Answers the question the chapter has been deferring: how to find a root when none is obvious. The Rational Zero Theorem reduces an infinite search to a finite list of candidates; synthetic division tests them cheaply; and the Fundamental Theorem of Algebra guarantees the count is exact over the complex numbers. Closes with the conjugate pair theorem and Descartes' Rule of Signs.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 3 — Polynomial and Rational Functions
§3.6 Zeros of Polynomial Functions, pp. 375-391
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 375-391 — the pages these objectives are drawn from
Warm-up
You have a division machine and nothing to feed it. This section is the missing half.
Discussion prompt
A cubic has no obvious factorisation. You could test numbers one at a time — but which numbers, and how would you know when to give up?
Hint: How many numbers are there to try, and is there any reason to prefer some?
Answer:
Testing at random is hopeless: there are infinitely many numbers and no reason to stop. Even restricting to whole numbers leaves infinitely many.
But if a root is a fraction in lowest terms, its numerator and denominator turn out to be forced to divide two particular coefficients. That cuts the candidates down to a short finite list.
So the search becomes finite, and each test is one synthetic division. If none of the list works, there are no rational roots at all — which is itself a definite answer rather than a failure, and that is exactly what makes the theorem worth having.
Concept
If a polynomial with integer coefficients has a rational root, then in lowest terms its numerator divides the constant term and its denominator divides the leading coefficient.
Rational Zero Theorem — For a polynomial with integer coefficients, any rational root written in lowest terms has numerator dividing the constant term and denominator dividing the leading coefficient. It supplies a finite candidate list, not the roots themselves.
\[ \text{root } = \pm\frac{\text{factor of the constant}}{\text{factor of the leading coefficient}} \]
It is worth being precise about what the theorem does and does not promise. It does not say any candidate is a root — most are not. It says every rational root appears somewhere on the list, so testing the whole list settles the question one way or the other.
Figure (svg): The Rational Zero Theorem shown as a fraction: factors of the constant term over factors of the leading coefficient, with a worked list of candidates enumerated beneath
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 375-380
Section
Section 1
Concept
The candidate list is built from the factors of two coefficients: the constant term supplies numerators and the leading coefficient supplies denominators.
That last point is worth exploiting. A monic polynomial — one with leading coefficient 1 — has only the factors of its constant term as candidates, so the list is short and every entry is a whole number. Many textbook problems are arranged to be monic for exactly this reason.
Figure (svg): The Rational Zero Theorem shown as a fraction: factors of the constant term over factors of the leading coefficient, with a worked list of candidates enumerated beneath
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 375-381
Picture it
Two coefficients, their factors, and every quotient of one by the other.
Figure (svg): The Rational Zero Theorem shown as a fraction: factors of the constant term over factors of the leading coefficient, with a worked list of candidates enumerated beneath
Eight candidates for a cubic that could have had infinitely many. And if all eight fail, the conclusion is definite: no rational roots exist.
Worked example
Two lists of factors, then every quotient.
\[ \text{List the candidate rational roots of } 2x^3-5x^2-4x+3. \]
Find the factors of the constant
Why: Three is prime.
\[ +- 1, +- 3 \]
Find the factors of the leading coefficient
Why: Two is prime.
\[ 1, 2 \]
Form every quotient
Why: Each numerator over each denominator.
\[ +- 1, +- 3, +- \frac{1}{2}, +- \frac{3}{2} \]
Count them
Why: Four values, each with two signs.
Figure (svg): The Rational Zero Theorem shown as a fraction: factors of the constant term over factors of the leading coefficient, with a worked list of candidates enumerated beneath
\[ \pm 1, \; \pm 3, \; \pm\tfrac{1}{2}, \; \pm\tfrac{3}{2} \]
Verify: check one candidate
Why: Testing 3 by synthetic division: bring down 2, times 3 is 6, negative 5 plus 6 is 1; 1 times 3 is 3, negative 4 plus 3 is negative 1; negative 1 times 3 is negative 3, 3 plus negative 3 is 0. The remainder is zero, so 3 is a root and the list has delivered. Seven of the eight candidates are not roots, which is normal.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 376-378
Faded example
For the polynomial 3x cubed plus 2x squared minus 7x plus 4.
Fill in the blanks
\text4 3; \quad \text___ ___
Why: The constant 4 supplies the numerators, giving plus and minus 1, 2 and 4, and the leading coefficient 3 supplies the denominators, giving 1 and 3. The candidates are therefore plus and minus 1, 2, 4, one third, two thirds and four thirds — twelve in all.
Worked example
Leading coefficient 1 makes every candidate an integer.
\[ \text{List the candidates for } x^3-6x^2+11x-6. \]
Find the factors of the constant
Why: Six factors as 1, 2, 3 and 6.
\[ +- 1, +- 2, +- 3, +- 6 \]
Find the factors of the leading coefficient
Why: It is 1.
\[ \text{just } 1 \]
Form the quotients
Why: Dividing by 1 changes nothing.
Note the simplification
Why: No fractions arise at all.
Figure (svg): The solution to Worked example a monic polynomial shown as a ladder of expressions, one row per legal move
\[ \pm 1, \; \pm 2, \; \pm 3, \; \pm 6 \]
Verify: test the smallest candidates first
Why: Trying 1: one minus six plus eleven minus six is zero, so 1 is a root immediately. Small candidates are worth testing first because they are quicker to evaluate and, in constructed problems, more often the answer. The polynomial in fact factors as x minus 1, x minus 2 and x minus 3.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 378-380
Trap
\[ 2x^3+\dots+3: \quad \text{candidates} = \pm\frac{\text{factors of } 2}{\text{factors of } 3} = \pm 1, \pm 2, \pm\tfrac{1}{3}, \pm\tfrac{2}{3} \]
Form the fraction from the two coefficients
Why: Both coefficients are used and every quotient is formed.
The leading coefficient supplies the numerators and the constant the denominators.
It is the other way round: the constant supplies the numerators and the leading coefficient the denominators.
The correct candidates are plus and minus 1, 3, one half and three halves. The wrong list contains 2, which is not a root, and omits 3, which is.
Remember it by the monic case. When the leading coefficient is 1, the candidates are the factors of the constant — so the constant must be on top, since dividing by 1 leaves it there.
Prediction
A polynomial has leading coefficient 1 and constant term 12.
Predict first
How many candidate rational roots are there?
Correct: Twelve, being plus and minus each factor of 12.
Why: Twelve has six positive factors — 1, 2, 3, 4, 6 and 12 — and each appears with both signs, giving twelve candidates. The leading coefficient of 1 contributes no denominators, so every candidate is an integer. Forgetting the negative signs halves the list and can lose the actual root.
Sorting
For the polynomial 2x cubed plus x squared minus 5x plus 3.
Sort into buckets
Sort each number by whether the theorem lists it.
Socratic
The candidate list is guaranteed to be complete, and most of it is wrong.
Discussion prompt
What are the two things the theorem does not tell you?
Hint: Does it say any candidate works, and does it say anything about irrational roots?
Answer:
It does not say any candidate is actually a root. Most are not, and a polynomial may have no rational roots at all, in which case every candidate fails and that is the correct answer.
It says nothing about irrational or complex roots. A polynomial can have three irrational roots and pass through the whole candidate list without a single success — the square root of 2 is a root of x squared minus 2, and it is not on any candidate list.
So the theorem's contribution is precisely to make the rational search finite. That is a real contribution, since it converts an impossible search into a bounded one, but it is worth knowing its limits before spending a page testing candidates that were never going to work.
Section
Section 2
Concept
Each candidate is tested by synthetic division. A zero remainder identifies a root and simultaneously produces the reduced polynomial, so a success advances the problem twice.
The last point catches multiplicity. Once a root is divided out, the same value may still be a root of the quotient, and it will be if the multiplicity was more than one. Moving on to the next candidate immediately can miss the repeat.
Figure (svg): A flow showing the full root-finding process: list the candidates, test them by synthetic division, divide out any that work, and repeat until a quadratic remains
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 381-385
Picture it
This section supplies step one; the previous section supplied the rest.
Figure (svg): A flow showing the full root-finding process: list the candidates, test them by synthetic division, divide out any that work, and repeat until a quadratic remains
The loop runs until the quotient is a quadratic, at which point the quadratic formula finishes the job regardless of whether the remaining roots are rational.
Worked example
List, test, divide, finish.
\[ \text{Factor } f(x)=2x^3-5x^2-4x+3 \text{ completely.} \]
List the candidates
Why: From the previous section's work.
\[ +- 1, +- 3, +- \frac{1}{2}, +- \frac{3}{2} \]
Test until one works
Why: Three succeeds.
Divide it out
Why: Synthetic division by x minus 3.
\[ \text{quotient } 2 x ^{2} + x - 1 \]
Factor the quadratic
Why: Two numbers for the product and sum.
\[ (2 x - 1) (x + 1) \]
Figure (svg): A flow showing the full root-finding process: list the candidates, test them by synthetic division, divide out any that work, and repeat until a quadratic remains
\[ f(x)=(x-3)(2x-1)(x+1) \]
Verify: check the roots against the candidate list
Why: The roots are 3, one half and negative 1 — all three were on the candidate list, as the theorem promised. Note that testing every candidate individually would have found all three, but dividing after the first success made the remaining work a quadratic instead of two more cubic tests.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 382-384
Ranking
The full root-finding loop.
Put in order
Why: The list comes first because it is what makes the testing finite. Testing identifies a root, dividing exploits it, and the loop repeats until the quotient is a quadratic — at which point the formula finishes regardless of whether the remaining roots are rational.
Worked example
Retest the successful candidate on the quotient.
\[ \text{Factor } f(x)=x^3-3x^2+4 \text{ completely.} \]
List and test
Why: Candidates are the factors of 4; negative 1 works.
Divide it out
Why: Synthetic division by x plus 1.
\[ \text{quotient } x ^{2} - 4 x + 4 \]
Recognise the quotient
Why: It is a perfect square trinomial.
\[ (x - 2) ^{2} \]
Assemble
Why: Noting the multiplicity.
\[ (x + 1) (x - 2) ^{2} \]
Figure (svg): The solution to Worked example a repeated root shown as a ladder of expressions, one row per legal move
\[ f(x)=(x+1)(x-2)^2 \]
Verify: connect to the graph
Why: The root at 2 has multiplicity 2, so §3.4 predicts the graph bounces off the axis there rather than crossing. The root at negative 1 has multiplicity 1 and crosses. The multiplicities sum to 3, matching the degree. Had 2 been tested and divided out once without retesting, the second copy would have been missed.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 384-385
Error analysis
A student finds a root of a cubic and reports it.
Annotate
On: \( 2x^3-5x^2-4x+3: \quad 3 \text{ is a root} \;\Longrightarrow\; \text{the root is } 3 \)
A successful division delivers two things: a root and a reduced polynomial. Discarding the second means abandoning the problem partway through, and the degree is the reminder of how many roots are still owed.
Prediction
A quartic is being factored, and two roots have been found and divided out.
Predict first
What should happen next?
Correct: Use the quadratic formula on the remaining quadratic.
Why: Two divisions reduce a quartic to a quadratic, and the formula solves any quadratic including those with irrational or complex roots. Continuing to test candidates would only find rational roots and would miss the others. Testing the original again wastes the reductions already made.
Faded example
A cubic with a root at 2 divides to leave the quotient x squared minus x minus 6.
Fill in the blanks
f(x) = (x - 2)(x - 3)(x + 2)
Why: The quotient factors as x minus 3 times x plus 2, since negative 3 and 2 multiply to negative 6 and add to negative 1. The three roots are 2, 3 and negative 2. Note that the divided-out factor keeps its own bracket in the final answer — dropping it is a common slip.
Missing information
You have tested every candidate on the list and none is a root.
Discussion prompt
What does that tell you, and what would you do next?
Hint: What kind of roots does the theorem cover?
Answer:
It tells you definitively that the polynomial has no rational roots. That is a real conclusion rather than a failure — the list was guaranteed complete, so exhausting it settles the question.
It does not mean there are no roots. There may be irrational ones, or complex ones, and neither kind ever appears on the candidate list.
What to do next depends on the degree. A quadratic yields to the formula. A cubic with no rational roots has three irrational or complex ones and needs either a numerical method or the sign-change bracketing from §3.4. At this level, a problem with no rational roots is usually signalling that a different technique is intended.
Section
Section 3
Concept
Every polynomial of degree at least one has at least one complex root. Applying this repeatedly, a polynomial of degree n factors into exactly n linear factors over the complex numbers.
The theorem changes the character of every question in the chapter. Over the reals, 'how many roots' has an answer between zero and n and requires work to determine; over the complex numbers it is always exactly n, and the interesting question becomes what kind of roots they are.
Figure (svg): A card stating the Fundamental Theorem of Algebra, with a degree five polynomial shown factoring into five linear factors over the complex numbers, some real and some in conjugate pairs
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 385-388
Picture it
Only one of these five roots is real, and the theorem still counts five.
Figure (svg): A card stating the Fundamental Theorem of Algebra, with a degree five polynomial shown factoring into five linear factors over the complex numbers, some real and some in conjugate pairs
The two repeated complex conjugate pairs account for four of the five. Over the reals this polynomial would be described as having one root; over the complex numbers it has exactly five, as its degree requires.
Worked example
The total is fixed; the split between real and complex is not.
\[ \text{A degree } 5 \text{ polynomial has } 3 \text{ real roots. How many complex ones?} \]
Apply the theorem
Why: Five roots in total, with multiplicity.
\[ 5\text{ roots total} \]
Subtract the real ones
Why: Three are accounted for.
\[ 2\text{ remaining} \]
Note what those two must be
Why: Non-real roots pair up.
State the answer
Why: Two non-real roots, forming a pair.
\[ 2\text{ complex roots} \]
Figure (svg): A card stating the Fundamental Theorem of Algebra, with a degree five polynomial shown factoring into five linear factors over the complex numbers, some real and some in conjugate pairs
\[ 2 \text{ non-real roots, a conjugate pair} \]
Verify: check the parity
Why: Non-real roots come in pairs for a real polynomial, so their count must be even — and 2 is. Had the question said 4 real roots, the remaining count would be 1, which is odd and therefore impossible: a degree 5 real polynomial cannot have exactly 4 real roots. The parity constraint does genuine work.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 386-387
Prediction
A degree 7 polynomial is factored over the complex numbers.
Predict first
How many roots does it have, counted with multiplicity?
Correct: Exactly 7.
Why: The Fundamental Theorem makes the count exact over the complex numbers, provided roots are counted with multiplicity. Over the reals the honest statement would be 'at most 7', which is §3.3's bound — the difference between the two is precisely what the complex numbers buy.
Worked example
Real irreducible quadratics split once complex numbers are allowed.
\[ \text{Factor } x^4-1 \text{ completely over the complex numbers.} \]
Factor as a difference of squares
Why: Twice.
\[ (x ^{2} - 1) (x ^{2} + 1) \]
Factor the first bracket
Why: Another difference of squares.
\[ (x - 1) (x + 1) (x ^{2} + 1) \]
Factor the last over the complex numbers
Why: Its roots are i and negative i.
\[ (x - i) (x + i) \]
Assemble
Why: Four linear factors.
\[ (x - 1) (x + 1) (x - i) (x + i) \]
Figure (svg): The solution to Worked example factor completely over the complex numbers shown as a ladder of expressions, one row per legal move
\[ (x-1)(x+1)(x-i)(x+i) \]
Verify: count and check the pairing
Why: Four linear factors for degree 4, as the theorem requires. The two non-real roots are i and negative i, a conjugate pair — as they must be, since the original polynomial has real coefficients. Over the reals the last quadratic could not be factored at all, which is exactly the limitation the complex numbers remove.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 387-388
Trap
\[ x^3+x = x(x^2+1) \;\Longrightarrow\; \text{one root, at } 0 \]
Factor and read off the real roots
Why: The quadratic factor has no real roots, so only the root at zero is counted.
The cubic is described as having exactly one root.
It has three roots, as its degree requires: 0, i and negative i. The quadratic factor has two complex roots even though it has no real ones.
The statement 'one root' is correct only if 'root' is restricted to real numbers, and it should be said so explicitly.
Say which number system you are counting in. Over the reals the answer is one; over the complex numbers it is three, and the Fundamental Theorem is a statement about the second.
Sorting
Some statements need the larger number system.
Sort into buckets
Sort each statement.
Faded example
A degree 6 polynomial with real coefficients has 2 real roots.
Fill in the blanks
6 - 2 = 4 \text2 ___ \text___
Why: Four non-real roots remain, and since they pair off there are two conjugate pairs. The count of non-real roots must be even for a real polynomial, and 4 is — a degree 6 polynomial with real coefficients cannot have exactly 3 real roots, because that would leave an odd number of non-real ones.
Socratic
The name claims a great deal.
Discussion prompt
What does the theorem guarantee that makes the whole chapter's approach work?
Hint: What would go wrong if some polynomials had no roots at all?
Answer:
It guarantees the factoring process always terminates in linear factors. Each root gives a factor, dividing it out lowers the degree, and the theorem promises the quotient still has a root — so the process cannot get stuck.
Without it, a polynomial might simply have no roots and be unfactorable, and there would be no way to know in advance which ones. The chapter's entire method — find a root, divide, repeat — would have no guarantee of finishing.
It is also the reason complex numbers are not optional. §3.1 looked like a detour into a strange number system; this theorem is why it was on the way. The theorem is false over the reals and true over the complex numbers, and that gap is what the earlier section was closing.
Section
Section 4
Concept
If a polynomial has real coefficients and a non-real root, then the conjugate of that root is also a root. Non-real roots therefore always come in pairs.
The fourth point explains what real factorisation can achieve. Over the reals, a polynomial factors into linear factors and irreducible quadratics — and each of those quadratics is exactly one conjugate pair multiplied together. Nothing worse than a quadratic is ever needed.
Figure (svg): Two complex roots plotted as mirror images across the real axis, with a note that a polynomial with real coefficients must contain both if it contains either
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 388-390
Picture it
Two roots that a real polynomial cannot separate.
Figure (svg): Two complex roots plotted as mirror images across the real axis, with a note that a polynomial with real coefficients must contain both if it contains either
Containing one of these forces containing the other. That is why the count of non-real roots is even, and why an odd degree cannot avoid a real root.
Worked example
One complex root is worth two.
\[ \text{A cubic with real coefficients has } 2+i \text{ as a root. Find the others.} \]
Apply the conjugate pair theorem
Why: The conjugate is also a root.
Count what is left
Why: Degree 3, two roots found.
Determine its kind
Why: Non-real roots pair, and both are used.
Note how to find it
Why: Divide out the quadratic from the pair.
\[ \text{divide by } x ^{2} - 4 x + 5 \]
Figure (svg): Two complex roots plotted as mirror images across the real axis, with a note that a polynomial with real coefficients must contain both if it contains either
\[ 2-i \text{ is a root; the third is real} \]
Verify: check the quadratic is real
Why: The pair multiplies to x minus 2 minus i, times x minus 2 plus i, which is x minus 2 all squared plus 1, giving x squared minus 4x plus 5 — real coefficients, as required, with discriminant 16 minus 20 which is negative. Dividing the cubic by this real quadratic leaves a real linear factor, giving the third root.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 389-389
Prediction
A polynomial with real coefficients has some non-real roots.
Predict first
What can be said about how many?
Correct: There is an even number of them.
Why: Non-real roots pair off with their conjugates, so they come two at a time and the total is even. This is what forces an odd-degree real polynomial to have a real root: an odd total cannot be made up entirely of pairs, so at least one root is left over and must be real.
Worked example
Real coefficients require the pairing to be respected.
\[ \text{Find a real polynomial of least degree with roots } 3 \text{ and } 1-2i. \]
Include the forced conjugate
Why: Real coefficients demand it.
Count the roots
Why: Three of them.
\[ ^\circ 3 \]
Multiply the conjugate pair
Why: Giving a real quadratic.
\[ x ^{2} - 2 x + 5 \]
Multiply by the real factor
Why: The root at 3.
\[ (x - 3) (x ^{2} - 2 x + 5) \]
Figure (svg): The solution to Worked example build a polynomial from its roots shown as a ladder of expressions, one row per legal move
\[ f(x)=(x-3)(x^2-2x+5) \]
Verify: check the coefficients came out real
Why: Expanding gives x cubed minus 5x squared plus 11x minus 15 — every coefficient real, as required. Omitting the conjugate would have given a quadratic with complex coefficients, which is a legitimate polynomial but not a real one. The pairing is what keeps the coefficients real.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 390-390
Trap
\[ f(x)=x-i \;\Longrightarrow\; \text{since } i \text{ is a root}, \; -i \text{ must be too} \]
Apply the conjugate pair theorem
Why: A non-real root is present, so its conjugate is added.
The polynomial is said to have both i and negative i as roots.
The theorem requires real coefficients, and this polynomial has a coefficient of negative i. So the pairing does not apply.
The polynomial has degree 1 and therefore exactly one root, which is i. Substituting negative i gives negative 2i, not zero.
Check the coefficients are real before invoking the pairing. It is a strong theorem and it earns its strength from that hypothesis, which is easy to overlook because almost every polynomial you meet satisfies it.
Faded example
A real quartic has roots 1, negative 2 and 3 plus i.
Fill in the blanks
\text- 3 conjugate i, \text___ ___ \text___
Why: Real coefficients force the conjugate 3 minus i to be a root as well, which accounts for all four roots of the quartic. Two real roots and one conjugate pair is a legitimate configuration; two real roots and a single non-real root would not be, since the non-real count must be even.
Sorting
Check whether the non-real roots pair off.
Sort into buckets
Sort each proposed root set for a polynomial with real coefficients.
Explain it to yourself
Section 3.3 gave a graphical argument; this section gives an algebraic one.
Discussion prompt
Explain why an odd-degree real polynomial must have a real root, using the pairing.
Hint: How many roots are there in total, and how do the non-real ones come?
Answer:
The Fundamental Theorem says there are exactly n roots for degree n, and n is odd here. The non-real roots come in conjugate pairs, so however many there are, that number is even.
An odd total cannot be made up entirely of even groups, so at least one root is left over — and a root that is not part of a non-real pair must be real.
This is a completely different argument from §3.3's, which used the graph running from below the axis to above it. Both are correct, and having two independent proofs of the same fact is a good sign that the fact is structural rather than accidental.
Section
Section 5
Concept
The number of positive real roots is the number of sign changes in the coefficients, or less than that by an even number. Substituting negative x gives the corresponding bound for negative roots.
The rule is most useful negatively. A polynomial with no sign changes has no positive roots at all, which removes half the candidate list before any testing begins — and that saving is often larger than anything the rule contributes by narrowing a count from three to one.
Figure (svg): A card stating the Fundamental Theorem of Algebra, with a degree five polynomial shown factoring into five linear factors over the complex numbers, some real and some in conjugate pairs
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 390-391
Picture it
The rule bounds the real ones; the theorem fixes the total.
Figure (svg): A card stating the Fundamental Theorem of Algebra, with a degree five polynomial shown factoring into five linear factors over the complex numbers, some real and some in conjugate pairs
The even-number reduction is the conjugate pairing again: each pair of complex roots takes two away from what would otherwise have been real, which is why the count drops in twos.
Worked example
Read across the coefficients in descending order.
\[ \text{Bound the positive real roots of } f(x)=2x^3-5x^2-4x+3. \]
List the signs in order
Why: Plus, minus, minus, plus.
\[ +- - + \]
Count the changes
Why: Plus to minus, then minus to plus.
\[ 2\text{ changes} \]
Apply the rule
Why: That count, or less by an even number.
\[ 2\text{ or } 0 \]
Interpret
Why: Either two positive roots or none.
\[ 2\text{ or } 0\text{ positive roots} \]
Figure (svg): The solution to Worked example count the sign changes shown as a ladder of expressions, one row per legal move
\[ 2 \text{ or } 0 \text{ positive real roots} \]
Verify: check against the actual roots
Why: The roots are 3, one half and negative 1, so there are exactly 2 positive ones — matching the upper option. Note that the rule alone could not distinguish 2 from 0; it narrows the possibilities without deciding between them, which is what makes it a bound.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 390-391
Faded example
For the polynomial x to the fourth, minus x cubed, plus x squared, minus 1.
Fill in the blanks
\text3 +, -, +, -: \quad 1 \text___ ___ \text___ ___ \text___
Why: Reading across, the sign changes three times. So there are 3 positive roots, or fewer by an even number, which gives 1 as the only other possibility. Note there is no x term, so the coefficient sequence has only four entries — a missing term contributes no sign and is simply skipped.
Worked example
Substitute negative x first, then count again.
\[ \text{Bound the negative real roots of the same polynomial.} \]
Substitute negative x
Why: Odd powers change sign.
\[ -2 x ^{3} - 5 x ^{2} + 4 x + 3 \]
List the signs
Why: Minus, minus, plus, plus.
\[ -- + + \]
Count the changes
Why: Only one, from minus to plus.
\[ 1\text{ change} \]
Apply the rule
Why: One, or less by an even number — but zero is not reachable.
\[ \text{exactly } 1 \]
Figure (svg): The solution to Worked example bound the negative roots shown as a ladder of expressions, one row per legal move
\[ \text{exactly } 1 \text{ negative real root} \]
Verify: check why the count is exact here
Why: Reducing 1 by an even number would give negative 1, which is impossible, so 1 is the only option and the bound becomes a count. This is when the rule is at its most informative: an odd number of sign changes equal to 1 pins the answer down exactly. The actual root is negative 1, confirming it.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 391-391
Error analysis
A student bounds the negative roots.
Annotate
On: \( f(x)=x^3-2x^2+x-6: \quad 3 \text{ sign changes} \;\Longrightarrow\; 3 \text{ negative roots} \)
One substitution separates the two questions. Counting on the original bounds the positive roots and counting on the substituted version bounds the negative ones, and using one count for both is the standard error.
Prediction
A polynomial's coefficients are all positive.
Predict first
What does the rule tell you?
Correct: There are no positive real roots at all.
Why: With no sign changes there are zero positive real roots, and zero cannot be reduced further. This is the rule at its most valuable: it eliminates every positive candidate from the search list in one glance, which is a larger saving than narrowing a count usually provides.
Sorting
One substitution separates the two questions.
Sort into buckets
Sort each task by which polynomial's signs you count.
Edge cases
The rule says the count may be less by an even number.
Discussion prompt
Why an even number rather than any number?
Hint: What replaces a pair of real roots when they are not real?
Answer:
Because non-real roots come in conjugate pairs. If a polynomial has fewer real roots than the sign changes suggest, the missing ones have become complex — and they can only go missing two at a time.
So the shortfall between the sign-change count and the actual number of positive roots is exactly twice the number of conjugate pairs that displaced them, which is necessarily even.
This connects three of the section's ideas at once: the Fundamental Theorem fixes the total, the pairing theorem constrains how non-real roots arrive, and Descartes' rule inherits the even-number reduction from that constraint. The rule's odd-looking proviso is not arbitrary; it is the pairing showing up again.
Comparison
Fill the blanks from memory. Four results, each answering a different part of the same question.
Comparison matrix
| what it tells you | what it does not | |
|---|---|---|
| Rational Zero Theorem | a finite list of candidate rational roots | whether any of them is actually a root |
| Factor Theorem | that a root gives a factor, so the degree drops | how to find the root |
| Fundamental Theorem | exactly n roots over the complex numbers | what the roots are, or how many are real |
| conjugate pair theorem | non-real roots come in pairs | anything when coefficients are not real |
| Descartes' Rule | a bound on the positive and negative real roots | which numbers the roots are |
None of these finds a root by itself. Together they bound the search, make each test cheap, and guarantee the process terminates — which is enough to factor any polynomial whose roots are rational.
Pattern
Six steps, combining this section with the previous one.
Step 6's count is the completeness check. A degree 5 polynomial owes five roots with multiplicity, and anything less means a factor is still unfactored or a repeat was missed.
OpenStax Algebra and Trigonometry 2e, §5.5 Zeros of Polynomial Functions §5.5
Check
Constant on top.
Check your understanding
Which of these is NOT a candidate rational root of 3x^3 + 2x^2 - x + 4?
Answer: A
Why: Candidates are factors of the constant 4 over factors of the leading coefficient 3. Three does not divide 4, so it cannot be a numerator, and 3 over 1 is not on the list. The other three all have numerators dividing 4 and denominators dividing 3.
Check
Count over the complex numbers.
Check your understanding
A degree 4 polynomial with real coefficients has exactly 2 real roots. How many non-real roots does it have?
Answer: A
Why: The Fundamental Theorem gives four roots in total, two of which are real, leaving two non-real ones. Since the coefficients are real, those two must be conjugates of each other.
Check
Count on the right polynomial.
Check your understanding
For f(x) = x^3 + 2x^2 + 5x + 4, how many positive real roots can there be?
Answer: A
Why: Every coefficient is positive, so there are no sign changes and therefore no positive real roots. Any positive input gives a sum of positive terms, which cannot be zero — the rule agrees with the direct argument here.
Real world
Turning an infinite search into a finite one is a move that recurs across mathematics and computing.
Discussion prompt
The Rational Zero Theorem does not find roots — it only rules candidates out. Why is that still valuable?
Hint: How long would an exhaustive search take without it?
Answer:
Without it the search space is infinite, and no exhaustive method can terminate. With it the space is finite and usually small, so exhaustive testing becomes a viable strategy rather than an impossible one.
This pattern — constrain the search space, then search exhaustively — is one of the most common in computing. Constraint propagation in a solver, pruning in a game tree, and type checking all work this way.
It is also why a negative result here is genuinely informative. Exhausting the list proves no rational root exists, which is a theorem about the polynomial rather than a failure to find one. Search methods that cannot bound their space can never make that kind of claim.
Commit first
State your confidence along with your answer.
Predict first
Every candidate on a polynomial's rational root list has been tested and none is a root. What follows?
Correct: It has no rational roots, though it may have irrational or complex ones.
Why: The list is guaranteed to contain every rational root, so exhausting it proves there are none. It says nothing about irrational or complex roots — the square root of 2 is a root of x squared minus 2 and appears on no candidate list. The Fundamental Theorem still guarantees the full complement of complex roots.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate what the Rational Zero Theorem promises and what it does not, and why it is worth using anyway.
Hint: Is it a root-finder or a search-narrower?
Answer:
It promises that every rational root is somewhere on the list. It does not promise that any of them is a root — most are not — and it says nothing at all about irrational or complex roots.
It is worth using because it converts an infinite search into a finite one. Testing every number is impossible; testing eight candidates is an afternoon's work at worst, and with synthetic division each test is a few seconds.
The strongest point is that it makes failure informative. If the whole list fails, you have proved there are no rational roots — a definite conclusion, not a dead end. A search with no bound on its space could never establish that.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The first is a one-line fact that costs the whole problem when inverted, so it is worth fixing firmly. The second is the section's actual technique and is where the sustained work lies; the third and fourth are the results that make the technique guaranteed to succeed.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write a cubic with a leading coefficient other than 1. Build its candidate list, showing which coefficient supplies numerators and which supplies denominators. Then draw the loop: test, divide, repeat, finish with the quadratic — and beside it note what the Fundamental Theorem guarantees about how many roots you should end up with.
If your loop ends at a quadratic rather than continuing to test candidates, you have the point at which this section hands over to a technique you have had since Algebra 1.
Recap
Five things, and together they factor any polynomial whose roots are rational.
| if you remember one thing | it should be this |
|---|---|
| about the candidate list | constant on top, leading coefficient underneath |
| about the loop | divide out a success and continue on the quotient |
| about the count | exactly n roots over the complex numbers, with multiplicity |
| about non-real roots | with real coefficients they always come in pairs |
Section 3.7 allows a polynomial in a denominator, which introduces the first functions in the course whose graphs have breaks — and the asymptotes that describe them.
OpenStax, Precalculus, §3.6 Zeros of Polynomial Functions §3.6, pp. 375-391 — everything on these slides traces back here
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