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I recently took an online test for my math class and was asked the question "What type of polynomial is $\frac{x}{x+5}$." The supposed correct answer was "not a polynomial" because there is a variable in the denominator. However, can't this expression be simplified into $\frac 15$, therefore making it a polynomial? I appreciate any answers, and I'm not particularly good at math so if this has an obvious answer, I apologize.

Behrouz Maleki
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Skyaots
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  • You can't cancel like that. e.g., ${2\over 2+3}\ne{1\over 1+3}$. – David Mitra Jul 12 '16 at 21:34
  • Why do you choose to simplify it to $\frac 15$, but not to $\frac 05$? – peterwhy Jul 12 '16 at 21:37
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    If you meant to write $\frac{x}{5x}$ instead of $\frac{x}{x+5}$, then it differs between contexts whether you should consider it to be the same as the constant polynomial $\frac15$; see for example https://math.stackexchange.com/questions/1738278/is-fracx2xx1-a-polynomial – hmakholm left over Monica Jul 12 '16 at 21:40
  • "However, can't this expression be simplified into 1/5" No. It absolutely can not be simplified to that at all. if x = 0 then 0/5 ne 1/5. If x = 1 then 1/6 ne 1/5, if x = 2 then 2/7 ne 1/5 if x = 3 then 3/8 ne 1/5, etc. etc. etc. – fleablood Jul 12 '16 at 22:05
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    In short, it's a rational function, not a polynomial. It's not a polynomial because it cannot be simplified any further. Just like a fraction is a rational number and not an integer when in fully reduced form. – scott Jul 13 '16 at 22:25
  • The easiest way to show that this is not a polynomial is that the limit for $x$ tending to $\pm \infty$ is $1$ which is impossible for a polynomial , unless for the constant polynomial $1$ , which we obviously do not have either. – Peter Apr 14 '21 at 16:13

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This expression can't be simplified to $\frac{1}{5}$ because there is no common factor in the numerator and denominator. This would actually be called an "Improper rational function". If we let $P(x)$ and $Q(x)$ be two polynomials then an improper rational function is a function $f(x) = \frac{P(x)}{Q(x)}$ such that $\operatorname{deg}(P) \geq \operatorname{deg(Q)}.$ In your case, $x$ has degree equal to $x + 5$, so this is an improper rational function.

Hope this helps.

Edward Evans
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Polynomials can be defined as expressions of the form

$$ a_n x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_{2}x^{2} + a_1 x + a_0 $$

where $a_0,a_1,a_2,\ldots,a_n$ are constants, and $x$ is a variable (raised to a different non-negative integer power in each term). So, for example, if $a_3 = 4$, $a_2 =1$, $a_1=0$, and $a_0=-7$, then the corresponding polynomial would be $4x^3+x^2-7$. No matter what you do, $\frac{x}{x+5}$ cannot be written in this form, since it will involve negative exponents (recall that $x^{-1} = \frac1x$).

Another way that you can think about it is in terms of continuity. To put it simply, function is continuous if it can be drawn as a single, unbroken curve. For example, our polynomial function from before, $f(x)=4x^3+x^2-7$, is continuous. In fact, it has been proven that all polynomials are continuous. So, to figure out if $g(x)=\frac{x}{x+5}$ is a polynomial, we can consider its continuity; if it is not a continuous function, then it cannot be a polynomial. Plugging in $-5$, we see that $g$ is not continuous at all points, since

$$g(-5)=\frac{5}{-5+5}=\frac50,$$

and $5/0$ is undefined, so $g$ cannot be a polynomial. (Note that this method does not work in the reverse—"if a function is not a polynomial, then it is not continuous"—because, for example, $h(x)=\sin (x)$ is not a polynomial, but is continuous.)

DooplissForce
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You can also use the notion of derivative, in order to show that $f(x) = \frac{x}{x+5}$ is not a polynomial.

Indeed, if $P(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+ \ldots+a_{1}x+a_{0}$ is a polynomial of $\deg(P)=n$, then by differentiating $n+1$ times we obtain $$\frac{d^{n+1}}{d x^{n+1}} P(x) = 0$$

And note that $f^{(n)}(x)$ is not identically zero for any $n$.

hyperkahler
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