Rational Zero Theorem

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

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