#### Answer

See the explanation below.

#### Work Step by Step

Suppose that f(x) represents a polynomial equation of degree n.
According to the Fundamental Theorem of Algebra, the equation has at least one complex root; we’ll call it ${{c}_{1}}$.
According to the Factor Theorem, we know that $\left( x-{{c}_{1}} \right)$ is a factor of f(x).
Equating equation (1) equal to 0, we get
$f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right).....\left( x-{{c}_{n}} \right)=0$
Then, ${{a}_{n}}\left( x-{{c}_{1}} \right)=0$
This implies that either ${{a}_{n}}=0$ or $(x-{{c}_{1}})=0$.
So, the order of the given polynomial $f\left( x \right)$ in the equation (1) has to be at least 1. Similarly,
\[\] \[\begin{align} & \left( x-{{c}_{2}} \right)=0 \\
& \left( x-{{c}_{3}} \right)=0 \\
& . \\
& . \\
& . \\
& \left( x-{{c}_{n}} \right)=0 \\
\end{align}\]
Hence, the roots of the given polynomial in equation (1) can be ${{c}_{2}},{{c}_{3}},......,{{c}_{n}}$.
Thus, for a polynomial of $n^{th}$ order, there are n roots.