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 