Question 431581
in general, using the fundamental theorem of algebra, you can conclude that any polynomial of degree n has n solutions (including multiplicity(order)) of the roots.

ex: g(x) = x^2 + 1. This has no solutions in real numbers, but by the fund thm of algebra, they have 2 solutions in complex numbers; namely x=i and x=-i, where i = sqrt(-1)

ex: p(x) = x^2+2i+1. This has 1 solution in the complex numbers, x=i, but it has order 2, therefore it satisfies the fund thm of algebra

can you now name how many solutions your functions will have in complex numbers?