# Fundamental theorem of algebra simple proof for rewriting with roots

My question is very basic, as I do not understand the concept of rewriting a (complex) polynomial into a product of terms using the roots of the polynomial.

I have encountered the fundamental theorem of algebra many times, but never in a concise way. It either stated that a polynomial contains at least one root/a polynomial of degree d has at most d roots/a polynomial can be rewritten into a product. The last one, even though all statements are nonequivalent, interests me the most.

I have seen many proofs relying on the fact that if you solve the roots of any given polynomial (including imaginary roots if present), you can rewrite it into a product with factors $(x-z)$ if $x=z$ is a root. So for instance, a very accessible example would be rewriting $x^2-3x+2=(x-2)(x-1)$. But when higher powers come into the game or even an infinite series (Euler's proof of Basel's problem) I can impossibly work the polynomials out in order to check if it is true.

How could you prove to me in reasonable elementary terms that a polynomial can be written in its "summation" form as well as its "product" form using the roots? When I look up proof of fundamental theorem of algebra, they seem to focus on proving the presence of roots rather than this rewriting. I do not even know for sure if this is regarded as the fundamental theorem of algebra, maybe it is something completely different from what I think it is.

EDIT: Additional question: why do we use the roots, that is to say when $P(x)=0$, to rewrite and not solutions for $P(x)=1$ or $P(x)=-2$ for instance? Why is the root specially suited for this?

As far as I know, fundamental theorem algebra is in only one version, and others you stated are corollary to this:

Fundamental Theorem of Algebra: Any non-constant polynomial with one variable in complex coefficients have at least one root.

The fact that a polynomial $f(z)$ in complex variables can be written as

$$f(z)=(z-\alpha)g(z)$$

for some complex polynomial $g$ and a root $\alpha$ of $f$ is a result of division algorithm, and nothing else. It is not a part of fundamental theorem of algebra. The division algorithm in this case, is stated as

Division algorithm: For any complex polynomials $f,g$ there exist complex polynomials $q,r$ such that

$$f(z)=q(z)g(z)+r(z) \quad \text{or simply written as} \quad f=qg+r$$

where $\deg g < \deg r$ or $r=0$

Now you can see from this easily that if we set $g(z)=z-\alpha$ then we must have $r=0$ so acclaimed factorization is valid.

There is another misunderstanding here. The polynomials, by definition, cannot have infinite power. The degree of polynomial, which is equal to highest power in the polynomial, must be finite for any polynomial. Stuff like

$$1+z+z^2+z^3+\cdots$$

is not regarded as a polynomial, but a power series

So fundamental theorem of algebra does not assert anything about "infinite" polynomial and further, as stated above division algorithm only apply to "polynomials", and so it factoring infinite series by its roots is invalid. That is precisely where Euler's proof of Basel problem is invalid (non-rigorous) in the scope of modern mathematics though the proof can be modified so that it is riogorous in modern sense too.

For your final question, look at how we arrived to conclude that $r=0$ using division algorithm. That is the part of reason by we are woring with $P(x)=0$ case, and another reason would be that simply $0$ is such a nice and powerful number to work with (it gives $0$ when multiplied to ANY number, is also an identity in addition etc..). Really, why would you work with other numbers when you have this nice number $0$?

Also, assuming fundamental theorem of algebra, we can show that following:

Corollary Any complex polynomial of degree $d$ has precisely $d$ roots.

Proof If $d=0$, then we are done. Otherwise, let $f(z)=\sum_{n=0}^{d} a_n z^n$. Fundamental theorem of algebra asserts that it has at least one root, say $\alpha$. Thus by division algorithm, we can factor it so that

$$f(z)=(z-\alpha)g(z)$$ where $\deg g=d-1$ (this is consequence of $\deg(fg)=\deg f + \deg g$). We apply FTA repeatedly, and since $d$ is finite, we should stop after some finite steps. Then we have factored $f$ into

$$f(z)=(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_d)$$

So $f$ has exactly $d$ roots.

$+$If you know some bits of ring theory, then division algorithm holds if the ring of polynomials(or in fact ANY RING) is a Euclidean domain. This is by definition of Euclidean algorithm.

The factorisation of polynomials relies on the following theorem, which is a consequence of the fact there exists a euclidean division in rings of polynomials over a field (such as $\mathbf R$ or $\mathbf C$).

Theorem: Let $P(x$ be a polynomial over a field $K$, $\alpha \in K$. Then $P(\alpha)=0$ if and only if $P(x)$ is divisible by $x-\alpha$.

Proof. Let's divide $P(x)$ by $x-\alpha$. This means there exist polynomials $Q(x), R(x)$ such that: $$P(x)=Q(x)(x-\alpha )+R(x),\quad R(x)=0\enspace\text{or}\enspace \deg R(x)<\deg(x-\alpha).$$

As $\deg(x-\alpha)=1$, the condtion means $R(x)$ is a constant polynomial, we'll denote simply $R$. Then observe from the equality above that $R=P(\alpha )$. Hence the remainder is $0$ if and only if $\alpha$ is a root of $P(x)$.

Now if you have $r$ distinct roots $\alpha _1,\dots,\alpha _r$ are roots of $P(x)$, this polynomial is divisible separately by $x-\alpha _1, \dots, x-\alpha _r$. These are coprime, hence $P(x)$ is divisible by $(x-\alpha _1)\dotsm(x-\alpha _r)$.

It is also true if there are ‘multiple’ roots, but the proof is more technical. The Fundamenral Theorem of Algebra asserts that a polynomial of degree $d$ has exactly $d$ roots over $\mathbf C$, counting their multiplicities, and hence splits as a product of $d$ linear factors.

Concerning the last question, (i) It is easy to solve equations of the form $f(x)=0$ when $f(x)$ is a product of factors, because such a product is $0$ if and only if one of the factors is $0$. You have nothing similar for $f(x)=1$.

(ii) Anyway, $P(x)=1$ can be rewritten as $P(x)-1=0$.

1. Rewriting polynomials

You are most probably used to writing polynomials in the form $$P=a_0x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n$$ This is a special that singles out a very particular point on the real line (or more generally on the ring of definition, e.g. the rationals $\mathbb Q$, or the reals $\mathbb R$, or the complex numbers $\mathbb C$, or the integers $\mathbb Z$, or the modular integers $\mathbb Z/n\mathbb Z$; any ring). The particular point singled out by the above form of the polynomial is the point $0$. It is singled out in the sense that it is very easy to compute the value of the polynomial at $0$: $P(0)$ is simply the constant term $a_n$ (in fact this form of the polynomial also makes it easy to compute any derivative of the polynomial at $0$; the above form is the Taylor series of the polynomial centered at $0$).

There are however other forms into which we may rewrite polynomials, one for each element of the ring of definition. For example, if we wanted to center the polynomial around $1$, we would want to write the polynomial $P$ not in terms of $x$, but in terms of $x-1$. This is not conceptually difficult, but it is computationally intensive: we observe that $$P=a_0x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n=a_0((x-1)+1)^n+a_1((x-1)+1)^{n-1}+\dots+a_{n-1}((x-1)+1)+a_n$$ and then we would have to do all the multiplications $$((x-1)+1)^n=(x-1)^n+n(x-1)^{n-1}+\dots+n(x-1)+1$$ to get in the end the form $$P=b_0(x-1)^n+b_1(x-1)^{n-1}+\dots+b_{n-1}(x-1)+b_n$$

This form is equally as special as the one we started with, only this form makes it easy to compute not the value $P(0)$ but the value $P(1)=b_n$.

1. Factoring polynomials

To determine why a polynomial can be written multiplicatively in terms of its roots, consider what happens when $x=z$ is a root. The form of the polynomial centered at $z$ has to be $P=c_0(x-z)^n+c_1(x-z)^{n-1}+\dots+c_{n-1}(x-z)+c_n$, and the value at $z$ is $P(z)=c_n$. Since $z$ is a root of the polynomial, this value is $0$, and so in fact $P=c_0(x-z)^n+\dots+c_{n-k}(x-z)^k$ where $k$ is the multiplicity of the root $z$.

We can then factor to obtain $P=(x-z)^kQ$ where $$Q=c_0(x-z)^{n-1}+c_1(x-z)^{n-2}+\dots+c_{n-2}(x-z)+c_{n-k}$$ Notice that $Q$ has smaller degree than $P$. If you now find a root of $Q$, you can again do the factoring; iterating this process yields $$P=(x-z_1)^{k_1}(x-z_2)^{k_2}\dots(x-z_r)^{k_r}$$ where $k_1+k_2+\dots+k_r=n$ is the degree $n$ of the polynomial $P$.

This does crucially depend on the existence of roots of any non-constant polynomial over our ring of definition.

1. Uniqueness

There is a separate and subtle question of whether there is only one product form of the polynomial $P$. Above we showed how to compute one product form of the polynomial, but that form seemingly depends on the order in which we find the roots. As long as our ring of definition does not have zero-divisors, that is, the product of non-zero elements is non-zero, then the roots $z_1,\dots,z_r$ above are the only roots of the polynomial since plugging anything else in gives a product of non-zero elements which would be non-zero. Similarly, one can check that the multiplicities also agree.

1. Power/Infinite series

The above rewriting trick does not exactly work for infinite series, because you would have to deal with an infinite sum of $1$s in the constant term, one coming from each $((x-1)+1)^n$. Nevertheless, it is still true that any complex-differentiable function can be written as an infinite series centered around any point; it's just that getting from one point to another requires more complicated tools than basic algebra (e.g. analytic continuation).