Is there a polynomial function $P(x)$ with real coefficients that has an infinite number of roots? What about if $P(x)$ is the null polynomial, $P(x)=0$ for all x?

**Answer**

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.

**Attribution***Source : Link , Question Author : Карпатський , Answer Author : Milo Brandt*