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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 '15 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 '15 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. $\endgroup$ – darij grinberg Feb 7 '15 at 1:43
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    $\begingroup$ @Joffan Constant polynomials are always considered to be polynomials. $\endgroup$ – Bill Dubuque Feb 7 '15 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 '15 at 8:33

11 Answers 11

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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. $\endgroup$ – Matthew Levy Feb 7 '15 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. $\endgroup$ – Milo Brandt Feb 7 '15 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 '15 at 2:26
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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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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 $\endgroup$ – Карпатський Feb 7 '15 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 '15 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. $\endgroup$ – Matthew Levy Feb 8 '15 at 20:06
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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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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 $\endgroup$ – Карпатський Feb 7 '15 at 1:29
  • $\begingroup$ yes, as Matt mentions in his answer the zero polynomial isnt usually considered. $\endgroup$ – Tim Raczkowski Feb 7 '15 at 1:35
  • $\begingroup$ Why this was downvoted? $\endgroup$ – Konstantinos Gaitanas Feb 7 '15 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. $\endgroup$ – Steve Jessop Feb 7 '15 at 17:46
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The only such polynomial is identically zero.

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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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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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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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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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