Solutions to $z^3 - (b+6) z^2 + 8 b^2 z - 7+b^2 = 0, b\in \mathbb R, z \in \mathbb C$ $z_1 = 1+i$ is a given solution.
I guess what I have to find is $z_2$ and $z_3$ in
$(z - (1 + i))(z - z_2)(z-z_3) = z^3 - (b+6) z^2 + 8 b^2 z - 7+b^2$.
I tried to divide the polynomial by $(z - (1 + i))$, but that didn’t seem to work because of the $b$. According to the Complex conjugate root theorem $z_2 = \overline{z_1} = 1 - i$ is a solution too and somebody mentioned that it’s a hint that all coefficients are real. But I still don’t know how to proceed. What am I missing?
 A: Consider the cubic polynomial
$$\begin{equation*}
P(z)=z^{3}+Az^{2}+Bz+C
\end{equation*}\tag{1},$$
where the coefficients $A,B$ and $C$ are real numbers. If we denote its
roots by $z_{1},z_{2}$ and $z_{3}$, then it factors as
$$\begin{eqnarray*}
P(z) &=&\left( z-z_{1}\right) \left( z-z_{2}\right) (z-z_{3}) \\
&=&z^{3}-\left( z_{1}+z_{3}+z_{2}\right) z^{2}+\left(
z_{1}z_{2}+z_{2}z_{3}+z_{1}z_{3}\right) z-z_{1}z_{2}z_{3}.
\end{eqnarray*}\tag{2}$$
The constant term is
$$\begin{equation*}
P(0)=C=-z_{1}z_{2}z_{3}
\end{equation*}.$$
In the present case $A=-(b+6)$, $B=8b^{2}$ and $C=-7+b^{2}$. Since $z_{1}=1+i$ is a given solution, then $z_{2}=\overline{z}_{1}=\overline{1+i}=1-i$ is another solution, as you concluded. We thus have $z_{1}z_{2}=\left( 1+i\right) \left( 1-i\right)=2$ and
$$\begin{equation*}
-7+b^{2}=-2z_{3}
\end{equation*}\tag{3},$$
whose solution is
$$\begin{equation*}
z_{3}=\frac{7-b^{2}}{2}.
\end{equation*}\tag{4}$$
Since $P(z_1)=P(z_2)=0$, we have
$$\begin{eqnarray*}
&&\left( 1+i\right) ^{3}-(b+6)\left( 1+i\right) ^{2}+8b^{2}\left( 1+i\right)
-7+b^{2} \\
&=&-9+9b^{2}+i\left( -10-2b+8b^{2}\right)=0, 
\end{eqnarray*}\tag{5}$$
$$\begin{eqnarray*}
&&\left( 1-i\right) ^{3}-(b+6)\left( 1-i\right) ^{2}+8b^{2}\left( 1-i\right)
-7+b^{2} \\
&=&-9+9b^{2}+i\left( 10+2b-8b^{2}\right)=0, 
\end{eqnarray*}\tag{6}$$
which means that $b$ satisfies the system
$$\begin{equation*}
\left\{ 
\begin{array}{c}
-9+9b^{2}=0 \\ 
10+2b-8b^{2}=0.
\end{array}
\right. 
\end{equation*}\tag{7}$$
The solution of $(7)$ is $b=-1$. Using $(4)$ we find $$z_{3}=3.\tag{8}$$
A: You have two unknowns here, the third root which we call $r$, and the variable $b$ in your equation. You can find two equations in these two unknowns to solve for both $r$ and $b$.
First, the $z^2$ coefficient is always $-1$ times the sum of the roots, so you have
$$b + 6 = (1 + i) + (1 - i) + r = 2 + r$$
This reduces to $r = b + 4$. Next, the the constant term is equal to $(-1)^3$ times the product of the roots, so you have
$$b^2 - 7 = -(1 + i)(1-i)r = -2r$$
You now can solve these two equations for $b$ and $r$. I substitute $r = b + 4$ in the last equation, and get
$$b^2 - 7 = -2b - 8$$
This reduces to the quadratic equation $b^2 + 2b + 1 = 0$, which is solved only by $b = -1$. Then $r = b+ 4 = 3$ is your remaining root.
