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.
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).
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):
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).
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:- There are no rational roots. This is possible. But teachers don't usually ask you to find roots when they can't be found.
- 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.
- 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.