Does anyone know of a relationship between the number of zeros (complex and real) and the degree of a polynomial? Specifically, if a polynomial has a double root, is it compensated with complex zeros?
[Math] The number of zeros in a polynomial
polynomialsroots
Related Solutions
Hint: $\alpha$ is a root of $f(x)$ if and only if $(x-\alpha)$ divides $f(x)$.
The complex numbers $\mathbb{C}$ are algebraically closed (which is what your teacher probably meant).
This mean that a polynomial $p(x)$ of degree $n$with coefficients in $\mathbb{C}$ always "factors completely" as \begin{equation} p(x)=a(x-\alpha_{1})\dots(x-\alpha_{n}), \end{equation} where $a,\:\alpha_{1},\dots,\:\alpha_{n}\in\mathbb{C}$ and the roots of course can be repeated.
I don't know where you got the information that an odd degree polynomial has at most two turning points:
Actually the maximum number of turning points of a degree $n$ polynomial is $n-1$ and the one in the picture indeed has four of them, and five real roots.
Generalizing, if $n$ is any positive integer, the polynomial $$ (x-1)(x-2)\dots(x-n) $$ has $n$ distinct roots.
Best Answer
The fundamental theorem of algebra states that an $n^{th}$ degree polynomial has exactly $n$ roots, provided you count them with their multiplicity.
This is equivalent to saying that any polynomial has a root.
Because if you divide a polynomial of the $n^{th}$ degree by the monomial $x-r$ (where $r$ is a root), you get an $(n-1)^{th}$ polynomial, which has a root, and you can continue until $n=1$ (this is called deflating the polynomial).
Said differently, any polynomial can be factored as the product of $n$ binomials. $$x^3-3x^2+2x=(x-0)(x-1)(x-2),$$ $$x^5-3x^4+4x^3-4x^2+3x-1=(x-1)(x-1)(x-1)(x-i)(x+i),$$ $$x^4+8x^3+26x^2+40x+25=(x+2+i)(x+2+i)(x+2-i)(x+2-i).$$