$

so $$\boxed{2m^2k(1+{2\over m}+{1\over k})=-2}\implies m^2(2m+3)=-1$$

Since $m^2\mid -1$ we have $m^2=1\implies m=-1$ and $k=2$.

The rest should be easy now.

Answer 3

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}