# Polynomials and rational functions

• Jul 12th 2010, 03:25 AM
Glitch
Polynomials and rational functions
The question asks "State whether the function is a polynomial, a rational function (but not a polynomial), or neither."

This is the function:

$F(x) = \frac{x^3 - 3x^{3/2} + 2x}{x^2 - 1}$

It looks rational, but apparently the answer is 'neither'. Why is this the case? Thanks.
• Jul 12th 2010, 03:31 AM
running-gag
Hi

This is due to the exponent 3/2
To get a rational function you must have only integer exponents
• Jul 12th 2010, 03:32 AM
Glitch
Ahh, thanks!
• Jul 12th 2010, 03:33 AM
Glitch
Oh, also, does a polynomial have to have to have integer exponents?
• Jul 12th 2010, 03:37 AM
running-gag
Yes
• Jul 12th 2010, 03:50 AM
running-gag
Some examples

$F(x) = x^3 - 3x^{3/2} + 2x$ is NOT a polynomial

$F(x) = x^3 - 3x^2 + 2x$ is a polynomial

$F(x) = \frac{x^3 - 3x^{3/2} + 2x}{x^2 - 1}$ is NOT a rational function

$F(x) = \frac{x^3 - 3x^2 + 2x}{x^2 - 1}$ is a rational function

Now a more tricky example

$F(x) = \frac{x^3 - 3x^2 + 2x}{x - 1}$ is a rational function

But since $x^3 - 3x^2 + 2x$ is equal to 0 when x=1, you can factor out x-1

$x^3 - 3x^2 + 2x = (x-1)(x^2 - 2x)$

Therefore $F(x) = \frac{x^3 - 3x^2 + 2x}{x - 1} = \frac{(x-1)(x^2 - 2x)}{x - 1} = x^2-2x$ which is a polynomial

Actually $\frac{x^3 - 3x^2 + 2x}{x - 1}$ and $x^2-2x$ are not totally equal since the first one is defined for all real except 1 whereas the second one is defined for all real

The 2 functions are equal except on x=1