I have a problem that asks me to find a polynomial $P(x)$ so that $P(3)$ is 9.

Now I can say with certainty that $P(x)$ can be $x^2$. This is a second degree polynomial.

But what about functions such as $\frac{1}{x^2+3}$, are these not polynomials? If not, why?

P.S. I seem to be having troubles with the math tags. Some help on that would be appreciated as well.

  • 1
    $\begingroup$ Use dollar signs either side of mathematics. For more information, see the guides here. $\endgroup$
    – Shaun
    Oct 31, 2013 at 21:44
  • $\begingroup$ You could use the polynomial $P(x) = 9$, or $P(x) = 9+(x-3)^{1001}$. $\endgroup$
    – copper.hat
    Oct 31, 2013 at 21:45
  • $\begingroup$ Polynomials in an indeterminate $x$ (normally) don't have terms of negative powers of $x$. They're defined that way. Functions do. $\endgroup$
    – Shaun
    Oct 31, 2013 at 21:46
  • $\begingroup$ Also $P(x)=3x$ will do, or $P(x)=4x-3$. There are infinitely many polynomials with the property, in any (positive) degree. $\endgroup$
    – egreg
    Oct 31, 2013 at 21:47
  • $\begingroup$ @Shaun Functions don't have to have negative powers at all. $\endgroup$ Oct 31, 2013 at 22:16

2 Answers 2


A polynomial (in one variable, over the real numbers) is by definition an expression of the form

$f(x) = a_0 + a_1x+a_2x^2 + \cdots + a_nx^n$

for some non-negative integer $n$ and real numbers $a_i,\ i=0,1,\dotsc, n$.

Note that the variable $x$ must have non-negative powers; things like $r(x) = \frac{1}{x^2+3}$ are, by definition, not polynomials in $x$ since they include negative powers of $x$. (Properly speaking, it contains a negative power of the polynomial $x^2-3$, and hence cannot be written in the same form as $f(x)$ above.)

Of course, the numerator and denominator of $r$ are both polynomials; expressions of this form $\frac{p}{q}$ where $p,q$ are polynomials are called rational functions.

  • $\begingroup$ Nice explanation! :) $\endgroup$
    – 1233dfv
    Oct 31, 2013 at 21:52
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    $\begingroup$ I don't really see a "negative power of $x$" in $1/(x^2+3)$. The only negative power is of $x^2+3$, not of $x$ itself. $\endgroup$ Oct 31, 2013 at 21:53

Polynomials are finite formal sums $\sum_{i = 0}^n a_i x^i$, where $a_i$ belongs to some ring (if you need some idea of what a ring can be, the real numbers are a ring, the integers are a ring, and the rationals are a ring). By definition, there are no negative powers of the indeterminate $x$ involved (likewise, there are no reciprocals of expressions involving $x$).

Technically, a polynomial is not a function, although you can consider it to be one if you take an element $r$ in the original ring and "plug it in" (let $x = r$ and compute). However, if you interpret a polynomial as a function in that manner, two different polynomials can give the same function. For example, consider the field $\Bbb Z/p\Bbb Z$ (i.e. integers modulo $p$). The polynomials $0$ and $x^p - x$ define the same function by Fermat's little theorem, but they are not the same polynomial.

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    $\begingroup$ While all of this is fine, the technical nature of this answer I fear is much too advanced for the level of question. $\endgroup$ Oct 31, 2013 at 22:21
  • $\begingroup$ Annoying precision got the better of me once again! :P $\endgroup$
    – Stahl
    Oct 31, 2013 at 23:03
  • $\begingroup$ To add to this, $\endgroup$
    – Dylan Yott
    Oct 31, 2013 at 23:39

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