Is There A Polynomial That Has Infinitely Many Roots?

Question

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 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.

Answer 2

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.

Answer 3

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 $\mathbb{H}[x]$.

Answer 4

That can't happen. The number of roots can't exceed the degree which is finite.

Answer 5

You don't need to get into noncommutative rings to provide a polynomial with infinite roots, let $R=\mathbb Z[s]/(2s)$ and $p(x)=\overline s(x^2+x)$. Then all integers are roots of $f$.

Answer 6

Hint $\ $ If $\,0\ne f\in \Bbb C[x]\,$ has distinct roots $\,a_1,\ldots\,a_n\,$ then $\,f(x) = (x\!-\!a_1)\cdots (x\!-\!a_n)\, g(x)\,$ for $\,0\ne g\in\Bbb C[x],\,$ by inductively applying the Factor Theorem. Comparing degrees on both sides shows that $\,f\,$ has at most $\,\deg f\,$ roots.

Note $ $ The above proof works for polynomials over any field (or any integral domain, i.e. a commutative ring where $\,ab=0\,\Rightarrow\,a=0\,$ or $\,b=0).\,$ It may fail over more general coefficient rings, e.g. $\rm\,x^2\!-\!1\,$ has $\,4\,$ roots $\,\pm1,\,\pm3\pmod 8.\,$ And nonzero polynomials can have infinitely many roots over noncommutative fields, e.g. $\,x^2+1\,$ over the quaternions.

Answer 7

If $c$ is a root of $P(x)$ then $P(x)=(x-c)Q(x)$ for some polynomial $Q(x)$ of lower degree. The degree can't keep getting lower forever.

[This assumes the degree of $P(x)$ is at least $1$.]

Answer 8

The only such polynomial is identically zero.

Answer 9

By Gauss's fundamental theorem of algebra a polynomial has number of roots equal to its degree, where roots are counted with multiplicities. So in order to have infinite roots we should consider asymptotically infinite degree polynomial. Let us consider a three term recurrence, $$ P_{n+1}(z) = (z^3-z)P_{n}(z)-P_{n-1}(z).$$ With initial conditions $$P_0(z) = 1 $$ and $$P_1(z) = z^3-z+1.$$ As $$ \lim_{n-> \infty} P_{n+1}(z) $$ will have infinite roots. Note that all the coefficients of this polynomial are real. Asymptotically the roots fall on the projection of the polynomial $$ z^3-z = r $$ on complex plane. Here $$r \in (-2i, 2i)$$ and $z$ is any complex number. We can observe this in case of spectrum of tridiagonal k Toeplitz matrices.

Answer 10

Let $f(x)$ be a non-constant polynomial and $\deg(f) = n \geq 1$. If $f(x)$ has infinitely many different roots, then $f(x)$ has $n+1$ different roots $a_1, \cdots, a_n, a_{n+1}$. Since $f(a_1) = 0$, $f(x) = (x-a_1) g_1(x)$ for a polynomial $g_1(x)$ whose degree is $n-1$. Since $f(a_2) = 0$ and $a_2 \neq a_1$, $g_1(a_2) = 0$. So, $g_1(x) = (x-a_2) g_2(x)$. So, $f(x) = (x-a_1)(x-a_2)g_2(x)$. If we continue like this, finally we have $f(x) = (x-a_1) \cdots (x-a_n) g_n(x)$. Since $\deg(f) = n$, $\deg(g_n) = 0$. This means $g_n(x)$ is non-zero constant $b$. So, $f(x) = b (x-a_1) \cdots (x-a_n)$. Since $a_{n+1} \neq a_i$ for $i \in \{1, \cdots, n\}$, $f(a_{n+1}) \neq 0$. This is a contradiction.

Answer 11

As comments have noted above, on a field a non-constant polynomial of degree $n$ has at most $n$ distinct roots. The quaternions, which are not a field, provide what you're looking for. For example, there are infinitely many square roots of $-1$. If $z=bi+cj+dk$ is a quaternion with $b,c,d$ real so that $b^2+c^2+d^2=1$, then $z^2+1=0$.

Answer 12

A non zero polynomial can also have all its base set as roots!

Have a look at Math Counterexamples.