I can perform operations on polynomials. I can add, multiply, and find their roots. Despite this, I cannot define a polynomial.

I wasn't in the advanced mathematics class in 8th grade, then in 9th grade I skipped the class and joined the more advanced class. This question isn't about something I don't understand; it's something I missed.

My classes have not covered what really a polynomial is. I can generate one, but not define one. The internet has yielded incomplete definitions: "Consisting of multiple terms" or "A mathematical expression containing 2 or more terms and variables."

Take the following expressions for example:

$2x^2-x+12-2x^2+x-12$. Consists of multiple terms, but can also be expressed as $0$. Is zero a polynomial?

What about $x^{-1}$? $x^{-1}$ have -1 zeroes?">I've been told this one isn't a polynomial, but I don't understand why.

Is $x^2+x+1-x^{-1}-x^{-2}-x^{-3}$? a polynomial? It contains both positive and negative exponents?

tl;dr: What actually is the mathematical definition of a polynomial? Is $0$ a polynomial, and why isn't $x^{-1}$ a polynomial under this definition?

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    $\begingroup$ $0$ is a polynomial of degree $0$ as is any other constant. A polynomial is any linear combination of nonnegative integer powers of an indeterminate. $\endgroup$
    – Qudit
    Commented Mar 14, 2017 at 0:15
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    $\begingroup$ @Qudit 0 is a polynomial, but there is some question of how to define its "degree" in the most useful way, or if it should be defined at all. It may depend on what you want the "degree" for. See math.stackexchange.com/questions/1796312 $\endgroup$
    – David K
    Commented Mar 14, 2017 at 0:28
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    $\begingroup$ Your question about $x^{-1}$ can be interpreted in two ways: first, why is $x^{-1}$ not a polynomial by its form; second, why is $f(x) = x^{-1}$ not a polynomial function? The first question is answered below: the exponent $-1$ is not a nonnegative integer, as required in the definition of a polynomial. The second question means: is it possible that for some polynomial $P(x)$, we have $x^{-1} = P(x)$ for all values of $x$. The answer is no, for the simple reason that $x^{-1}$ is not defined when $x = 0$, whereas the polynomial $P(x)$ must be. But one could still ask, is it possible... $\endgroup$
    – user49640
    Commented Mar 14, 2017 at 2:32
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    $\begingroup$ According to Serge Lang, a polynomial with coefficients in a ring $A$ is the same thing as a finitely-supported function from the monoid of natural numbers (including zero of course!) into $A$. $\endgroup$
    – Dorebell
    Commented Mar 14, 2017 at 2:46
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    $\begingroup$ @Qudit : Actually, the degree of the constant polynomial $0$ is often defined to be $-\infty$, and isn't usually defined to be zero. That's done so that the formula $\operatorname{deg}(pq)=\operatorname{deg}(p) + \operatorname{deg}(q)$ holds even if $p$ or $q$ is $0$. $\endgroup$
    – MPW
    Commented Mar 17, 2017 at 20:03

19 Answers 19


A polynomial (in one variable) is an expression of the form $$ p(x) = a_0+a_1x+a_2x^2+\ldots+a_nx^n$$ where the coefficients $a_i$ are some kind of number (or more generally they're elements of a Ring). The exponents $1,2,\ldots n$ must all be integers.

Unless we've been silly and $a_n=0,$ $n$ is called the degree of the polynomial. We can formalize this by defining the largest $n$ such that $a_n\ne0$ as the degree.

Notice that constants are allowed. $p(x) = 3$ is a zero-th degree polynomial.

You asked about zero. Yes, $p(x) =0$ is considered to be a polynomial. However, you'll notice that there is a problem with the definition of degree here since there is no coefficient that is nonzero. The degree of the zero polynomial is thus undefined.

This allows us to say that if we multiply two polynomials $w(x)=p(x)q(x)$ with $p$ of degree $n$ and $q$ of degree $m,$ then $w$ has degree $n+m.$ (Notice how the zero polynomial would mess this up if its degree were defined to be zero like the other constants.)

You're right that simplification is important. The $x$ is just a symbol and we can always "combine like terms" $$ a_lx^l+b_lx^l= (a_l+b_l)x^l.$$ We always combine all the terms together and simplify in order to get an expression into the form above with only one term for each power before we do things like consider the degree.

Notice we can add two polynomials according to the simplification rule and get a polynomial as a result. This is a good reason to consider zero to be a polynomial... it allows the sum of two polynomials to always be a polynomial. Likewise we can multiply two polynomials according to the the distributive property, the rule $$ (a_mx^m)(a_lx^l) = a_ma_l x^{m+l},$$ and the additive simplification rule. The result will be another polynomial.

Yes, the exponents all need to be positive. Of course other expressions are possible but they aren't called polynomials. Terms like $x^{-3}$ are considered part of the family of rational functions (or as a commenter noted, the Laurent polynomials, not to be confused with the (unqualified) polynomials). This is just a definition and thus somewhat arbitrary (though good definitions are important for organization). It's just like saying $-4$ is an integer but not a natural number. It's true by definition, and yes, a bit arbitrary, but nonetheless useful and a nearly universal convention.

EDIT As Paul Sinclair pointed out in the comments, there are also polynomials in multiple variables. For instance $$p(x,y) = A + Bx + Cy +Dx^2+Exy+Fy^2$$ is the general degree two polynomial in two variables. The degree of a term is just the sum of the degrees with respect to the individual variables. So a term like $3xy$ has degree two and a term like $3x^4y^5z$ would have degree $4+5+1=10.$ The degree of a polynomial is the degree of its highest-degree term with nonzero coefficient.

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    $\begingroup$ This may sound pedantic, but I would call the polynomial $p$, $p(x)$ is just a number (or some other thing depending on your ring, etc). $\endgroup$ Commented Mar 14, 2017 at 2:30
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    $\begingroup$ @YoTengoUnLCD Agree there's a distinction and potential ambiguity between the polynomial in $R[x]$ and its image under evaluation at an arbitrary point $x\in R$. But to me, $p(x)$ could refer to either. $x$ is not a ring element on the right hand side so it doesn't need to be on the left. Usually see this distinction handled by having a character other than $x$ be standard notation for an arbitrary element of $R$ (say it's $\alpha$) and writing $f(\alpha)$ when evaluation is intended. I guess for me the notation $p = a_0+a_1 x$ feels imbalanced, but everyone's mileage varies. $\endgroup$ Commented Mar 14, 2017 at 3:05
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    $\begingroup$ I think it's pretty common to say that the zero polynomial has degree $-\infty$ to restore the additivity of degree under multiplication. $\endgroup$
    – Danu
    Commented Mar 14, 2017 at 10:39
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    $\begingroup$ More accurately, what you've described is a polynomial of one variable. There are of course polynomials of many variables, which alas only Ethan Bolker has mentioned, and he only as an "abstraction". $\endgroup$ Commented Mar 14, 2017 at 16:15
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    $\begingroup$ To prevent some confusion: When you say " a polynomial is an expression of the form $p(x)=a0+a_1 x \cdots$" someone could believe the "expression" includes the equal sign - but the "expression" is not the equation but the RHS. $\endgroup$
    – leonbloy
    Commented Mar 17, 2017 at 11:51

There are lots of good answers here and they are all essentially correct, even though they are different! I will try to contribute another, which is somewhat more abstract than the others. I normally wouldn't try this for a high school student, but your very good question deserves different kinds of answers. Maybe this one will help.

It's the "what actually is" in your question that I want to address. In mathematics at a more advanced level you don't think as much about what something "is" as you do about how it "behaves". (The same is true in object oriented programming languages = you say you're studying computer science. If you're learning Java you know about this.)

To manipulate polynomials (which you know how to do) all you really need to know is the sequence of coefficients. We'll assume for the moment that those coefficients are ordinary numbers. It's useful to start those coefficients with the constant term. since the degree (which is the place that holds the last nonzero coefficient) isn't fixed. So the polynomial $$ 8x^3 + 5x + 7 $$ is "really just" the sequence $$ (7, 5, 0, 8) $$ or, if you like $$ (7, 5, 0, 8, 0, 0, \ldots) $$ where the zeroes go on forever.

What "really just" means there is that if you know the sequences of coefficients for two polynomials you can calculate out the sequence for their sum. Just add the sequences element by element. You can also calculate their product. It's a little harder to write down the algorithm, but you can figure it out if you understand how writing a polynomial the high-school way with powers of $x$ makes the multiplication automatic.

You can even divide one polynomial by another as long as you're willing to allow yourself a remainder (and allow fractions for the coefficients). You may in fact have learned how to do that and called it "synthetic division".

You can also "evaluate" a polynomial at a number $n$ when you know its coefficients.

What all this means in practice is that you don't need "$x$" or its powers to think about polynomials. The "variable" just helps to keep the polynomial arithmetic straight. And that's so useful that we almost always write polynomials with an $x$ rather than as a sequence of coefficients.

Finally, this abstract view lends itself to further abstraction! All you need to know to manipulate polynomials (written as sequences) is how to add and multiply the coefficients. So the coefficients themselves might be polynomials. So, for example, you can think of $$ 4x^2y^3 + 6xy^3 - 2xy^2 $$ as "a polynomial in $x$ whose coefficients are polynomials in $y$": $$ (0, -2y^2 + 6y^3 , 4y^3) = ((0), (0, 0, -2, 6), (0, 0, 0, 4)) $$ or as "a polynomial in $y$ whose coefficients are polynomials in $x$". (You write that one.)

The coefficients can even be matrices, when you learn what matrices are and how to add and multiply them.

Further thoughts:

You can think of the algorithms for addition and multiplication you learned a long time ago as like the arithmetic of polynomials, only more complicated. When you "collect like powers of $x$" in a polynomial, you just add up what you see. When you "collect like powers of $10$" in ordinary arithmetic you have to simplify further by "carrying", so replacing, say, $21 + 7 \times 10$ by $1 + 9 \times 10$.

If you relax the requirement that the coefficients be $0$ from some point on then you are dealing with a (formal) power series, traditionally written $$ a_0 + a_1 x + a_2 x^2 + \cdots = \sum_{n=0}^\infty a_n x^n . $$ You can add these and multiply them with the usual polynomial rules. They are "formal" power series because trying to evaluate them by substituting a value for $x$ is much more subtle than it is for polynomials. You'll study that in calculus. (And formal power series have uses that don't depend on evaluation.)

Then you can decide allow a few terms with negative powers, like $$ 4x^{-3} + 7x^{-1} + \text{ an ordinary formal power series} . $$ These are called "Laurent series"; they come up when you study functions of a complex variable. You have lots of nice mathematics to look forward to.

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    $\begingroup$ I see that @CarlMummert wrote essentially the same answer while I was writing mine. So you have two explanations for the same idea. That's often useful. $\endgroup$ Commented Mar 14, 2017 at 1:01
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    $\begingroup$ @hello_there_andy I think you misunderstand. I wasn't accusing CarlMummert. If anytning I was apologizing for seeming to duplicate his fine answer. I'm pretty sure he and I agree that both answers may be helpful. $\endgroup$ Commented Mar 15, 2017 at 12:17
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    $\begingroup$ @CarlMummert See my response above to hello_there_andy. $\endgroup$ Commented Mar 15, 2017 at 12:18
  • $\begingroup$ +1 just for explicitly mentioning the gem that is 'synthetic division' - oh how grateful I was to my high school teacher for dispensing with cumbersome writing out of indeterminates (if it's what I think you mean)! A cute first algorithm to learn as a kid! $\endgroup$
    – Mehness
    Commented Nov 24, 2018 at 6:31
  • $\begingroup$ Typo fixed per here $\endgroup$ Commented Jan 18, 2020 at 18:32

Note: in this answer I will try to motivate the definition which is used in more advanced contexts such as "abstract algebra". This may go beyond what is in a typical pre-algebra book, but I hope it will show how the mathematics community has found a way to come up with a workable definition, even it is less obvious at first.

It is hard to define polynomials because there is a tension between several of their key properties, which don't quite agree:

  1. A polynomial can be written as an expression in the form $a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$ for some $n \geq 0$ and some choice of coefficients $a_0, \ldots, a_n$.

  2. The sum of two polynomials is a polynomial. The product of two polynomials is a polynomial. Overall, the collection of polynomials is the smallest collection that includes all the numbers, $x$, and is closed under addition and multiplication.

  3. The expressions $(x+1)(x-1)$ and $x^2-1$ determine the same polynomial.

If we want to use something like (1) as a definition, we end up with the issue that $x$ and $2x$ are defined to be polynomials, but $x + 2x$ is a polynomial according to (2) but is not literally in the form shown in (1). So we have to define a "simplification" operation.

If we want to use something like (2) as a definition, then we still have the issue of defining when two polynomials are equal, as (3) points out.

In general, although it is tempting to define polynomials in terms of "expressions", this causes more trouble than it is worth. So it is common in more advanced texts to define polynomials as follows:

A polynomial (over the real numbers) is a sequence of real numbers $(a_i : i \in \mathbb{N})$ in which at most finitely many of the terms are nonzero. Two polynomials are equal when they are the same sequence.

So $(2,1,0,0,\ldots)$ and $(0,1,3,0,0,\ldots)$ are polynomials according to this definition. Of course, the "polynomial" $(2,1,0,0,\ldots)$ is meant to stand for $2 +x$, and $(0,1,3,0,0,\ldots)$ stands for $x + 3x^2$. But in these definitions we do not define the polynomials in terms of the expressions. Rather, we view the expressions as nothing more than notation - shorthand - for the sequences which are actually polynomials.

We continue the definition by defining addition of polynomials using the formula $(a_n) + (b_n) = (a_n + b_n)$.

Multiplication is defined in a way analogous to the Cauchy Product: $(a_n)(b_n)$ is defined to be the sequence $(c_n)$ where $$ c_k = \sum_{i=0}^k a_i b_{k-i}. $$ This is exactly the formula you would discover if you multiply polynomials in the usual, pre-algebra style.

In this way, the collection of polynomials in the variable $x$ is identified with the ring $\mathbb{R}[x]$, which is also defined as the set of finitely-supported sequences of reals with the operations shown above. These definitions of the operations take care of simplification automatically, so we do not need to worry about "unsimplified" polynomials in the formal definition.

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    $\begingroup$ good job remembering that n must be finite! otherwise, you have a power series. also, IIRC, a polynomial with more than one indeterminate would be a "multinomial". $\endgroup$ Commented Mar 14, 2017 at 23:23
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    $\begingroup$ For the benefit of the OP, Chapter 4 of Lang's Undergraduate Algebra contains further reading on polynomials in the context of abstract algebra. $\endgroup$ Commented Mar 15, 2017 at 3:18
  • $\begingroup$ Motivation-wise, the more interesting feature IMO is that polynomials can be associated with algebraic numbers (numbers which can be derived from rational numbers via addition, multiplication and exponentiation): for any algebraic number a, there is a unique rational-coefficient polynomial p for which p(a)=0, and all other polynomials with that property are multiples of p. So in a (very sloppy) sense polinomials describe possible ways in which the rational numbers can be extended. This concept turns out to generalize well to other number systems (such as integers modulo some prime). $\endgroup$
    – Tgr
    Commented Mar 15, 2017 at 9:53
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    $\begingroup$ "Rather, we view the expressions as nothing more than notation - shorthand - for the sequences which are actually polynomials." - If you define $x$ as the sequence $0,1,0,0,0\ldots$ (assuming the coefficient ring has, or can be given, a 1), the expressions are actually valid for the polynoleum ring. $\endgroup$ Commented Mar 16, 2017 at 10:38
  • $\begingroup$ Also, your definition (2) is interesting, except that the coefficients only need to be elements of a ring, and are not necessarily "numbers". In a ring, elements have additive inverses and there is an additive identity, so two polynomials p1 and p2 are equal if p1 + (-p2) = 0. $\endgroup$ Commented Mar 16, 2017 at 23:48

Added: 15/12/2018

Although I still think the ideas in this answer are great, in retrospect the exposition is lacking. As one commenter says, this answer would be infinitely more useful if it actually explained things rather than just stating stuff. Consequently, I would request that someone edit or totally rewrite it to make the answer more comprehensible. If there's any takers, please comment below. If there are no takers, I might try myself, although I'm not sure where to even start.

the exposition and lack of explanatory

The other answers do a great job of giving a non-technical explanation. For users of the website who are a little further along in their studies, here's a fairly technical answer.

Philosophically speaking, I think that the concept polynomial with coefficients in $R$ somehow "is" the endofunctor $U \circ F : \mathbf{Set} \rightarrow \mathbf{Set}$, where $U$ is the forgetful functor $R\mathbf{Alg} \rightarrow \mathbf{Set}$ and $F$ is its left-adjoint. This ties in with Carl's answer, namely that:

The sum of two polynomials is a polynomial. The product of two polynomials is a polynomial. Overall, the collection of polynomials is the smallest collection that includes all the numbers, x, and is closed under addition and multiplication.

The reason this is a good description of polynomials is because:

  • Carl is being vague, and just emphasizing polynomials with integer coefficients
  • An object of $\mathbb{Z}\mathbf{Alg}$ is just a ring
  • The signature $(+,\times,0,1)$ is a sufficiently large to state the axioms of ring theory, so we just need closure under these operations (and Carl is being vague and not including $0$ and $1$.)

The reason this is an incomplete answer is because

  • it doesn't tell how to decide whether or not two polynomials are equal.

So, how do we decide whether or not two polynomials are equal? By applying the axioms of ring theory, of course! Two polynomials with integer coefficients are equal if, and only if, the axioms of ring theory can be used to prove they're equal. Otherwise, they're distinct. Seen from this vantage point, it's not too surprising that the category $\mathbb{Z}\mathbf{Alg}$ of rings has a direct connection to polynomials.

By the way, I think it's similarly the case that the concept $R$-linear combination "is" the endofunctor $U \circ F$, with $R\mathbf{Alg}$ is replaced by $R\mathbf{Mod}$. In fact, there's a whole dictionary of such things:

$R \mathbf{Alg} \mapsto \mbox{Polynomial with coefficients in $R$}$

$R \mathbf{Mod} \mapsto \mbox{$R$-linear combination}$

$\mathbf{Mon} \mapsto \mbox{Word}$

$\mathbf{Grp} \mapsto \mbox{Reduced Word}$

$\mathbf{PSet} \mapsto \mbox{Element}$

$\mathbf{SupLat} \mapsto \mbox{Subset}$

$\mathbf{Magma} \mapsto \mbox{Catalan Tree}$

etc. On the left we have concrete categories, and on the right we have the monads that they define, and are defined by. The technical concept that underlies this correspondence is that of a monadic adjunction. This is all well-known, of course, but I like reassuring myself that apparently abstract concepts give meaningful, coherent answers to the kinds of questions Year 9 students might ask their apparently-humble math tutor. This is the kind of thing that got me excited about mathematics in the first place :)

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    $\begingroup$ I could not resist voting up even though I suspect it is useless to the questioner. $\endgroup$
    – PJTraill
    Commented Mar 15, 2017 at 18:49
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    $\begingroup$ @PJTraill, thanks. I'd say you more than suspect, but I wrote this as a little treat for people like us :) $\endgroup$ Commented Mar 16, 2017 at 0:07
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    $\begingroup$ Your answer would be a lot more useful if you would explain instead of just stating. To start with, how are polynomials maps from sets to sets, rather than from rings to rings? $\endgroup$
    – celtschk
    Commented Mar 16, 2017 at 7:57
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    $\begingroup$ What is $\mathbb{Z}\textbf{Alg}$? (By the way, this is precisely the sort of thing that got me into math in the first place, too!) $\endgroup$ Commented Mar 16, 2017 at 15:55
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    $\begingroup$ @étale-cohomology, $\mathbb{Z}\mathbf{Alg}$ is the category of ring $\cong$ the category of $\mathbb{Z}$-algebras. Google the phrase "$R$-algebra" to learn more. $\endgroup$ Commented Mar 17, 2017 at 14:09

A polynomial in the indeterminate $x$ is an expression that can be obtained from numbers and the symbol $x$ by the operations of multiplication and addition.

$0$ is a polynomial, because it is a number.

Any positive integer power of $x$ is a polynomial, because you can get it by multiplying the appropriate number of $x$'s together (e.g. $x^3 = x \cdot x \cdot x$). But negative and non-integer powers of $x$ are not polynomials (e.g. $x^{-1}$ is not a polynomial), because those operations only give you positive integer powers of $x$.

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    $\begingroup$ One is introduced to polynomial functions in high school algebra, and later to formal polynomial expressions. The difference in simple terms is that polynomial functions take on values when argument $x$ (in the case of a single variable) is assigned a value, while formal polynomial expressions are not functions per se. $\endgroup$
    – hardmath
    Commented Mar 14, 2017 at 1:36
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    $\begingroup$ This is a good perspective that deserves to be represented higher in the answer sequence. But I do wish there were some gentle description of quotienting out the difference between expressions that can be proven to be equal based on ring axioms and arithmetic on the constants, and how each equivalence class then has exactly one member that's in standard form. $\endgroup$ Commented Mar 14, 2017 at 10:57
  • $\begingroup$ @hardmath... wrong. Not all highschools are equal. Any failure of the education-ing infrastructures causal to a lack of understanding, completely, the entirity of polynomials (in all associated fields), is not considered as a means to counter a fellow SM user... We are beyond the significance of our varied educational backgrounds here on this site. $\endgroup$ Commented Mar 15, 2017 at 4:12
  • $\begingroup$ I don't think "numbers" have anything to do with polynomials. As I remember, the $a_i$ only need to be elements of a ring. $\endgroup$ Commented Mar 16, 2017 at 23:51
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    $\begingroup$ @richard1941 True, but Travis seems to be in high school. I tried to pitch my answer to that level. $\endgroup$ Commented Mar 17, 2017 at 1:03

This isn't a definition suitable for pre-calculus, but I would say that a polynomial in a variable $x$ is anything whose $n^\text{th}$ derivative with respect to $x$ vanishes everywhere (i.e. is equal to zero everywhere), for some integer $n \geq 0$.

The nice thing about this definition is that it talks about how the polynomial behaves, rather than how you write it (so $\cos(2 \cos^{-1} x)$ is also a polynomial in $x$). It also generalizes appropriately to more abstract objects such as rings, functions, etc. as long as you define derivatives appropriately.

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    $\begingroup$ This is an intriguing definition, but in a sense backwards because the concept of derivatives basically amounts to “function can be locally approximated by a polynomial”. $\endgroup$ Commented Mar 18, 2017 at 11:50
  • $\begingroup$ @leftaroundabout: I think you're thinking about analyticity, not merely derivatives? Not every differentiable function is analytic... $\endgroup$
    – user541686
    Commented Mar 18, 2017 at 20:03
  • $\begingroup$ No, analyticity is another subject. An analytic function is equal to its Taylor expansion (in general an infinite series, i.e. not a polynomial), but every $n$-times differentiable function is approximated by a polynomial of degree $n$ corresponding to the truncated Taylor series (in some suitable $\varepsilon\mapsto\delta$ sense, which is how differentiability is defined in standard analysis). How well this approximation works has little to do with whether the function is analytic. $\endgroup$ Commented Mar 18, 2017 at 23:09
  • $\begingroup$ @leftaroundabout: no, doesn't make sense. Analyticity is what tells you that the Taylor series of the function converges to the function itself locally... i.e., it's what tells you the partial sums of the Taylor series approximate the function locally. If the Taylor series doesn't converge to the function locally then we don't say it's "approximating" it. But heck, if your function is continuous at $x_0$ then its value near $x_0$ is already "approximated" by $f(x_0) + 0$ too, by whatever hand-wavy definition you're using (what is it?) so is continuity now a polynomial concept too? C'mon... $\endgroup$
    – user541686
    Commented Mar 19, 2017 at 0:12
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    $\begingroup$ Furthermore there's literally no mention of or allusion to polynomials anywhere in the definition of a derivative, so I'm not sure how the concept is "backwards". It'd have been totally possible for derivatives to have been formulated first, with people subsequently realizing that polynomials always have uniformly vanishing derivatives. $\endgroup$
    – user541686
    Commented Mar 19, 2017 at 0:12

I will give you a rigorous definition.

Definition 1. A quadratic polynomial in the variable $x$ is an expression of the form $$ a x^2 + bx + c, $$ where $a$, $b$ and $c$ are real numbers and $a \not = 0$.

Example 1. Take $a=1$, $b=2$ and $c=0$. You can then see that $$ x^2 + 2x $$ is a quadratic polynomial.

More generally, we have the following definition of a polynomial (not necessarily quadratic).

Definition 2. A polynomial in the variable $x$ is either $0$ or an expression of the form $$ a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0, $$ where $n$ is a non-negative integer, $a_n, a_{n-1}, \dots, a_1, a_0$ are real numbers and $a_n \not = 0$. The non-negative integer $n$ is said to be the degree of the polynomial.

Example 2. The expression $x^{-1}$ is not a polynomial. While it is indeed an expression of the form $a_n x^{n}$, where $n = -1$ and $a_n = 1$, the integer $n$ is not positive, contradicting our definition.

Further remarks.

You can define the addition and the multiplication of polynomials in the way you are used to. This implies that $$ x+ 2x + 3x^2 = 3x^2 + 3x + 0 $$ and $$ (x-2)(x+2) = x^2 + 0x - 4 $$ are also polynomials, by definition.

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    $\begingroup$ Your definition 2, as currently stated, has a problem with the polynomial $0$. $\endgroup$ Commented Mar 14, 2017 at 8:52
  • $\begingroup$ Yes, thanks for the remark. The degree of $0$ is usually undefined or taken as $-\infty$. $\endgroup$
    – Olivier
    Commented Mar 14, 2017 at 13:07
  • $\begingroup$ I agree, but we must make sure that $0$ still satisfies the definition for a polynomial. It is fine that the definition of the degree breaks down. But we must agree that $0$ is a polynomial. $\endgroup$ Commented Mar 14, 2017 at 13:26
  • $\begingroup$ Corrected, thank you. $\endgroup$
    – Olivier
    Commented Mar 14, 2017 at 13:35
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    $\begingroup$ @richard1941 We are not talking about algebraic and transcendental numbers. There is really no problem about using real coefficients; you can find the ring $\mathbb{R}[x]$ used in all introductory abstract algebra books. Algebraic and transcentental numbers are defined with respect to $\mathbb{Z}[x]$. See here. $\endgroup$
    – Olivier
    Commented Mar 14, 2017 at 23:48

The simplest answer: a polynomial is a linear combination of a finite number of monomials.
See Wikipedia for monomial; also binomial and trinomial.

As the Wikipedia Monomial article says in the lead, in some contexts monomial may have negative integer exponents (for example in Laurent polynomials).

For ordinary polynomials (with positive exponents) a degree of a polynomial is the highest exponent among all monomial terms (those actually present in a polynomial, i.e. with non-zero coefficients) in case of one-variable polynomials, or a highest sum of exponents in case of multi-variable polynomials.

  • $2x^7+5x+2$ is of degree $7$ (which is the highest one among $7$, $1$ and $0$)
  • $3qt^3+5q^2 + t$ is of degree $4$ (which is the highest among $1+3$ from $q^\color{red}1t^\color{red}3$, $2$ and $1$)

This is a non-simple question, unfortunately. Polynomials can be defined in very abstract and potentially incomprehensible terms.

Formally, a polynomial in one variable--say $x$--with real coefficients can be defined as an expression that can be equivalently expressed as a real linear combination of finitely-many terms of the form $x^n$ (where $n$ is a non-negative integer, and $x^0:=1$).

$0$ is a polynomial, since it can be written (for example) as $0x^0.$ However, $x^2+x+1-x^{-1}-x^{-2}-x^{-3}$ is not a polynomial, since it has negative exponents. Neither is $\sqrt{x}$ a polynomial, since it has non-integer exponents. Neither is $1+x+x^2+x^3+\cdots$ a polynomial, since it cannot be expressed in finitely-many non-$0$ terms. On the other hand, the following is a polynomial: $-1+(x-x)+(x^2-x^2)+(x^3-x^3)+\cdots.$ In particular, it is equivalent to the (constant) polynomial $-1x^0.$

  • $\begingroup$ The series now is a series of parenthetical expressions, which does converge (very quickly indeed!). Now the question is, as with several of the other definitions, is a polynomial defined by a particular form or is it defined by what it evaluates to? $\endgroup$
    – David K
    Commented Mar 14, 2017 at 3:34
  • $\begingroup$ @David: I contend that a polynomial is more of a formal thing. One can indeed define a polynomial function, but there needn't be any notion of evaluation to have a polynomial. For that matter, it needn't "look like" a polynomial to "act like" a polynomial. One could instead define polynomials to be real sequences whose terms are eventually zero (all but finitely many terms are $0$). We can then define addition term-wise, multiplication in a more complicated fashion. $\endgroup$ Commented Mar 14, 2017 at 17:08
  • $\begingroup$ @CameronBuie I find the formal definition very attractive. It seems, however, not very compatible with writing things like $-1+(x-x)+(x^2-x^2)+\cdots,$ unless that expression is simply to be interpreted as the sequence $(-1,0,0,\ldots).$ $\endgroup$
    – David K
    Commented Mar 14, 2017 at 17:15
  • $\begingroup$ @David: More like $\langle -1,1-1,1-1,1-1,\dots\rangle,$ but this amounts to the same thing, of course. $\endgroup$ Commented Mar 14, 2017 at 17:31
  • $\begingroup$ @CarlMummert The polynomials (in $x$ over a ring $R$) form a ring, don't they? The product of two polynomials is a polynomial, isn't it? So $(x+1)(x-1)$ is the product of the polynomials $x+1$ and $x-1$ in the ring of polynomials, and is equal to the polynomial $x^2-1.$ $\endgroup$
    – bof
    Commented Mar 15, 2017 at 22:45

A polynomial is any element of a free extension of a ring (which in this answer is taken to mean "commutative ring with a multiplicative identity"). Thus, a polynomial can only be defined with respect to a given ring, say the ring $R$. The simplest free extension of $R$ is generated by augmenting $R$ with a single free element, say $x$, and is denoted by $R[x]$. Here free means that the elements of $R[x]$ are unconstrained by any condition apart from the ring axioms and any particular condition on the elements of $R$. Every element of $R[x]$ can be written in the form $\sum_{k=0}^n a_kx^k$, where $n\in \Bbb N$ and $a_k\in R$ for $k=0,...,n$, with the usual operations of addition and multiplication for such elements. In this context, the element $x$ is often called a variable.

Generally a ring can be freely extended by any number of variables, even infinitely many; the elements of such extensions are still called polynomials; and the resulting rings are called polynomial rings. As an example, we have the polynomial ring $R[x,y,z]$ in three variables.

Often the base ring is $\Bbb R$. In this case, note that the ordered-field structure of $\Bbb R$ does not extend to $\Bbb R[x]$, although division of elements of $\Bbb R[x]$ by nonzero elements of $\Bbb R$ is still definable. Another common example is $\Bbb C[z]$, where the name of the variable is $z$, rather than $x$, by convention. Other base rings often encountered are $\Bbb Z$ and $\Bbb Q$.

Added note: It may be asked why we need to have such an abstract definition of a polynomial. Indeed, for each of the familiar rings $\Bbb Z$, $\Bbb Q$, $\Bbb R$, and $\Bbb C$, the related polynomial rings are isomorphic to the corresponding rings of polynomial functions; for example, we could identify the element $x^8-2x^6+x^4$ in $\Bbb R[x]$ with the polynomial function $x\mapsto x^8-2x^6+x^4$ on $\Bbb R$. Sadly this doesn't work in general. In the case of the "clock arithmetic" ring $\Bbb Z_{12}$, the polynomial function $x\mapsto x^8-2x^6+x^4$ on $\Bbb Z_{12}$ is indistinguishable from the zero function, although the polynomial $x^8-2x^6+x^4$ is a perfectly good member of $\Bbb Z_{12}[x]$ in its own right.

  • $\begingroup$ Free extension of a commutative ring I presume? Otherwise you have to modulo some commutators out. Obviously this only really matters when looking at multivariate poly rings. $\endgroup$ Commented Mar 18, 2017 at 9:13
  • $\begingroup$ Thank you, @AliCaglayan. I have made this explicit now. $\endgroup$ Commented Mar 18, 2017 at 21:20

A polynomial is a mathematical expression (as opposed to an equation) where all terms are either added or subtracted from each other (if there is more than one term), each term contains some real number constant, and each term contains a variable with a non-negative power. You cannot have infinitely many terms. The number one is a polynomial. Likewise, zero is a polynomial. Any term with a negative powered variable invalidates the entire expression from being a polynomial.

Edit: In regards to the expression that simplifies to zero, both the original expression and zero are polynomials. The expression with negative powers is not a polynomial. If you had an expression with negative powers that simplified to zero, my understanding is that the unsimplified expression is not a polynomial, but the simplified expression, 0, is a polynomial.

Edit 2: No you cannot have infinitely many terms.

  • $\begingroup$ I think you meant "you cannot have infinitely many terms". Separately, the powers should be distinct, or else $x+x$ is a polynomial. $\endgroup$ Commented Mar 14, 2017 at 0:12
  • 2
    $\begingroup$ @UniqueWorldline No, you can't have infinitely many terms. If you do, it's not a polynomial, it's a power series. $\endgroup$ Commented Mar 14, 2017 at 0:19
  • 4
    $\begingroup$ @Sentinel135: if it has infinitely many terms, it isn't a polynomial. $\endgroup$ Commented Mar 14, 2017 at 0:22
  • 1
    $\begingroup$ @Sentinel135: en.wikipedia.org/wiki/Formal_power_series $\endgroup$ Commented Mar 14, 2017 at 0:38
  • 2
    $\begingroup$ @Sentinel135 It's a matter of definitions: is the word "polynomial" defined as allowing infinite summations or not? And the answer is that the standard definition of the word "polynomial" does not allow infinite summations. The definitions given on Wikipedia agree with the five of us. Do you have a source which says that a polynomial can include an infinite summation? $\endgroup$ Commented Mar 14, 2017 at 4:38

One way (Off the top of my head) to resolve the problem of of $x^{-1}$ not being a polynomial and $0$ being one is that all polynomials are the result of integrating $0$ a finite number of times.

  • 1
    $\begingroup$ But wouldn't that also rule out 1? $\endgroup$
    – Travis
    Commented Mar 14, 2017 at 21:49
  • 1
    $\begingroup$ Or any number other than 0, for that matter? $\endgroup$
    – Travis
    Commented Mar 14, 2017 at 21:50
  • 3
    $\begingroup$ @Travis 1 you can get by integrating 0 once and setting the constant of integration to 1. This answer can be made perfectly rigorous, although essentially the same point can be made by defining polynomials as infinitely differentiable functions which become constantly zero after differentiating a finite number of times. $\endgroup$ Commented Mar 15, 2017 at 3:56

For finitely many variables $x_i$ comprising a vector $\mathbb{x}$, define $\mathbb{x}^\boldsymbol{\alpha}:=\prod_i x_i^{\alpha_i}$. Such an expression, multiplied by a constant called the coefficient, is a monomial of degree $\left| \alpha\right| :=\sum_i \alpha_i$.

A polynomial is a sum of finitely many monomials with non-zero coefficients. The zero polynomial is the case where the number of such monomials is zero. The polynomial's degree is the supremum of the monomials' degrees, so the zero polynomial has degree $-\infty$. Any non-zero polynomial has at least one monomial, and among these some monomial has maximum degree, and if there is exactly one of these its coefficient is the leading coefficient. It is customary to write a polynomial as a sum over monomials of degree at most its degree, so for non-zero polynomials in one variable a unique non-zero leading coefficient exists.


A small remark to the role of $x$ in the spirit of the answer of @EthanBolker and @CarlMummert.

A representation of $x$:

We already know according to the given answers a polynomial \begin{align*} a_0+a_1x+a_2x^2+\cdots a_nx^n \end{align*} can be represented by the coefficients $a_0,\ldots, a_n$ as tuple with infinite many elements \begin{align*} (a_0,a_1,a_2,\cdots,a_n,0,0,\cdots) \end{align*} whereby all but finitely many elements are zero.

Question: But, what about the role of $x$ and why can we add and multiply $x$ with polynomials in more or less the same way as we can add and multiply the coefficients (i.e. the elements of the ring)?

Let's consider elements of $\mathbb{R}$ as coefficients of a polynomial and let's take e.g. \begin{align*} p(x)=7+5x+8x^3 \end{align*} We can represent this polynomial as \begin{align*} (7,5,0,8,0,0,\ldots) \end{align*}

We now pick out the special element $(0,1,0,0,\ldots)$, denote it with $$x:=(0,1,0,0,\ldots)$$ and using the Cauchy-product $\sum_{k=0}^n a_kb_{n-k}$ in order to multiply these tuples we can write \begin{align*} (7,5,0,8,0,\ldots)&=(7,0,0,0,0,\ldots)+(0,5,0,0,0,\ldots)+(0,0,0,8,0,\ldots)\\ &=(7,0,0,0,0,\ldots)+(5,0,0,0,0,\ldots)\cdot \color{blue}{x}+(8,0,0,0,0,\ldots)\cdot \color{blue}{x^3}\tag{1} \end{align*}

The right-hand side of (1) shows that all elements $a\in\mathbb{R}$ can be represented as \begin{align*} (\color{blue}{a},0,0,0,\ldots) \end{align*} while the indeterminate $x$ has a specific representation \begin{align*} (0,\color{blue}{1},0,0,0,\ldots) \end{align*} which is zero at the first coordinate but one at the second contrary to all other elements of the ring. In fact $x$ is an element of an extension ring in which all elements of the ring can be embedded.

This element $x$, called indeterminate or transcendental element has the following three properties

  • $x\cdot 1=1\cdot x =x$

  • $ax=xa\qquad\qquad\qquad \text{for all } a\in\mathbb{R}$

  • $a_0+a_1x+\ldots+a_nx^n=0 \quad(a_i\in\mathbb{R}) \qquad\Longleftrightarrow\qquad a_i=0,i=0,1,\ldots,n$

These properties of $x$ are fundamental and enables customary calculation with polynomials.


Normally we define a polynomial such that it can be written as $\sum_{i=0}^n a_ix^i$ for some $a_i\in \mathbb R$ where $i,n\in \mathbb N$. This is the reason why $x^{-i}$ isn't a polynomial. though it can be treated as a composition between a function and a polynomial.

The other reason is that when you start dealing with $\sum^n_{i=0}\frac{a_i}{x^i}$ you start to lose properties that all polynomials share. Like for instance $P(x)$ doesn't exist for $x=0$.


You ask if $x^{-1}$ is a polynomial and other answers are saying it is not. That's okay, but... you should look up the term "Laurent polynomial."


What is a polynomial? How about the definition usually found in Pre-Calculus texts:

A polynomial of degree $n$ is any function of the form

$$ p(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \ldots + a_{1}x + a_{0},$$

where $n$ is a non-negative integer and the $a_{i}$ are real numbers for $i \in \{0, 1, \ldots, n \}$.


This definition presented in ncatlab.org is pretty helpful:

Let $R$ be a commutative ring. A polynomial with coefficients in $R$ is an element of a polynomial ring over $R$ . A polynomial ring over $R$ consists of a set $X$ whose elements are called “variables” or “indeterminates”, and a function $X\to R[X]$ to (the underlying set of) a commutative $R$ -algebra that is universal among such functions, so that $R[X]$ is the free commutative $R$ -algebra generated by $X$ ; a polynomial is then an element of the underlying set of $R[X]$ .

The link to the original article is here.


Polynomial is an object on some particular algebra $\mathbb{A}$ that can be created with addition and multiplication of elements of $\mathbb{A}$.

If the particular algebra is also a field $\mathbb{F}$, then we can have a nice form for the polynomial, e.g. $a_1 x a_2 x a_3=a_1 a_2 a_3 x^2$, where $a_i,x \in \mathbb{F}$.


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