Why does degree determine the amount of zeros?

Question

We just learned about complex numbers in my math class and I have a question. Why does the degree of a polynomial equal the amount of zeros it has?

The degree of $f(x) = x^3 - x^2 + x - 1$ is $3$, but there is only $1$ real zero, $x=1$.

There are 3 complex zeros, $x=1$,$i$,$-i$, which equals the degree number. I just don't understand why there isn't a case where a fifth-degree polynomial has the zeros $x=1$,$i$,$-i$ but none other. Why should it have to have five zeros?

I asked my teacher and she said, "polynomials are closed under the complex numbers," but I don't know what that means. ._.

Answer 1

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.

Answer 2

Well, you need to understand the notion of roots I think. Each degree $n$ polynomial (let's call it $P$) has $n$ complex roots, but they can repeat. For a very simple example, we have that only the number $2$ is a root of $P(X)=X^2-4X-4$, but the point is, it's a double root, because $2$ is also a root of $\frac{P(X)}{X-2} = X-2$.

It's not difficult to see that $P(X)=X^5-X^4+2X^3-2X^2+X-1$ has only three roots $1, \pm\mathrm{i}$ in the sense that $$P(x)=0\iff x=0\mathrel\vee x=\mathrm{i}\mathrel\vee x=-\mathrm{i},$$ but still is of degree five; the trick is that both $\mathrm{i}$ and $-\mathrm{i}$ are double roots, making five roots if you consider multiplicites.

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For a proof that each degree $n$ polynomial has $n$ roots, see the Fundamental theorem of algebra.

Answer 3

@m_CCTM @Shoutre @yo'

One can give another representation which has been used by Gauss for one of his proofs of the fundamental theorem of algebra (see graphics below).

Let us write polynomial equation $z^3-z^2+z-1=0$ under the form:

$$(x + i y)^3 - (x + i y)^2 + (x + i y) - 1=0.$$

and expand it :

$$R(x,y) + i I(x,y)=0 \ \ \text{where}$$

$R(x,y)=-1 + x - x^2 + x^3 + y^2 - 3xy^2 \ \ \text{and} \ \ I(x,y)=y - 2xy + 3x^2y - y^3.$

The 3 curves with equations

• $R(x,y)=0 \Rightarrow (1) y=\pm \sqrt{(1 - x + x^2 - x^3)/(3x-1)}$ and

• $I(x,y)=0 \Rightarrow (2) y=0 \ or \ (3) y=\sqrt{1 - 2x + 3x^2}$

are represented below. Their intersections are roots $z=1$, $z=\pm i$.

This kind of vizualisation, in my opinion, is one of the most convincing ones.

Notice the fact that the different branches of (1), (2) and (3) draw a "star" with multiple "arms".

Gauss has shown (around 1800) that, necessarily, the curves generated by this process intersect in $n$ points if the polynomial has degree $n$ (it may happen for example that 3 of these curves intersect in the same point : in such a case, this point is said to have multiplicity 2, etc.).

I encourage you to draw these curves for a fourth degree polynomial, for example.