If a polynomial function, written in descending order of the exponents, has integer coefficients, then any rational zero must be of the form ± p/ q, where p is a factor of the constant term and q is a factor of the leading coefficient.
Example 1
Find all the rational zeros of
f ( x) = 2 x 3 + 3 x 2 – 8 x + 3
According to the rational zero theorem, any rational zero must have a factor of 3 in the numerator and a factor of 2 in the denominator.
The possibilities of p/ q, in simplest form, are 
These values can be tested by using direct substitution or by using synthetic division and finding the remainder. Synthetic division is the better method because if a zero is found, the polynomial can be written in factored form and, if possible, can be factored further, using more traditional methods.
Example 2
Find rational zeros of f(x) = 2 x 3 + 3 x 2 – 8 x + 3 by using synthetic division.
Each line in the table that follows represents the “quotient line” of the synthetic division.
p/q
2
3
–8
3
| 1 | 2 | 5 | –3 | 0 | 1 is a zero.
| –1 | 2 | 1 | –9 | 12 |
| 
| 2 | 4 | –6 | 0 | 
is a zero.
| 
| 2 | 2 | –9 | 
|
| 3 | 2 | 9 | 19 | 60 |
| –3 | 2 | –3 | 1 | 0 | –3 is a zero.
| 
| 2 | 6 | 1 | 
|
| 
| 2 | 0 | –8 | 15 |
The zeros of f ( x) = 2 x 3 + 3 x 2 – 8 x + 3 are 1, 
, and –3. This means
f (1) = 0, 
, and f (–3) = 0
The zeros could have been found without doing so much synthetic division. From the first line of the chart, 1 is seen to be a zero. This allows f ( x) to be written in factored form using the synthetic division result.
f ( x) = 2 x 3 + 3 x 2 – 8 x + 3 = ( x – 1)(2 x 2 + 5 x – 3)
But 2 x 2 + 5 x – 3 can be further factored into (2 x – 1)( x + 3) using the more traditional methods of factoring.
2 x 2 + 5 x – 3 = ( x – 1)(2 x – 1)( x + 3)
From this completely factored form, the zeros are quickly recognized. Zeros will occur when 