Algebra II: Polynomials
The Rational Zeros Theorem
Roots of a Polynomial
A root or zero of a function is a number that, when plugged in for the variable, makes the function equal to zero. Thus, the roots of a polynomial P(x) are values of x such that P(x) = 0 .
The Rational Zeros Theorem
The Rational Zeros Theorem states:
If P(x) is a polynomial with integer coefficients and ifis a zero of P(x) ( P(
) = 0 ), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x) .
We can use the Rational Zeros Theorem to find all the rational zeros of a polynomial. Here are the steps:
- Arrange the polynomial in descending order
- Write down all the factors of the constant term. These are all the possible values of p .
- Write down all the factors of the leading coefficient. These are all the possible values of q .
- Write down all the possible values of
. Remember that since
factors can be negative,
and
-
must both be included.
Simplify each value and cross out any duplicates.
- Use synthetic division to determine the values of
for which
P(
) = 0
. These are all the rational roots of
P(x)
.
Example: Find all the rational zeros of
P(x) = x
3 -9x + 9 + 2x
4 -19x
2
.
- P(x) = 2x 4 + x 3 -19x 2 - 9x + 9
- Factors of constant term: ±1 , ±3 , ±9 .
- Factors of leading coefficient: ±1 , ±2 .
- Possible values of
:
±
,
±
,
±
,
±
,
±
,
±
.
These can be simplified to:
±1
,
±
,
±3
,
±
,
±9
,
±
.
- Use synthetic division:
, and
3
.
We can often use the rational zeros theorem to factor a polynomial. Using synthetic division, we can find one real root a and we can find the quotient when P(x) is divided by x - a . Next, we can use synthetic division to find one factor of the quotient. We can continue this process until the polynomial has been completely factored.
Example (as above): Factor
P(x) = 2x
4 + x
3 -19x
2 - 9x + 9
.
As seen from the second synthetic division above,
2x
4 + x
3 -19x
2 -9x + 9÷x + 1 = 2x
3 - x
2 - 18x + 9
. Thus,
P(x) = (x + 1)(2x
3 - x
2 - 18x + 9)
. The second term can be divided synthetically by
x + 3
to yield
2x
2 - 7x + 3
. Thus,
P(x) = (x + 1)(x + 3)(2x
2 - 7x + 3)
. The trinomial can then
be factored into
(x - 3)(2x - 1)
. Thus,
P(x) = (x + 1)(x + 3)(x - 3)(2x - 1)
. We can see that this solution is correct because the four rational roots
found above are zeros of our result.





