Question

Granted that a certain cubic equation has the root 2 and no real root different

from 2, does it have two imaginary roots?​

Answer 1

Step-by-step explanation:

Find the real root of the equation z3+z+10=0 given that one complex root is 1–2i.

I've realized that the roots are (1−2i),(1+2i), and a real number we'll call a.

So using the theorem got me (z−1−2i)(z−1+2i)(z−x).

No idea on where to go next.

Answer 2

Find the real root of the equation z3+z+10=0 given that one complex root is 1–2i.

I've realized that the roots are (1−2i),(1+2i), and a real number we'll call a.

So using the theorem got me (z−1−2i)(z−1+2i)(z−x).

No idea on where to go next