# Why does degree determine the amount of zeros?

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. ._.

• Look up the root-factor theorem: If $a$ is a root of $f$, then we can write $f(x) = (x-a)g(x)$ for some other polynomial $g$ of degree $\deg(f) - 1$. Ultimately, this implies we can write $\displaystyle f(x) = c \prod_k (x-a_k)$, where $\{a_k\}$ is the collection of roots of $f$ and $c$ is some constant. – Kaj Hansen Jun 7 '16 at 19:23
• Do you believe that every polynomial has at least one root? If so, the proof that it has as many roots as is the degree should be straightforward IIRC. – yo' Jun 7 '16 at 19:26
• As an aside, you can have a 5th degree polynomial with fewer than $5$ unique roots. For your case, consider $f(x) = (x-1)^3(x-i)(x+i)$. – Kaj Hansen Jun 7 '16 at 19:27
• This is the fundamental theorem of algebra. – user223391 Jun 7 '16 at 19:29
• The keyword here is "multiplicities." When counting the roots, for instance, $(x-5)^2(x+4)$ has two uniques roots, one (5) with multiplicity 2 and one (-4) with multiplicity 1 -- so a total of "$2+1=3$ (non-unique) roots." – Clement C. Jun 7 '16 at 19:30

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 $$p(x)=a(x-\alpha_{1})\dots(x-\alpha_{n}),$$ where $a,\:\alpha_{1},\dots,\:\alpha_{n}\in\mathbb{C}$ and the roots of course can be repeated.

• That's not quite answering the question: if I interpret it correctly, the main issue is about the multiplicities. – Clement C. Jun 7 '16 at 19:29
• I can be wrong, but I really think it is more elementary than that. The phrase "I just don't understand why there isn't a case where a fifth-degree polynomial has the zeros $x=1$, $x=i$, $x=−i$ but none other" doesn't seem to refer to multiplicity. But, of course, you can be right. – Shoutre Jun 7 '16 at 19:35
• I think he/she really is wondering why in $\mathbb{C}$ we don't have polynomials of degree (say) $3$ with only one root, which is "stuff that used to happen in $\mathbb{R}$", in the sense that it could really have only one root (in contrast to all the roots begin equal). That comparison leads me to strongly suspicion that my answer adresses that issue. – Shoutre Jun 7 '16 at 19:38
• After the edit, I think it does address both aspects anyway. – Clement C. Jun 7 '16 at 19:59
• Surely made it clear. Thanks for commenting. – Shoutre Jun 7 '16 at 20:00

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.

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

@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.