Question

According to the fundamental Theorem of algebra, which polynomial function has exactly 11 roots?

Answer 1

Explanation:

The Fundamental Theorem of Algebra states the following:

For any polynomial of degree n we will have n roots.

So the general form of the equation of a polynomial is:

$A \ \mathbf{polynomial \ function} \ of \ x \ with \ degree \ n \ is \ given \ by:\\ \\ f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots +a_{2}x^{2}+a_{1}x+a_{0} \\ \\ where \ n \ is \ a \ nonnegative \ integer \ and \ a_{n}, a_{n-1}, \ldots a_{2}, a_{1}, a_{0} \\ with \ a_{n}\neq$

Since our polynomial will have exactly 11 roots, then the equation will have the following form:

$f(x)=a_{11}x^{11}+a_{10}x^{10}+\ldots +a_{2}x^{2}+a_{1}x+a_{0}$