Last updated on July 4th, 2025
The rational root theorem is also known as rational zero theorem, and rational zero test. This theorem is applied to polynomial equations for finding their rational roots. However, note that not all polynomials have rational zeroes.
A number that can be written as a fraction of two integers where the denominator is not zero is rational. Fractions, decimals, and positive and negative whole numbers can all be rational numbers. Some examples of rational numbers are 2, 0.28, 7/25, and -3.5.
The rational root theorem states: for the root of a polynomial to be a rational number, the numerator and denominator must be factors of the constant term and leading coefficient, respectively. The leading coefficient is the coefficient of the term that has the highest power of the variable.
For a polynomial:
P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀,
aₙ denotes the leading coefficient and a₀ denotes the constant term,
According to the rational root theorem:
If p/q (in lowest terms) is a rational root of the polynomial, then:
p must be a factor (divisor) of the constant term a₀, and
q must be a factor (divisor) of the leading coefficient aₙ.
Any possible rational solution to the polynomial must be formed by dividing a factor of the constant term by a factor of the leading coefficient.
Therefore, possible rational roots of p/q are of the form ( factors of a0 ) / (factors of an).
There are two key conditions established by the rational root theorem to find rational zeros:
1. Leading Coefficient Condition: An equation’s leading coefficient should be divisible by the denominator of the rational solution. If p/q is a polynomial’s root, then the denominator must divide the leading coefficient equally. Let’s say 8x2 is the leading coefficient. In this case, the denominator q must divide 8.
2. Constant Term Condition: The constant term must be divisible by the numerator of the fraction.The numerator p must divide the constant term, which is the number at the end (with no variable).
For example, if 6 is a constant, then it must be divisible by p.
The rational root theorem is used to narrow down rational solutions to polynomial equations.
Statement: A rational number p/q is the root of a polynomial P(x) = anxn + an−1xn−1 + ⋯ + a1x + a0,
where all coefficients ai are integers, then:
p (numerator) divides the constant term a0
q(denominator) divides the leading coefficient an
Proof: Assume p/q as a rational zero of a given polynomial P(x), and that 𝑞 ≠ 0
Then, P(p/q) = 0
Substituting p/q into the polynomial, we get:
an(p/q)n + an-1(p/q)n-1 + … + a1(p/q) + a0,= 0
Now, to eliminate denominators, we should multiply by qn in both sides:
Anpn + an−1pn−1q + an−2pn−2 q2 + ⋯ + a1pqn−1 + a0qn = 0
Consider this as equation 1
Next, we should subtract a0qn from both sides of equation 1 to prove that p is a factor of a0
an(p)n + an-1(p)n-1(q)n + … + a2(p)2(q)n-1 + a1(p)(q)n-1 = -a0qn
Consider this as equation 2
p is a factor of every term on the left side. It is given that the left-hand side is equal to the right-hand side, so p is also a factor of right and side. Given that p and q have no factor in common, q will be a factor of a0,
Thus, p is a factor of a0,
Similarly, to prove q is a factor of an, we should subtract anpn from both sides of equation 1
an-1(p)n-1(q) + an-2(p)n-2(q)2 + … + a0qn = -an(p)n
q is a factor of every term on the left side. It is given that the left-hand side is equal to the right-hand side, so q is also a factor of the right-hand side. Given that p and q have no factor in common, p will be a factor of an,
So, it is proved that p is a factor of an,
The first step is to write the polynomial in standard form. Then, we can identify the leading coefficient and the constant term. Let’s take an example and explore it step-by-step:
P(x) = 2x3 − 3x2 − 8x + 3
From the above-mentioned polynomial, we know that the constant term (a0) is 3. We also know that the leading coefficient (an) is 2.
We should now write down all factors of both the constant and the leading coefficient. In this case, it is 3 and 2 respectively.
Factors of 3 are ±1, ±3
Factors of 2 are ±1,±2
The next step involves the formation of all possible rational roots (±p/q)
All combinations of p/q are 1/1,3/1,1/2,3/2.
These are the potential rational roots.
Next, we need to test each value by substituting it into the polynomial. If the result is zero, the number is a rational root.
Try x = 1,
P(1)= 2(1)3 - 3(1)2 - 8(1) + 3 = 2 - 3 - 8 + 3 = -6. Thus x=1 is not a root.
Keep testing until you find values that make P(x)=0
Once a root is found, use polynomial division to divide the polynomial by the corresponding linear factor. Doing so simplifies the polynomial.The process is repeated on the resulting polynomial until all rational zeros are found, or until it can be solved by other methods like the quadratic formula. If no rational roots are found for a given polynomial, then it means that it can have irrational or complex roots.
The rational root theorem might seem purely theoretical, but it has practical applications in various fields. Some of them are mentioned below:
1. Engineering & Physics
When solving polynomial equations that model physical systems, such as electrical circuits or mechanical vibrations, the theorem helps identify possible rational solutions efficiently.
2. Computer Science & Cryptography
In fields like algorithm design and encryption, polynomials play a crucial role in structuring and solving complex problems. The theorem aids in identifying possible rational solutions quickly.
3. Economics & Finance
Polynomial equations appear in financial models, such as calculating compound interest. Sometimes, these models use polynomials with rational coefficients; in such cases, the rational root theorem can identify solution values that are easier to verify.
4. Signal Processing
In digital signal processing, polynomials are used to design filters and analyze waveforms. Whenever polynomial equations represent these systems, the theorem is used to identify possible rational roots.
5. Structural Design & Architecture
Engineers use polynomial equations to model stress distributions and load-bearing calculations in structures. The theorem helps in identifying rational solutions for stability analysis.
The rational root theorem is an integral part of solving for rational roots. Students may overlook some details while applying the theorem. Here are some common mistakes to keep in mind and how to avoid them:
Find the rational roots of f(x) = x cube - 4x square + x + 6
Rational roots are x = 2, 3, -1
Step 1: Use the rational root theorem to write down all possible rational roots
The Rational Root Theorem states:
Possible rational roots = ± (factors of constant term) / (factors of leading coefficient)
Possible rational roots are ±1, ±2, ±3, ±6
Step 2: Try to find rational roots with the help of synthetic division or substitution method
Test x=1
f(1) = 13- 4(1)2+1+6=1-4+6=4 . Not a root.
Test x=2
f(2) = 8 − 16 + 2 + 6 = 0.Yes, x = 2 is a root.
Use synthetic division to factor:
2 | 1 -4 1 6
| 2 -4 -6
------------------
1 -2 -3 0
So, f(x) = (x − 2) ( x2 − 2x− 3)
Factor x2 − 2x − 3 = (x − 3) ( x + 1)
Find the rational roots of f(x) = 2x cube + 3x square -2x - 3
Rational roots are x=1, -3/2, -1
Step 1: Rational Root Candidates
Possible rational roots are ±1 , ±3, ±1/2, ±3/2
Step 2: Try Rational Roots
Try x=1
f(1 )= 2 + 3 − 2 − 3 = 0 Yes, root found.
1 | 2 3 -2 -3
| 2 5 3
-----------------
2 5 3 0
So, f(x) = (x − 1) (2x2 + 5x + 3)
Factors of 2x2 + 5x + 3 are (2x + 3) (x + 1)
Find the rational roots of f(x) =x3 - 4x2 + 5x - 2
Rational roots are x = 1, 1, 2. x = 1 has multiplicity 2, meaning that it occurs twice in the factored form of the polynomial.
Step 1: Possible values of p/q
a0=−2a → factors: ±1, ±2
an=1a → factors: ±1
⇒ Possible roots: ±1, ±2
Step 2: Try these values:
f(1) =1 − 4 + 5 − 2 = 0
. It is a root.
Use synthetic division to factor:
(x3 − 4x2 + 5x − 2) ÷ (x − 1) = x2 − 3x + 2
Factor further:
x2 − 3x + 2 = (x − 1)(x − 2)
Find rational roots of f(x) = 2x cube + 3x square - 8x - 3
This polynomial equation has no rational roots.
Step 1:
a0=−3: ±1, ±3
an=2: ±1, ±2
⇒ potential rational roots: ±1, ±3, ±1/2, ±3/2
Step 2: Try values:
f(1)= 2 + 3 − 8 − 3 = −6, Not a root.
f(3) = 2(27) + 3(9) − 8(3) − 3 = 54 + 27 − 24 − 3 = 54 Not a root.
Try f(−1): -2 + 3 + 8 - 3 = 6. Not a root
Try f(-3) = -54 + 27 + 24 - 3 = -6. Not a root
Try f(1/2) = 2(1/8) + 3(1/4) - 8(1/2) - 3 = 1/4 + 3/4 - 4 - 3 = 1 - 7 = -6
Try f(-1/2) = -1/4 + 3/4 + 4 - 3 = 1.5
None of the rational roots work.
Find rational roots of f(x)= x3 + x2 - 4x - 4
Rational roots are x = -1, 2, -2
Step 1:
a0=−4: ±1, ±2, ±4
an= 1: ±1
⇒ Possible roots: ±1, ±2, ±4
Step 2: Try:
f(1) = 1 + 1 − 4 − 4 = −6 . Root not found.
f(−1) = −1 + 1 + 4 − 4 = 0 . This is the root.
Divide by x+1:
x3 + x2 − 4x − 4 = ( x + 1 ) ( x2 − 4 )
Further factor:
x2 − 4 = ( x − 2 ) ( x + 2 )
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.