No. Complex zeros always come in conjugate pairs. So if a+bi is one zero, then a-bi is also a zero.
The fundamental theorem of algebra says
"Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers."
If you want to know how many complex root a given polynomial has, you might consider finding out how many real roots it has. This can be done with Descartes Rules of signs
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Sort of... but not entirely. Assuming the polynomial's coefficients are real, the polynomial either has as many real roots as its degree, or an even number less. Thus, a polynomial of degree 4 can have 4, 2, or 0 real roots; while a polynomial of degree 5 has either 5, 3, or 1 real roots. So, polynomial of odd degree (with real coefficients) will always have at least one real root. For a polynomial of even degree, this is not guaranteed. (In case you are interested about the reason for the rule stated above: this is related to the fact that any complex roots in such a polynomial occur in conjugate pairs; for example: if 5 + 2i is a root, then 5 - 2i is also a root.)
Yes - but only if the domain is restricted. Normally the domain is the whole of the real numbers and over that domain it must have at least one real zero.
A polynomial is a type of algebraic expression. They differ in the number of terms that contain variables. An algebraic expression has at least 1 variable, while a polynomial has multiple terms with variables in it.
Nope not all the rational functions have a horizontal asymptote
It is a function. If the graph contains at least two points on the same vertical line, then it is not a function. This is called the vertical line test.