Question 238361
The possible rational roots are all the rational numbers, positive and negative, which can be formed using a factor of the constant term (at the end) over a factor of the leading coefficient.<br>
This function's constant term is -5 and its leading coefficient if 3. These are both prime numbers so the number of possible rational roots is small: 1/1, 5/1, 1/3, and 5/3 (both positive and negative).<br>
Synthetic Division is a relatively quick and easy way to determine which, if any of these possible roots actually are roots. Here's how to test to see if 1/1 (aka 1):
<pre>
1 |   3   -8   -5   16   -5
---        3   -5  -10    6
     -----------------------
      3   -5  -10    6    1
Since the remainder (the "1" in the lower right above) is not 0, 1 is not a root for f(x).
</pre>
There is a lot of trial and error in this process. (Don't forget to try the possible negative rational roots, even if both the constant term and the leading coefficient are positive.) I have tried the possible rational roots and I was not able to find any. This means one of the following:<ul><li>There are no rational roots. This is possible. But teachers don't usually ask you to find roots when they can't be found.</li><li>There is an error in the function as you've written it. Please double check the problem and, if there was an error, resubmit a corrected function.</li><li>I have made a calculation error when I tried the various possible roots. This is possible, even though I checked my work. So if the function you have given here is correct, I would go ahead and try to find them on your own.</li></ul>