<PRE><B>Question 709227</B>
When a polynomial (this is not a rational function by the way) is written in standard form with the terms listed from highest degree to lowest (like yours), the possible rational roots are all the ratios, positive and negative, that can be formed using a factor of the constant term (at the end) in the numerator over a factor of the leading coefficient in the denominator.&lt;br&gt;
Your constant term is 33. Its factors are 1, 3, 11 and 33.&lt;br&gt;
Your leading coefficient is 9. Its factors are 1, 3 and 9.&lt;br&gt;
The possible rational roots of this polynomial are all the possible ratios, positive and negative, we can form using a factor of 33 (1, 3, 11 or 33) on top over a factor of 9 (1, 3 or 9):
&lt;u&gt;+&lt;/u&gt;1/1, &lt;u&gt;+&lt;/u&gt;3/1, &lt;u&gt;+&lt;/u&gt;11/1, &lt;u&gt;+&lt;/u&gt;33/1,
&lt;u&gt;+&lt;/u&gt;1/3, &lt;u&gt;+&lt;/u&gt;3/3, &lt;u&gt;+&lt;/u&gt;11/3, &lt;u&gt;+&lt;/u&gt;33/3, 
&lt;u&gt;+&lt;/u&gt;1/9, &lt;u&gt;+&lt;/u&gt;3/9, &lt;u&gt;+&lt;/u&gt;11/9, &lt;u&gt;+&lt;/u&gt;33/9
which reduce to:
&lt;u&gt;+&lt;/u&gt;1, &lt;u&gt;+&lt;/u&gt;3, &lt;u&gt;+&lt;/u&gt;11, &lt;u&gt;+&lt;/u&gt;33,
&lt;u&gt;+&lt;/u&gt;1/3, &lt;u&gt;+&lt;/u&gt;1, &lt;u&gt;+&lt;/u&gt;11/3, &lt;u&gt;+&lt;/u&gt;11, 
&lt;u&gt;+&lt;/u&gt;1/9, &lt;u&gt;+&lt;/u&gt;1/3, &lt;u&gt;+&lt;/u&gt;11/9, &lt;u&gt;+&lt;/u&gt;11/3
Removing the duplicates:
&lt;u&gt;+&lt;/u&gt;1, &lt;u&gt;+&lt;/u&gt;3, &lt;u&gt;+&lt;/u&gt;11, &lt;u&gt;+&lt;/u&gt;33, &lt;u&gt;+&lt;/u&gt;1/3, &lt;u&gt;+&lt;/u&gt;11/3, &lt;u&gt;+&lt;/u&gt;1/9, &lt;u&gt;+&lt;/u&gt;11/9&lt;br&gt;
So there are 16 possible rational roots to f(x).</PRE>