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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?

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    $\begingroup$ According to Fundamental Theorem of Algebra, any polynomial function of degree $n$ can have at most $n$ roots. en.wikipedia.org/wiki/Fundamental_theorem_of_algebra $\endgroup$
    – Henry
    Feb 7, 2015 at 1:21
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    $\begingroup$ Since half my comment has now made it into the question, I'll just add a little to say I don't think that either $P(x) := 0$ or $P(x) := k$ a constant qualify as polynomials in many senses. It's one of those cases when we'll count them as polynomials if it's convenient, then disqualify them when they complicate some process. $\endgroup$
    – Joffan
    Feb 7, 2015 at 1:35
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    $\begingroup$ @Joffan: This is rarely helpful. It is like saying that the zero vector is not a vector. $\endgroup$ Feb 7, 2015 at 1:43
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    $\begingroup$ @Joffan Constant polynomials are always considered to be polynomials. $\endgroup$ Feb 7, 2015 at 4:03
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    $\begingroup$ Here's some trivia: Euler thought of $\sin x$ as a "polynomial" with an infinite number of roots to guess at this correct identity. $\endgroup$
    – Casey Chu
    Feb 7, 2015 at 8:33

12 Answers 12

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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.

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    $\begingroup$ It does not work if we have zero-divisors though. $\endgroup$ Feb 7, 2015 at 1:34
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    $\begingroup$ @Matthew Correct, which is why this proof only works for fields, which necessarily have no zero divisors (since if $ab=0$ for non-zero $b$, then $a=abb^{-1}=0b^{-1}=0$). It does not extend to the case of rings where it is, for the reason you note, not necessarily true. $\endgroup$ Feb 7, 2015 at 1:39
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    $\begingroup$ This holds in any integral domain. $\:$ One could see that abstractly by embedding in the field of fractions, but it's more direct to just observe that your proof goes through for them. $\;\;\;\;$ $\endgroup$
    – user57159
    Feb 7, 2015 at 2:26
  • $\begingroup$ @MatthewLevy he said that for zero divisors , its trivially true that it has less than n roots which is here is 0 and 0<n? $\endgroup$
    – Orion_Pax
    May 16 at 18:10
  • $\begingroup$ @Orion_Pax: You are commenting on a post that is over 7 years old; addressing a user that hasn't been on the site for over 2 years; and your statement is false. Nobody in this post said that the result is "trivially true" when the ring has zero divisors. In fact, the opposite: it is stated several times that if the ring has zero divisors, then the result is not necessarily true. $\endgroup$ May 16 at 21:15
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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.

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  • $\begingroup$ Thank you for your answer. What about if P(X) is the null polynomial? P(X)=0 for all x $\endgroup$ Feb 7, 2015 at 1:28
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    $\begingroup$ No, fundamental theorem of algebra says something quite different. What applies here is the Bezout's theorem, instead (which, along with the fundamental theorem of algebra, implies that a complex polynomial decomposes into linear terms; incidentally, it is much more elementary, for example, I had been told a proof of it in high school classroom, in stark contrast to the fundamental theorem of algebra, whose proper proof I have not seen until second year undergraduate complex analysis and differentiable manifods courses, I think). $\endgroup$
    – tomasz
    Feb 8, 2015 at 2:07
  • $\begingroup$ You have to do a little bit of work, but yet it does come directly from The FTA. Obviously, there are many roots to the same answer in math and it is all connected. $\endgroup$ Feb 8, 2015 at 20:06
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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 $\mathbb{H}[x]$.

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  • $\begingroup$ Great. The answer of the above question is in general false for arbitrary Rings. Take a Commutative Ring with Trivial multiplication then any element is root of f(x)=x². Cool right? $\endgroup$
    – MathCosmo
    Dec 25, 2021 at 17:58
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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 $f$.

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

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  • $\begingroup$ Thank you for your answer. What about if P(X) is the null polynomial? P(X)=0 for all x $\endgroup$ Feb 7, 2015 at 1:29
  • $\begingroup$ yes, as Matt mentions in his answer the zero polynomial isnt usually considered. $\endgroup$ Feb 7, 2015 at 1:35
  • $\begingroup$ Why this was downvoted? $\endgroup$ Feb 7, 2015 at 8:33
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    $\begingroup$ Probably because it fell right into the questioner's "trap" (intended by the questioner or otherwise), and misstated the fundamental theorem of algebra by leaving out the special case. $\endgroup$ Feb 7, 2015 at 17:46
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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.

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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 $1$.]

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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.

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The only such polynomial is identically zero.

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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 $z^2+1=0$.

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

Have a look at Math Counterexamples.

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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.

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