What is Descartes Rule of Signs?
Descartes rule of signs states that the number of positive real roots of a polynomial $f(x)$ is either equal to the number of sign changes between consecutive nonzero coefficients, or less than that by an even number. Applying the same count to $f(-x)$ gives the possible number of negative real roots.
Written out, the rule has two parts:
Positive roots: count the sign changes in $f(x)$ as you read the coefficients in order. The number of positive real roots is that count, or that count minus 2, minus 4, and so on, down to 0 or 1.
Negative roots: form $f(-x)$, count its sign changes, and the number of negative real roots is that count or less by an even number.
The "less by an even number" wording is there because real roots that are not present show up as complex conjugate pairs, and each pair removes two from the count.
Variable Glossary:
Symbol | Meaning |
|---|---|
$f(x)$ | the polynomial, written in descending powers |
$f(-x)$ | the polynomial with $x$ replaced by $-x$, used for negative roots |
sign change | a place where consecutive nonzero coefficients have opposite signs |
degree $n$ | the total number of roots, counting complex roots and multiplicity |
How Do You Apply Descartes Rule of Signs?
The procedure is mechanical once you write the polynomial in standard form. Take $f(x) = x^3 - 3x - 2$:
Write $f(x)$ in descending powers and list the nonzero coefficient signs. Here: $+, -, -$ (the $x^2$ term is missing, so skip it).
Count sign changes in $f(x)$. From $+$ to $-$ is one change; from $-$ to $-$ is none. Total: 1 sign change, so there is exactly 1 positive root.
Form $f(-x)$: $(-x)^3 - 3(-x) - 2 = -x^3 + 3x - 2$. Signs: $-, +, -$.
Count sign changes in $f(-x)$. From $-$ to $+$ is one; from $+$ to $-$ is one. Total: 2 changes, so there are 2 or 0 negative roots.
Account for complex roots. The degree is 3, so total roots = 3. The possible splits are recorded in a chart.
The possible-roots chart
For $f(x) = x^3 - 3x - 2$ (degree 3, with 1 positive root and 2 or 0 negative roots), the chart of possibilities is:
Positive roots | Negative roots | Complex roots | Total |
|---|---|---|---|
1 | 2 | 0 | 3 |
1 | 0 | 2 | 3 |
Each row must sum to the degree, 3. (In fact $f(x) = (x-2)(x+1)^2$, so the first row is the true case: 1 positive root and 2 negative roots.)
Examples of Descartes Rule of Signs
Example 1
Find the possible number of positive real roots of $f(x) = x^4 - 3x^2 + 2x - 1$.
Signs of the nonzero coefficients: $+, -, +, -$ (the $x^3$ term is missing). Count the changes: $+\to-$, $-\to+$, $+\to-$ — three sign changes.
So the number of positive real roots is 3 or 1.
Final answer: 3 or 1 positive real roots.
Example 2
Find the possible number of negative real roots of $f(x) = x^3 + 2x^2 + 5x + 4$.
This is where the most common error appears, so watch it first.
Wrong attempt. A student counts the sign changes of the original $f(x)$ — all coefficients are positive, so zero changes — and concludes there are no negative roots. But for negative roots you must use $f(-x)$, not $f(x)$. The original tells you about positive roots only.
Correct. Form $f(-x)$:
$$f(-x) = (-x)^3 + 2(-x)^2 + 5(-x) + 4 = -x^3 + 2x^2 - 5x + 4$$
Signs: $-, +, -, +$. Count changes: $-\to+$, $+\to-$, $-\to+$ — three changes.
So the number of negative real roots is 3 or 1.
Final answer: 3 or 1 negative real roots.
Example 3
Determine the possible number of positive and negative roots of $f(x) = x^5 + 4x^4 - 3x^2 + x - 6$.
Positive: signs are $+, +, -, +, -$ (the $x^3$ term is missing). Changes: $+\to+$ none, $+\to-$ one, $-\to+$ one, $+\to-$ one — three changes. Positive roots: 3 or 1.
Negative: $f(-x) = -x^5 + 4x^4 - 3x^2 - x - 6$, signs $-, +, -, -, -$. Changes: $-\to+$ one, $+\to-$ one, then $-\to-$, $-\to-$ none — two changes. Negative roots: 2 or 0.
Final answer: positive roots 3 or 1; negative roots 2 or 0.
Example 4
Find the exact number of positive roots of $f(x) = x^2 + x + 1$.
Signs: $+, +, +$. Zero sign changes, so there are exactly 0 positive roots. Likewise $f(-x) = x^2 - x + 1$ has signs $+, -, +$ — two changes, so 2 or 0 negative roots. Since a degree-2 polynomial with no positive and no rational structure here has discriminant $1 - 4 = -3 < 0$, both roots are complex.
Final answer: 0 positive roots (and the two roots are a complex conjugate pair).
Example 5
Build the complete roots chart for $f(x) = x^4 - 1$.
Positive: signs $+, -$ (only $x^4$ and the constant are nonzero). One change, so exactly 1 positive root.
Negative: $f(-x) = x^4 - 1$, same signs $+, -$. One change, so exactly 1 negative root.
Degree is 4, so total roots = 4. With 1 positive and 1 negative, the remaining 2 are complex:
Positive | Negative | Complex | Total |
|---|---|---|---|
1 | 1 | 2 | 4 |
Final answer: 1 positive, 1 negative, 2 complex roots. (Indeed $x^4 - 1 = (x-1)(x+1)(x^2+1)$.)
Example 6
Use the rule to bound the complex roots of $f(x) = x^3 - x^2 + x - 1$.
Positive: signs $+, -, +, -$. Three changes, so 3 or 1 positive roots. Negative: $f(-x) = -x^3 - x^2 - x - 1$, signs $-, -, -, -$. Zero changes, so 0 negative roots.
Degree is 3. With 0 negative roots, the chart is:
Positive | Negative | Complex | Total |
|---|---|---|---|
3 | 0 | 0 | 3 |
1 | 0 | 2 | 3 |
Final answer: either 3 positive and 0 complex, or 1 positive and 2 complex. (Here $f(x) = (x-1)(x^2+1)$, so the second row is true: 1 positive, 2 complex.)
The Mathematicians Behind Descartes Rule of Signs
René Descartes (1596–1650, France) published the rule in his 1637 work La Géométrie, the appendix that founded analytic geometry and introduced the convention of using letters near the end of the alphabet for unknowns. He stated the sign-change count as a practical aid for locating a polynomial's roots.
Carl Friedrich Gauss (1777–1855, Germany) later sharpened the rule, proving that when the number of positive roots is less than the sign-change count, it is always less by an even number — the precise form used today. His work on the fundamental theorem of algebra explains why the leftover roots arrive as complex conjugate pairs.
Why This Rule Earns Its Keep
Descartes rule of signs is the WHY behind narrowing a root search before any heavy computation. On its own it does not find roots — its job is to fence off the search space. Paired with the rational root theorem, which lists candidate rational roots, the sign rule tells you how many of each sign to expect, so you stop testing positive candidates once you have found the maximum the rule allows.
It also reveals the presence of complex roots without computing them: subtract the maximum possible real roots from the degree and the remainder must be complex, arriving in conjugate pairs. This connects to finding the zeros of a function and, for the simplest case, the roots of a quadratic equation, where the discriminant does the same job the sign rule does for higher degrees.
Where Students Trip Up On Descartes Rule of Signs
Mistake 1: Using $f(x)$ to count negative roots
Where it slips in: Right after counting positive roots, students reuse the same sign sequence for negative roots instead of forming $f(-x)$.
Don't do this: Concluding $x^3 + 2x^2 + 5x + 4$ has zero negative roots because $f(x)$ has no sign changes.
The correct way: Negative roots always come from the sign changes of $f(-x)$, the polynomial with $x$ replaced by $-x$. The first-instinct error is to skip the substitution and reuse the original signs — the habit that fixes it is writing $f(-x)$ out explicitly every time, flipping the sign of every odd-power term.
Mistake 2: Counting zero-coefficient terms as sign changes
Where it slips in: When a power is missing (coefficient 0), students either count a phantom change or get confused about whether to skip it.
Don't do this: Treating the missing $x^3$ in $x^4 - 3x^2 + 2x - 1$ as a sign in the sequence.
The correct way: Only nonzero coefficients count. Skip any term with a zero coefficient entirely and compare each nonzero coefficient to the next nonzero one. The memorizer who lines up all the powers including the missing ones miscounts here every time.
Mistake 3: Treating the count as exact instead of a maximum
Where it slips in: A student reports "3 sign changes means exactly 3 positive roots" and forgets the "or less by an even number" clause.
Don't do this: Claiming $x^4 - 3x^2 + 2x - 1$ has exactly 3 positive roots.
The correct way: The sign-change count is the maximum. The true number is that count, or 2 fewer, or 4 fewer, down to 0 or 1. Each step of 2 represents a complex conjugate pair replacing two real roots. The build-the-chart habit forces you to list every allowed possibility rather than committing to one.
A "the count is only an upper bound" lesson has a hard-edged engineering cousin. Confusing a bound with a guarantee is exactly the failure mode behind several safety incidents where a designed maximum was read as a certainty. The discipline Descartes rule teaches — stating what is possible rather than asserting what is — is the same care that keeps an estimate from being mistaken for a fact.
Practice Questions on Descartes Rule of Signs
For each, count the sign changes in $f(x)$ for positive roots and in $f(-x)$ for negative roots, then state the possibilities. Answers follow.
Find the possible number of positive real roots of $f(x) = x^3 + 2x^2 + x + 5$.
Find the possible number of negative real roots of the same $f(x) = x^3 + 2x^2 + x + 5$.
Find the possible number of positive real roots of $f(x) = x^4 - x^3 + x^2 - x + 1$.
Build the full roots chart for $f(x) = x^3 - 1$, with every row summing to 3.
State the possible positive and negative roots of $f(x) = 2x^4 - 5x^3 - x + 6$.
How many positive roots does $f(x) = x^2 + 3x + 5$ have, and what does that imply about its roots?
Answers
Answer to Question 1: signs $+, +, +, +$ — zero sign changes, so 0 positive roots.
Answer to Question 2: $f(-x) = -x^3 + 2x^2 - x + 5$, signs $-, +, -, +$ — three changes, so 3 or 1 negative roots.
Answer to Question 3: signs $+, -, +, -, +$ — four changes, so 4, 2, or 0 positive roots.
Answer to Question 4: positive: signs $+, -$, one change, so 1 positive root; negative: $f(-x) = -x^3 - 1$, signs $-, -$, zero changes, so 0 negative roots; the remaining 2 are complex. Chart: 1 positive, 0 negative, 2 complex, total 3. (Indeed $x^3 - 1 = (x-1)(x^2+x+1)$.)
Answer to Question 5: positive: signs $+, -, -, +$ — two changes, so 2 or 0 positive; negative: $f(-x) = 2x^4 + 5x^3 + x + 6$, signs $+, +, +, +$ — zero changes, so 0 negative roots.
Answer to Question 6: signs $+, +, +$ — zero sign changes, so 0 positive roots; the discriminant $9 - 20 = -11 < 0$, so both roots are a complex conjugate pair.
Key Takeaways
Descartes rule of signs bounds the number of positive and negative real roots from sign changes in the coefficients.
Positive roots come from $f(x)$; negative roots come from $f(-x)$ — always form $f(-x)$ separately.
The sign-change count is a maximum; the true number is that count or less by an even number.
Complex roots fill the gap between the maximum real roots and the degree, always in conjugate pairs.
Only nonzero coefficients count; skip missing terms when reading sign changes.
A Practical Next Step
Work through these three to solidify your understanding:
Find the possible number of positive real roots of $f(x) = x^4 + x^3 - 2x + 1$.
Find the possible number of negative real roots of the same polynomial using $f(-x)$.
Build the full roots chart for $f(x) = x^3 - 1$, with every row summing to 3.
If you get stuck on negative roots, come back to Mistake 1 and write $f(-x)$ out term by term. At Bhanzu, our trainers teach this rule alongside the rational root theorem, so students see how the sign count and the candidate list work together to corner a polynomial's roots. Want a live trainer to work through more problems on Descartes rule of signs? Book a free demo class.
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