College Algebra

Descartes' Rule of Signs

Last updated: March 2026 · Advanced

Before you start

You should be comfortable with:

The Rational Root Theorem tells you which rational numbers could be zeros. Descartes’ Rule of Signs tells you how many positive and negative real zeros are possible — before you test a single candidate. This information narrows your search dramatically.

The Rule for Positive Real Zeros

Given a polynomial $f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ with real coefficients:

The number of positive real zeros is either equal to the number of sign changes in $f(x)$ , or less than that by an even number.

A sign change occurs when consecutive nonzero coefficients (in standard form, written in descending order of degree) have opposite signs. Zero coefficients are skipped.

Worked Example 1: Counting Sign Changes in $f(x)$

Count the sign changes in $f(x) = 2x^5 - 3x^4 + x^3 + 4x^2 - x - 6$ .

Write the signs of the coefficients in order:

$+, -, +, +, -, -$

Count sign changes:

$+$ to $-$ : change (1)
$-$ to $+$ : change (2)
$+$ to $+$ : no change
$+$ to $-$ : change (3)
$-$ to $-$ : no change

There are 3 sign changes. By Descartes’ Rule, the number of positive real zeros is $3$ or $3 - 2 = 1$ .

The Rule for Negative Real Zeros

To find the possible number of negative real zeros, apply the same rule to $f(-x)$ .

The number of negative real zeros is either equal to the number of sign changes in $f(-x)$ , or less than that by an even number.

Worked Example 2: Counting Sign Changes in $f(-x)$

Using the same $f(x) = 2x^5 - 3x^4 + x^3 + 4x^2 - x - 6$ , compute $f(-x)$ :

$f(-x) = 2(-x)^5 - 3(-x)^4 + (-x)^3 + 4(-x)^2 - (-x) - 6$

$= -2x^5 - 3x^4 - x^3 + 4x^2 + x - 6$

Signs: $-, -, -, +, +, -$

Sign changes:

$-$ to $-$ : no change
$-$ to $-$ : no change
$-$ to $+$ : change (1)
$+$ to $+$ : no change
$+$ to $-$ : change (2)

There are 2 sign changes. The number of negative real zeros is $2$ or $0$ .

Building the Possibilities Table

Since $f(x)$ has degree 5, it has exactly 5 zeros (counting multiplicity, including complex zeros). Complex zeros of a polynomial with real coefficients always come in conjugate pairs — so the number of complex zeros is always even.

Combine the information:

Positive	Negative	Complex	Total
3	2	0	5
3	0	2	5
1	2	2	5
1	0	4	5

This table tells us the polynomial has either 1 or 3 positive real zeros, either 0 or 2 negative real zeros, and the rest are complex. It cannot have exactly 2 positive zeros or exactly 1 negative zero.

Worked Example 3: Complete Analysis

Find the possible combinations for $g(x) = x^4 + 2x^3 + 3x^2 + 4x + 5$ .

Sign changes in $g(x)$ : All coefficients are positive: $+, +, +, +, +$ . Zero sign changes.

Result: $g$ has 0 positive real zeros.

Sign changes in $g(-x)$ :

$g(-x) = x^4 - 2x^3 + 3x^2 - 4x + 5$

Signs: $+, -, +, -, +$ . Four sign changes.

Result: $g$ has 4, 2, or 0 negative real zeros.

Negative	Complex	Total
4	0	4
2	2	4
0	4	4

Since all coefficients of $g(x)$ are positive and $x^4, x^3, x^2, x$ are all positive for $x > 0$ , no positive $x$ can make $g(x) = 0$ , confirming 0 positive zeros directly.

Worked Example 4: Polynomial with Missing Terms

Analyze $h(x) = x^3 + 1$ .

Sign changes in $h(x)$ : Coefficients of $x^3, x^2, x, 1$ are $+, 0, 0, +$ . Skip zeros: $+, +$ . Zero sign changes.

Positive real zeros: 0.

$h(-x) = -x^3 + 1$ . Signs (skipping zeros): $-, +$ . One sign change.

Negative real zeros: Exactly 1.

Positive	Negative	Complex	Total
0	1	2	3

Indeed, $h(x) = (x + 1)(x^2 - x + 1)$ , with one negative real zero at $x = -1$ and two complex zeros from $x^2 - x + 1 = 0$ .

Combining with the Rational Root Theorem

Descartes’ Rule is most powerful when used together with the Rational Root Theorem. The strategy:

Descartes’ Rule tells you how many positive/negative zeros are possible
Rational Root Theorem lists the candidates
Skip unnecessary testing — if Descartes says 0 positive zeros, do not test any positive candidates

Worked Example 5: Efficient Zero-Finding

Find all rational zeros of $f(x) = x^4 + 5x^3 + 5x^2 - 5x - 6$ .

Descartes’ Rule on $f(x)$ : Signs: $+, +, +, -, -$ . One sign change.

Positive zeros: Exactly 1.

Descartes’ Rule on $f(-x) = x^4 - 5x^3 + 5x^2 + 5x - 6$ : Signs: $+, -, +, +, -$ . Three sign changes.

Negative zeros: 3 or 1.

Rational candidates: $\pm 1, \pm 2, \pm 3, \pm 6$ .

Since there is exactly 1 positive zero, test positive candidates first:

$f(1) = 1 + 5 + 5 - 5 - 6 = 0$ . Found it: $x = 1$ .

We now know the only positive rational zero is $1$ . Synthetic division:

$\begin{array}{c|ccccc} 1 & 1 & 5 & 5 & -5 & -6 \\ & & 1 & 6 & 11 & 6 \\ \hline & 1 & 6 & 11 & 6 & 0 \end{array}$

Quotient: $x^3 + 6x^2 + 11x + 6$ . Since all positive zeros are accounted for, we only test negative candidates on this cubic.

$f(-1): -1 + 6 - 11 + 6 = 0$ . So $x = -1$ is a zero.

Divide: $x^2 + 5x + 6 = (x + 2)(x + 3)$ .

Zeros: $x = 1, -1, -2, -3$ (1 positive, 3 negative — matching the table).

Edge Cases and Cautions

Zero Coefficients

When a coefficient is zero (the term is missing), skip it when counting sign changes. Do not count it as a sign change.

Example: $f(x) = x^5 - 4x^3 + x$ . The coefficients are $1, 0, -4, 0, 1, 0$ .

Nonzero coefficients in order: $+, -, +$ . Two sign changes, so 2 or 0 positive real zeros.

The Rule Counts Multiplicity

A zero of multiplicity $m$ counts as $m$ zeros. If $x = 2$ is a zero of multiplicity 3, it accounts for 3 of the sign-change count.

The Rule Does Not Apply to Complex Coefficients

Descartes’ Rule requires real coefficients. It does not apply to polynomials with complex coefficients.

Real-World Application: Stability Analysis

In control systems engineering, the stability of a system depends on whether all roots of the characteristic polynomial have negative real parts. Descartes’ Rule provides a quick first check:

For a system with characteristic polynomial $p(s) = s^3 + 4s^2 + 5s + 2$ :

$p(s)$ : signs $+, +, +, +$ — zero sign changes, so zero positive real roots.

$p(-s) = -s^3 + 4s^2 - 5s + 2$ : signs $-, +, -, +$ — three sign changes, so 3 or 1 negative real roots.

This tells us all real roots are negative (on the stable side of the complex plane). Combined with Routh-Hurwitz criteria, engineers confirm full stability.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: Use Descartes’ Rule to determine the possible number of positive and negative real zeros of

f(x) = 3x^4 - x^3 + 2x^2 - x + 1

$f(x)$ signs: $+, -, +, -, +$ . Four sign changes.

Positive real zeros: 4, 2, or 0.

$f(-x) = 3x^4 + x^3 + 2x^2 + x + 1$ . Signs: $+, +, +, +, +$ . Zero sign changes.

Negative real zeros: 0.

Answer: 4, 2, or 0 positive zeros; 0 negative zeros.

Problem 2: Build the complete possibilities table for

g(x) = x^5 - 2x^4 + x^3 - x + 2

$g(x) = x^5 - 2x^4 + x^3 + 0x^2 - x + 2$ . Signs of nonzero coefficients: $+, -, +, -, +$ . Four sign changes.

Positive zeros: 4, 2, or 0.

$g(-x) = -x^5 - 2x^4 - x^3 + x + 2$ . Signs: $-, -, -, +, +$ . One sign change.

Negative zeros: 1.

With neg = 1, we need pos + 1 + complex = 5, with complex even:

Positive	Negative	Complex	Total
4	1	0	5
2	1	2	5
0	1	4	5

Answer: 4, 2, or 0 positive zeros; exactly 1 negative zero; 0, 2, or 4 complex zeros.

Problem 3: Descartes’ Rule says

f(x) = x^3 + x + 1

has 0 positive real zeros. Verify this by testing all Rational Root Theorem candidates.

$f(x)$ signs: $+, +, +$ . Zero sign changes, so 0 positive zeros.

Candidates: $\pm 1$ . $f(1) = 3 \neq 0$ . $f(-1) = -1 \neq 0$ . No rational zeros.

$f(-x) = -x^3 - x + 1$ : signs $-, -, +$ . One sign change: exactly 1 negative real zero.

The single real zero is negative and irrational (approximately $-0.6824$ ).

Answer: Confirmed — no positive rational zeros and no negative rational zeros (the real zero is irrational).

Problem 4: For

p(x) = x^6 - 1

, determine the possible real zeros using Descartes’ Rule, then factor completely over the reals.

$p(x)$ : signs $+, -$ . One sign change: exactly 1 positive zero.

$p(-x) = x^6 - 1$ : same as $p(x)$ . One sign change: exactly 1 negative zero.

Table: 1 positive, 1 negative, 4 complex zeros.

Factor: $x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x-1)(x^2+x+1)(x+1)(x^2-x+1)$ .

Real zeros: $x = 1$ and $x = -1$ . The quadratics $x^2 + x + 1$ and $x^2 - x + 1$ have no real roots (discriminants are $-3$ ).

Answer: Real zeros are $x = 1$ and $x = -1$ ; four complex zeros from the irreducible quadratics.

Problem 5: A polynomial

f(x)

of degree 4 has 2 sign changes in

f(x)

and 0 sign changes in

f(-x)

. List all possible combinations of positive, negative, and complex zeros.

Positive: 2 or 0. Negative: 0. Complex must be even.

Positive	Negative	Complex	Total
2	0	2	4
0	0	4	4

Answer: Either 2 positive and 2 complex, or 0 positive and 4 complex.

Key Takeaways

Descartes’ Rule counts sign changes in $f(x)$ to bound the number of positive real zeros
Substitute $-x$ and count sign changes to bound the number of negative real zeros
The actual count is either the sign-change count or less by an even number
Complex zeros come in conjugate pairs, so the number of complex zeros is always even
Build a possibilities table to see all valid combinations of positive, negative, and complex zeros
Use Descartes’ Rule to skip testing impossible candidates from the Rational Root Theorem
Zero coefficients are skipped when counting sign changes

Return to College Algebra for more topics in this section.

Next Up in College Algebra

The Rational Root Theorem (In Depth)Asymptote Analysis Basic Probability The Binomial Theorem

All College Algebra topics

Last updated: March 29, 2026