Algebra Help?

To answer your first question: no, a polynomial equation does not have to have at least one root. If you have a quadratic, both can be complex roots.

But if you have a third degree polynomial (aka a cubic equation), you do have to have at least one real root.

Why? Because complex roots always come in conjugate pairs. Since a cubic equation will have three roots, if you take out two complex roots, you are left with one root left. Since this cannot be complex since you don't have another root to have a conjugate pair with, it has to be real.

x³ = 1

is an example of a cubic that has one real solution (x = 1). It does have two complex solutions:

x = -(-1)^(1/3)
x = (-1)^(2/3)

These do end up being complex conjugate pairs, approximately equal to:

-0.5 ± 0.86603i

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An example of a polynomial with three roots can be derived starting with the roots that you want, then expand it into a cubic by reverse engineering the factors based on the roots. If I want my roots to be -1, 1, and 2. I'd have these factors:

(x + 1)(x - 1)(x - 2) = 0

now simplify that to get:

(x² - 1)(x - 2) = 0
x³ - 2x² - x + 2 = 0

And there you have a cubic with three real roots.