Last updated on July 16th, 2025
Descartes' rule of signs is an algebraic theorem that estimates the maximum number of positive and negative real roots of a polynomial equation. Over time, the rule has been extended to algorithms like Budan’s theorem and real-root isolation.
Descartes' rule of signs helps determine the number of positive and negative real roots by counting sign changes in the terms of the polynomial. It states that,
f(x) = anxn + an-1xn-1 + . . . + a1x + a0
Here, the single variable polynomial is written in standard form.
The rules applied are:
The number of positive real roots is equal to the sign changes in f(x) or less than that by an even number. For example, if there are 5 sign changes in f(x), then the number of positive roots will be:
5 in case all roots are real and distinct,
3 (5 - 2), or
1 (5 - 4)
To find the number of negative real roots, substitute -x into the polynomial and count the sign changes.
Descartes' rule of signs can be applied using the following steps:
Arrange the polynomials in standard form, with exponents, in descending order.
Count sign changes in f(x) to estimate the maximum number of real roots.
Substitute f(x) with f(-x) and count sign changes to estimate the maximum number of negative real roots.
To find the number of complex/Imaginary roots, subtract total real roots (positive and negative) from the degree of the polynomial. Complex roots = degree of polynomial - (no. of positive roots + no. of negative roots)
Let’s apply these steps to an example.
Question: Apply Descartes' rule of signs to estimate the number of positive, negative, and imaginary roots of the given polynomial: f(x) = 2x4 - 3x3 - 5x2 + 9x - 4
Step 1: The polynomial is already in standard form, so we will write it as is.
f(x) = 2x4 - 3x3 - 5x2 + 9x - 4
Step 2: Counting sign changes in f(x)
The signs of the coefficients are +, -, -, +, -
Sign changes:
- (1 change)
- - (0 change)
- + (1 change)
+ - (1 change)
There are 3 sign changes in f(x), so the total number of positive real roots is either 3 or 1, depending on the number of complex roots.
Step 3: Substitute -x in f(x) to find negative real roots
f(-x) = 2(-x)4 - 3(-x)3 - 5(-x)2 + 9(-x) - 4
Upon simplification, we get
2x4 + 3x3 - 5x2 - 9x - 4
Now counting sign changes:
+ + (0 changes)
+ - (1 change)
- - (0 changes)
- - (0 changes)
The number of negative real roots is equal to the number of sign changes, which is 1.
Step 4: For finding the number of imaginary roots:
The degree of the polynomial is 4, so the equation will have a total of 4 roots, including real, complex, and their multiplicity.
Complex roots = degree of polynomial - (no. of positive roots + no. of negative roots)
So, there can be two cases:
Case 1: Number of positive real roots = 3
Number of negative real roots = 1
Number of complex roots = 4 - (3 + 1) = 0
Or,
Case 2: Number of positive real roots = 1
Number of negative real roots = 1
Number of complex roots = 4 - (1 + 1) = 4 - 2 = 2
So,
Number of positive real roots = 3 or 1
Number of negative real roots = 1
Number of complex roots = 0 or 2
While Descartes' rule of signs does not give us the exact number of roots, we can create a chart with the possible number of real roots. A few things to keep in mind while constructing this chart are:
Positive Real Roots |
Negative Real Roots |
Complex (Imaginary) Roots |
Total Roots |
3 | 1 | 0 | 4 |
1 | 1 | 2 | 4 |
There are two generalizations of Descartes' rule of signs:
Non-real roots
According to the fundamental theorem of algebra, any polynomial of degree n has exactly n roots, whether real or complex.
Non-real roots = n - (positive real roots + negative real roots)
Fewnomial Theory
Khovanski’s Fewnomial theory extends Descartes' rule of signs to include functions involving transcendental terms like exponentials and logarithms, etc. It suggests that the number of real roots depends more on the number of terms than on the degree of the expression. So, according to the Fewnomial theory, even if an equation has a higher degree, it can have fewer real roots depending on the number of terms present in the equation.
Descartes' rule is extended using the method of linear fractional transformation. This idea forms the basis of Budan’s theorem and Budan-Fourier theorem.
These theorems allow us to estimate the number of real roots present within any specified interval. These extensions are widely used in fast computer algorithms for root calculations. Real-root isolation locates each root of a polynomial within a separate interval on the number line. It ensures that each interval contains exactly one real root and does not overlap with intervals of other roots.
Descartes' rule of signs is used to estimate the number of positive and negative real roots of a polynomial equation. Here are some real-life applications of the rule:
Given below is a summary of some common errors while applying Descartes’ rule of signs, along with their solutions for reference.
Find positive and negative real roots for the given polynomial: f(x) = x3 − 6x2 + 11x − 6
Possible positive real roots: 3, 1. There are no negative roots.
Positive roots: Sign changes: +x3 → −6x2 → +11x → −6
→ 3 sign changes, possible positive real roots can be: 3, 1
Negative roots: Substitute f(−x) = (−x)3 − 6(−x)2 + 11(-x) - 6
= -x3 - 6x2 - 11x - 6
→ No sign changes, so there are zero negative real roots
Apply Descartes' rule of signs to estimate positive and negative real roots of the polynomial f(x) = x4 + x3. - x -1
Positive real roots: 1 or 0, negative real roots: 3 or 1
Positive roots: sign changes: +x4 → + x3 → −x → −1
→ 1 sign change
Negative roots: f(−x) = x4 − x3 + x − 1
Sign changes: +x4 → −x3 → +x → −1
→ 3 sign changes
Determine the possible number of positive and negative real roots for the polynomial f(x) = x5 − 4x4 + 6x3 − 4x2 + x - 2 using Descartes' rule of signs
Positive real roots: 4, 2, or 0. Negative real roots: 0
Positive real roots;
f(x) = +x5, -4x4, +6x3, -4x2, +x, -2
Signs are +, -, +, -, +, -
There are 5 sign changes, so the possible number of real roots is 5, 3, or 1.
Negative real roots;
f(-x) = (−x)5 − 4(−x)4 + 6(−x)3 − 4(−x)2 + (−x) − 2
= =−x5 − 4x4 − 6x3 − 4x2 − x − 2
There are no sign changes, so there are 0 negative real roots.
Use Descartes' rule to find the maximum number of positive and negative real roots for the polynomial f(x) = x3 + x2 + x +1
There are zero positive roots and 3 or 1 negative roots.
Positive roots: All coefficients are positive, no changes in signs
Negative roots: f(−x) = −x3 + x2 − x + 1
Sign changes: −x3 → +x2 → −x → +1
→ 3 sign changes
Apply Descartes' rule of signs to the polynomial f(x) =x4 −5x2 + 4
positive roots
Count the sign changes in f(x)
Write the polynomial in standard form: f(x) = x4 + 0x3 - 5x2 + 0x + 4
Signs of coefficients are +, 0, -, 0, +
There are a total of 2 sign changes, f(x) = 2
By Descartes' rule of signs, this means that the number of positive real roots is either 2 or 0.
Count sign changes in f(-x)
f(-x) = (-x)4 - 5(-x)2 + 4 = x4 - 5x2 + 4 = f(x)
So f(-x) = f(x), which means the polynomial is even.
There are 0 sign changes in f(-x)
So, by Descartes' rule, the number of negative roots is zero.
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.
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