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251 LearnersLast updated on October 21, 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 roots.
A rational number is a number that can be expressed as a fraction of two integers, with a non-zero denominator. Fractions, decimals, and integers (positive or negative) are all rational numbers. Some examples of rational numbers are 2, 0.28, \(\frac{7}{25} \), 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:
\(P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \),
\(a_n\) denotes the leading coefficient and a0 denotes the constant term,
According to the rational root theorem:
If \(\frac{p}{q} \) (in the lowest terms) is a rational root of the polynomial, then:
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 \(\frac{p}{q} \) are of the form \(\frac{\text{factors of } a_0}{\text{factors of } a_n} \).
An example of how the rational root theorem will look like is given as,
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 \(\frac{p}{q} \) is a polynomial’s root, then the denominator must divide the leading coefficient equally.
For example, if the leading coefficient is 8, then the denominator q must be a factor of 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 \(\frac{p}{q} \) is the root of a polynomial \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \),
where all coefficients "\(a_i\)" are integers, then:
Proof: Assume \(\frac{p}{q} \) as a rational zero of a given polynomial P(x), and that 𝑞 ≠ 0
Then, P(\(\frac{p}{q} \)) = 0
Substituting \(\frac{p}{q} \) into the polynomial, we get:
\(a_n\left(\frac{p}{q}\right)^n + a_{n-1}\left(\frac{p}{q}\right)^{n-1} + \cdots + a_1\left(\frac{p}{q}\right) + a_0 = 0\)
Now, to eliminate denominators, we should multiply by \(q_n\) in both sides:
\(a_n p^n + a_{n-1} p^{n-1} q + a_{n-2} p^{n-2} q^2 + \cdots + a_1 p q^{n-1} + a_0 q^n = 0 \)
Consider this as equation 1
Next, we should subtract \(a_0q^r\) from both sides of equation 1, to prove that p is a factor of a0
\(a_n p^n + a_{n-1} p^{\,n-1} q + \dots + a_2 p^2 q^{\,n-2} + a_1 p q^{\,n-1} = -a_0 q^n \)
Consider this as equation 2
Since p and q are co-prime, p divides \(a_0\), and q divides \(a_n\)
Thus, p is a factor of \(a_0\),
Similarly, to show that q divides \(a_n\), subtract \(a^np^n\) from both sides of equation 1
\(a_{n-1} (p)^{n-1} (q) + a_{n-2} (p)^{n-2} (q)^2 + \dots + a_0 q^n = -a_n (p)^n \)
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 \(a_n\)
So, it is proved that p is a factor of \(a_n\).
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:
\(P(x) = 2x^3 - 3x^2 - 8x + 3 \)
In this polynomial, the constant term \(a_0\) is 3 and the leading coefficient \(a_n\) 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 (\(\pm \frac{p}{q} \))
All combinations of \(\frac{p}{q} \) are \(\frac{1}{1}, \frac{3}{1}, \frac{1}{2}, \frac{3}{2} \).
These are the potential rational roots.
Next, substitute each candidate value into the polynomial to test if it is a 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 factor it out from the polynomial. 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.
Here are some of the tips and tricks to master rational root theorem and its applications:
Start with simple numbers first. Usually, roots like 1,−1,2,−2 are more likely than complex fractions. Test small integers first; it saves time.
Use synthetic division. Once you have a list of possible roots, pick a candidate (say x = 2). Use synthetic division to see if it gives a remainder of 0. If remainder = 0 → root found!. Factor the polynomial and repeat for the smaller polynomial. Synthetic division is faster than substituting directly into the polynomial.
Look for patterns. Check for common factors in all terms (factor out first!). Check for sign patterns (Descartes’ Rule of Signs can give hints). This reduces the number of candidates to test.
Combine with factoring tricks. If a polynomial looks like a difference of squares or sum/difference of cubes, factor it first. This might reveal integer roots immediately before testing rational ones.
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:
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.
Find the rational roots of f(x) = x³ - 4x² + 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
\(f(1) = 1^3 - 4(1)^2 + 1 + 6 = 1 - 4 + 6 = 4\). 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, \(f(x) = (x − 2) (x^2 − 2x− 3)\)
Factor \(x^2 − 2x − 3 = (x − 3) (x + 1)\)
Find the rational roots of f(x) = 2x³ + 3x² -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, a root is found.
1 | 2 3 -2 -3
| 2 5 3
-----------------
2 5 3 0
So, \(f(x) = (x − 1) (2x^2 + 5x + 3)\)
Factors of \(2x^2 + 5x + 3\) 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 is a root of multiplicity 2, occurring twice in the factored form of the polynomial.
Step 1: Possible values of \(\frac{p}{q} \)
\(a_0 = −2a\) → factors: \(±1, ±2\)
\(a_n = 1a\) → factors: \(±1\)
⇒ Possible roots: \(±1, ±2\)
Step 2: Try these values:
\(f(1) = 1 − 4 + 5 − 2 = 0.\)
This confirms that it is a root
Use synthetic division to factor:
\((x^3 − 4x^2 + 5x − 2) ÷ (x − 1) = x^2 − 3^x + 2\)
Factor further:
\(x^2 − 3x + 2 = (x − 1) (x − 2)\)
Find rational roots of f(x) = 2x³ + 3x² - 8x - 3
This polynomial equation has no rational roots.
Step 1:
\(a_0 = −3: ±1, ±3\)
\(a_n = 2: ±1, ±2\)
⇒ 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
Let us try \(f\left(\frac{1}{2}\right) \)
\(f\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) + 3\left(\frac{1}{4}\right) - 8\left(\frac{1}{2}\right) - 3\\ = \frac{1}{4} + \frac{3}{4} - 4 - 3\\ = 1 - 7\\ = -6 \)
\(f\left(-\frac{1}{2}\right) = -\frac{1}{4} + \frac{3}{4} + 4 - 3 = 1.5 \)
None of the candidate rational roots satisfy the polynomial.
Find rational roots of f(x)= x³ + x² - 4x - 4
Rational roots are x = -1, 2, -2
Step 1:
\(a_0 = -4: \pm 1, \pm 2, \pm 4 \)
\(a_n = 1: \pm 1 \)
⇒ Possible roots: \(\pm 1, \pm 2, \pm 4 \)
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:
\(x^3 + x^2 - 4x - 4 = (x + 1)(x^2 - 4) \)
Further factor:
\(x^2 - 4 = (x - 2)(x + 2) \)
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.






