Question

Given that the complex number z = -2 + 7i is a root to the equation:

z³ + 6 z² + 61 z + 106 = 0

find the real root to the equation.

Answer 1

Since the complex number $z = -2 + 7i$ is a root of the equation $z^3 + 6 z^2 + 61 z + 106 = 0$ with real coefficients, the conjugate complex number $\overline{z} = -2 - 7i$ is also a root of the equation. By

Bézout's theorem, $(z+2-7i)(z+2+7i)=(z+2)^2+49=z^2+4z+53$ divides $z^3 + 6 z^2 + 61 z + 106$. It follows that $z^3 + 6 z^2 + 61 z + 106=(z^2+4z+53)(z+2)$. We conclude that $z+2=0$, and hence $z=-2$ is a real root of the equation $z^3 + 6 z^2 + 61 z + 106 = 0.$