Is There A Polynomial That Has Infinitely Many Roots?

Solution 1:

The only polynomial with infinitely many roots is $$P(x)=0.$$ You can prove this without appealing to the fundamental theorem of algebra. In particular, let us prove the following:

A polynomial of degree $n\geq 1$ has at most $n$ roots.

We prove this by induction. In the linear case, we obviously have that $P(x)=mx+b$ has exactly one root at $\frac{-b}m$. Next, suppose $P(x)$ is a polynomial of degree $n$. If it has no roots, it satisfies the hypothesis trivially. Otherwise, let $r$ be a root. One can, by using polynomial long division, determine that there is a degree $n-1$ polynomial $P_2$ such that $$P_2(x)\cdot (x-r)=P(x).$$ You can check that this condition is actually equivalent to saying that $r$ is a root, since, when doing polynomial long division, you'll find that the remainder is exactly $P(r)$.

However, by the zero-product law, this means that $P$ has a root exactly when either $x-r$ or $P_2(x)$ is $0$. By inductive hypothesis, $P_2(x)$ is $0$ for at most $(n-1)$ distinct values of $x$ and clearly $x-r$ is zero only at $r$. Thus, $P(x)$ can have at most $n$ zeros. This completes the proof.

Notice that this, unlike the fundamental theorem of algebra, holds in any field - that is, we only need multiplication, addition, and their inverses to be suitably well behaved.

Solution 2:

Not possible if the coefficients ring is an integral domain.

But it is possible in general, e.g. real algebra of quaternions $\mathbb{H}$. The polynomial $f(x) = x^2+1$ has infinitely many roots in $\mathbb{H}[x]$.

Solution 3:

No, the fundamental theorem of algebra tells us that there are at most $n$ roots in the complex plane $\Bbb C$ for $\deg(f)=n$. Since $\Bbb R$ is a subset of $\Bbb C$, this means $f$ has at most $n$ real roots. Usually one rules out the case $f\equiv 0$ when talking about such things.

If you want to think about analytic functions as "infinite degree" polynomials, then

$$\sin z:=\sum_{n=0}^{\infty}{\frac {z^{2n+1}}{(2n+1)!}}$$

has infinitely many roots on the real line, and $\sin z \not \equiv 0$.

In fact, in any integral domain $R$, the number of factors of $f\in R[x]$ is at most $\deg(f)$. $0$ is only a factor of the zero polynomial, so let's exclude it.