Roots of Polynomials: Integer, Rational, Multiplicity and Linear Factors
A root of a polynomial is a number r with p(r) = 0, and every root corresponds to a linear factor (x − r) of the polynomial. In brief: the rational root theorem says that any rational root p/q of a polynomial with integer coefficients has p dividing the constant term and q dividing the leading coefficient. For x3 − 2x2 − 5x + 6 the candidates are ±1, ±2, ±3, ±6, and the roots are 1, 3 and −2.
Finding all the roots of a polynomial means factoring it completely. The tools are the rational root test, the factor theorem and polynomial division, and the concept of multiplicity tells how often a root is repeated. This guide explains the search for integer and rational roots, how multiplicity is counted and how the number of roots relates to the degree.
How do you find integer roots?
If a polynomial with integer coefficients and leading coefficient 1 has an integer root, that root divides the constant term. The reason: for xn + … + a1x + a0 = 0, every term except a0 contains a factor x, so x must divide a0.
- List all (positive and negative) divisors of the constant term.
- Test each candidate using substitution or Horner’s scheme.
- For each root found, divide it out and continue with the quotient.
Example 1: x3 − 2x2 − 5x + 6
Divisors of 6: ±1, ±2, ±3, ±6. p(1) = 0, so 1 is a root. Dividing gives x2 − x − 6 = (x − 3)(x + 2). So the roots are 1, 3 and −2, and p(x) = (x − 1)(x − 3)(x + 2). The tests use the factor theorem.
How does the rational root test work?
For p(x) = anxn + … + a0 with integer coefficients, every rational root in lowest terms p/q satisfies: p divides a0 and q divides an. Candidates are all fractions ± (divisor of a0)/(divisor of an). The test gives candidates only, and each must be checked.
Example 2: 2x3 − 3x2 − 8x − 3
Divisors of the constant term −3: 1, 3. Divisors of the leading coefficient 2: 1, 2. Candidates: ±1, ±3, ±1/2, ±3/2. Test: p(−1) = −2 − 3 + 8 − 3 = 0 ✓. Divide by x + 1: the quotient is 2x2 − 5x − 3 = (2x + 1)(x − 3). Roots: −1, −1/2 and 3. So p(x) = (x + 1)(2x + 1)(x − 3).
Example 3: no rational roots
x3 − 2: candidates ±1, ±2. p(1) = −1, p(−1) = −3, p(2) = 6, p(−2) = −10. None is zero, so there are no rational roots. The only real root is ∛2, which is irrational.
What is the multiplicity of a root?
The multiplicity of a root r is the number of times (x − r) divides the polynomial. A root of multiplicity 1 is simple, of multiplicity 2 is a double root, and so on. A root has multiplicity at least 2 if p(r) = 0 and p′(r) = 0, where p′ is the derivative, a result from calculus.
Example 4
p(x) = x4 − 5x3 + 6x2 = x2(x2 − 5x + 6) = x2(x − 2)(x − 3). The root 0 has multiplicity 2, and 2 and 3 are simple. The total of multiplicities is 4, the degree.
Example 5: finding multiplicity by repeated division
Is 1 a root of x3 − 3x + 2? p(1) = 0. Divide: x2 + x − 2 = (x − 1)(x + 2). So p(x) = (x − 1)2(x + 2), and 1 is a double root.
How do roots give linear factors?
A polynomial of degree n with roots r1, …, rn (counted with multiplicity) can be written as p(x) = an(x − r1)(x − r2)…(x − rn). Over the real numbers, some factors can be irreducible quadratics such as x2 + 1. Over the complex numbers, every polynomial splits completely into linear factors, which is the fundamental theorem of algebra.
Relations between roots and coefficients are given by Vieta’s formulas. For a cubic x3 + bx2 + cx + d, the sum of the roots is −b, and the product is −d. In Example 1, the sum is 1 + 3 − 2 = 2 = −(−2) ✓ and the product is −6 = −6 ✓.
Candidates table
| Polynomial | Candidates | Roots found |
|---|---|---|
| x2 − 5x + 6 | ±1, ±2, ±3, ±6 | 2, 3 |
| x3 − 2x2 − 5x + 6 | ±1, ±2, ±3, ±6 | −2, 1, 3 |
| 2x3 − 3x2 − 8x − 3 | ±1, ±3, ±1/2, ±3/2 | −1, −1/2, 3 |
| x3 − 2 | ±1, ±2 | none rational |
| x3 − 3x + 2 | ±1, ±2 | 1 (double), −2 |
Where do these ideas come from?
Searching for roots is among the oldest tasks in algebra. The rational root theorem is often credited to the eighteenth century, and Descartes had stated rules about the signs of roots in the seventeenth. The fundamental theorem of algebra, first proved convincingly by Gauss in 1799, says that a polynomial of degree n has exactly n complex roots counted with multiplicity.
Where are they used today?
Roots of polynomials describe equilibrium points, natural frequencies of structures and break-even values. In control engineering, the stability of a system depends on the roots of its characteristic polynomial. Numerical software finds roots of high-degree polynomials with iterative methods, but the rational root test is the fastest way to find exact answers for small problems.
Which mistakes are common?
Forgetting negative candidates
Test both +r and −r for each divisor.
Using divisors of the wrong coefficient
Numerators come from the constant term and denominators from the leading coefficient.
Counting a double root once
A root that appears twice contributes twice to the total of degree n.
Assuming all roots are rational
The test finds only rational roots. After dividing out the roots found, the remaining quotient may have irrational or complex roots.