3.6Finding Polynomial Zeros
Apply the Rational Zero Theorem to list candidates, use Descartes' Rule of Signs, find all real and complex zeros, and write polynomials in fully factored form.
Finding all zeros of a polynomial — real and complex — is the central problem of algebra. The Fundamental Theorem of Algebra guarantees every degree-n polynomial has exactly n zeros (counting multiplicity).
Essential Question
How can we systematically find every zero — real and complex — of any polynomial with integer coefficients?
Lesson Overview
Every polynomial of degree n has exactly n zeros when we count complex zeros and repeated zeros with their multiplicity. The Fundamental Theorem of Algebra guarantees this. In this lesson we build a complete toolkit: the Rational Zero Theorem narrows our search to a finite list of candidates, Descartes' Rule of Signs tells us how many positive and negative real zeros to expect, synthetic division tests each candidate efficiently, and the Complex Conjugate Theorem ensures complex zeros always come in conjugate pairs.
Fundamental Theorem of Algebra
Every polynomial f(x) of degree n ≥ 1 has exactly n zeros in the complex number system, counting multiplicity. A degree-3 polynomial always has 3 zeros; a degree-4 always has 4, and so on.
Complex Conjugate Theorem
If a polynomial has real coefficients and a + bi (b ≠ 0) is a zero, then its conjugate a − bi is also a zero. Complex zeros always appear in conjugate pairs, so a real-coefficient polynomial of odd degree must have at least one real zero.
Rational Zero Theorem — f(x) = 2x³ − x² − 7x + 6
Descartes' Rule of Signs — f(x) = 2x³ − x² − 7x + 6
Strategy: Finding All Zeros
Worked Examples
List all possible rational zeros of f(x) = 3x³ + 5x² − 2x − 8.
Step 1: Identify a₀ = −8 and aₙ = 3.
Step 2: Factors of a₀ (p): ±1, ±2, ±4, ±8.
Step 3: Factors of aₙ (q): ±1, ±3.
Step 4: Form ±p/q: ±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3.
Use Descartes' Rule of Signs on f(x) = x⁴ − 3x³ + x² + 5x − 2.
Step 1: Count sign changes in f(x): + → − → + → + → − = 3 changes.
→ 3 or 1 positive real zeros.
Step 2: f(−x) = x⁴ + 3x³ + x² − 5x − 2.
Sign changes: + → + → + → − → − = 1 change.
→ exactly 1 negative real zero.
Step 3: Degree 4 → 4 total zeros.
Possible: (3 pos, 1 neg, 0 complex) or (1 pos, 1 neg, 2 complex).
Find all zeros of f(x) = x³ − 6x² + 11x − 6.
Step 1: Possible rational zeros: ±1, ±2, ±3, ±6.
Step 2: Test x = 1 by synthetic division:
1 | 1 −6 11 −6
| 1 −5 6
| 1 −5 6 0 ✓ (remainder 0)
Step 3: Quotient: x² − 5x + 6 = (x − 2)(x − 3).
Step 4: Zeros from quotient: x = 2, x = 3.
Find all zeros of f(x) = x⁴ − x² − 12.
Step 1: Treat as quadratic in u = x²: u² − u − 12 = 0.
Step 2: Factor: (u − 4)(u + 3) = 0.
Step 3: u = 4 → x² = 4 → x = ±2.
Step 4: u = −3 → x² = −3 → x = ±i√3.
Write a degree-4 polynomial with real coefficients whose zeros are x = 3 (multiplicity 2) and x = 1 + 2i.
Step 1: Complex Conjugate Theorem → x = 1 − 2i is also a zero.
Step 2: Factors: (x − 3)², (x − (1 + 2i)), (x − (1 − 2i)).
Step 3: Multiply conjugate pair:
(x − 1 − 2i)(x − 1 + 2i) = (x − 1)² + 4 = x² − 2x + 5.
Step 4: f(x) = (x − 3)²(x² − 2x + 5).
Step 5: Expand: (x² − 6x + 9)(x² − 2x + 5)
= x⁴ − 2x³ + 5x² − 6x³ + 12x² − 30x + 9x² − 18x + 45
= x⁴ − 8x³ + 26x² − 48x + 45.
Guided Practice
List all possible rational zeros of g(x) = 2x³ − 7x + 3.
Hint: Find all factors of the constant term (3) and all factors of the leading coefficient (2), then form every ±p/q ratio.
Apply Descartes' Rule of Signs to h(x) = x³ + 2x² − 5x − 6. How many positive and negative real zeros are possible?
Hint: Count sign changes in h(x) for positive zeros, then substitute −x and count sign changes in h(−x) for negative zeros.
Use synthetic division to verify that x = −2 is a zero of f(x) = x³ + 3x² − 4, then find the remaining zeros.
Hint: Set up synthetic division with −2 and the coefficients 1, 3, 0, −4. After confirming remainder = 0, factor the quotient quadratic.
Find all zeros of p(x) = x³ − 5x² + 8x − 6 given that x = 3 is a zero.
Hint: Divide p(x) by (x − 3) using synthetic division to get a quadratic, then use the quadratic formula on the quotient.
Write a degree-3 polynomial with real coefficients whose zeros are x = −1 and x = 2 + i.
Hint: Use the Complex Conjugate Theorem to identify the third zero, then multiply the three linear factors together.
Key Vocabulary
Zero of a polynomial
A value c such that f(c) = 0. Also called a root. Geometrically, zeros are the x-intercepts of the graph (for real zeros).
Example: f(x) = x² − 4 has zeros x = 2 and x = −2
Multiplicity
The number of times a factor (x − c) appears in the factored form. A zero of even multiplicity touches but does not cross the x-axis; odd multiplicity crosses.
Example: f(x) = (x − 1)²(x + 3) has x = 1 with multiplicity 2
Rational Zero Theorem
If f(x) has integer coefficients, every rational zero p/q (in lowest terms) satisfies: p divides the constant term a₀ and q divides the leading coefficient aₙ.
Example: f(x) = 2x³ − x − 6: candidates include ±1, ±2, ±3, ±6, ±1/2, ±3/2
Descartes' Rule of Signs
The number of positive real zeros equals the number of sign changes in f(x), or less by an even number. For negative zeros, apply the same rule to f(−x).
Example: f(x) = x³ − 2x + 1 has 2 sign changes → 2 or 0 positive real zeros
Complex Conjugate Theorem
If a polynomial has real coefficients and a + bi (b ≠ 0) is a zero, then a − bi is also a zero. Complex zeros always appear in conjugate pairs.
Example: If 3 + 2i is a zero, then 3 − 2i must also be a zero
Synthetic division
A shorthand algorithm for dividing a polynomial by a linear factor (x − c). Used to test rational zero candidates and reduce the polynomial's degree.
Example: Dividing x³ − 6x + 5 by (x − 1) using synthetic division
Check Your Understanding
Interactive Practice — 5 Questions
According to the Fundamental Theorem of Algebra, how many zeros does f(x) = x⁵ − 3x² + 1 have (counting multiplicity, including complex)?
For f(x) = 4x³ − 2x + 6, which of the following is a possible rational zero according to the Rational Zero Theorem?
How many sign changes are in f(x) = x⁴ − 2x³ − x² + 4x − 3?
A polynomial with real coefficients has zeros x = 2, x = −1, and x = 3 + i. What is the minimum degree of the polynomial?
After using synthetic division to verify x = 2 is a zero of f(x) = x³ − 5x² + 8x − 4, the quotient is x² − 3x + 2. What are the remaining zeros?
Independent Practice
Independent Practice
List all possible rational zeros of f(x) = x³ − 4x² + x + 6.
List all possible rational zeros of g(x) = 6x⁴ − 7x³ + 2x − 12.
Apply Descartes' Rule of Signs to f(x) = x³ + 3x² − x − 3. State the possible number of positive and negative real zeros.
Find all zeros of f(x) = x³ − 2x² − 5x + 6 using the Rational Zero Theorem and synthetic division.
Find all zeros of f(x) = 2x³ + x² − 7x − 6.
Common Mistakes
Forgetting to include ±1 and ±(constant term) in the list of p/q candidates.
Always include all factors of a₀ and aₙ, including 1 and the number itself. Missing a factor means missing a potential zero.
Concluding that a polynomial has exactly as many positive zeros as sign changes in f(x).
Descartes' Rule says the number of positive real zeros equals the sign-change count or less by an even number. It gives a maximum, not an exact count.
Forgetting to include the conjugate when a complex zero is given.
For real-coefficient polynomials, complex zeros always come in conjugate pairs. If a + bi is a zero, a − bi must also be listed.
Stopping after finding one zero and not fully factoring the reduced polynomial.
After each synthetic division, factor or apply the quadratic formula to the quotient until all zeros are found. The degree must reach 0.
Math Tips
Start Descartes' Rule before testing candidates — it tells you how many positive and negative zeros to look for, saving time.
Test simple integers (±1, ±2) first in synthetic division — they are the most common rational zeros in textbook problems.
Once you find one zero and reduce the polynomial to a quadratic, always use the quadratic formula rather than guessing — it handles complex roots automatically.
The Complex Conjugate Theorem is a shortcut: if you know one complex zero, you immediately know a second one for free.
Always verify your zeros by substituting back into the original polynomial or confirming the product of all linear factors equals f(x).