# 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?

2

Lenar Hoyt 2022-07-25 20:46:44

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