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