Polynomial Functions and Complex Zeros - AP Precalculus

SECTION 1

Historical Context & Motivation

For centuries, mathematicians encountered equations whose solutions seemed to lie beyond the real number line. Ancient Greek geometers could solve quadratics by completing the square, but when the discriminant turned negative, they dismissed the equation as having no solution. The problem persisted through the medieval period until Italian algebraists of the sixteenth century, working on cubic equations, found that even equations with three real roots sometimes required manipulation of square roots of negative numbers during intermediate steps. This uncomfortable necessity forced mathematicians to take imaginary numbers seriously — not as curiosities, but as essential tools for understanding polynomial behavior. The journey from reluctant acceptance to the elegant Fundamental Theorem of Algebra reshaped our understanding of what it means to solve an equation.

1545 • Cardano's Ars Magna

Gerolamo Cardano publishes general solutions to cubic and quartic equations, encountering √(−1) in intermediate calculations even when final answers are real — the first meaningful engagement with complex numbers.

1637 • Descartes Coins 'Imaginary'

René Descartes introduces the term "imaginary" for non-real roots in La Géométrie, and establishes the convention that polynomials of degree n should have n roots — foreshadowing the Fundamental Theorem.

1799 • Gauss Proves the Fundamental Theorem

Carl Friedrich Gauss provides the first rigorous proof that every non-constant polynomial with complex coefficients has at least one complex root, cementing the completeness of the complex number system.

1806 • Argand Diagram Published

Jean-Robert Argand publishes the geometric representation of complex numbers as points in a plane, giving mathematicians a powerful visual tool for understanding complex zeros and their relationships.

The central question that drives this topic is deceptively simple: given a polynomial of degree n, how many zeros does it have, and where do they live? On the real number line, a polynomial's graph may cross the x-axis fewer than n times — yet the Fundamental Theorem of Algebra guarantees exactly n zeros exist when we extend our search into the complex plane. Understanding this interplay between real and complex zeros is essential for fully factoring polynomials and interpreting their graphical behavior.

SECTION 2

Core Principles & Definitions

Before exploring the relationship between polynomial functions and complex zeros, we need a precise vocabulary. A polynomial function of degree n takes the form f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where the coefficients aₖ are real numbers and aₙ ≠ 0. A zero of f is any value c such that f(c) = 0; when c is real, it corresponds to an x-intercept on the graph. A complex number has the form a + bi where a and b are real numbers and i = √(−1). Real numbers are a subset of the complex numbers (with b = 0), so every real zero is also a complex zero.

Fundamental Theorem of Algebra

Every polynomial of degree n ≥ 1 with complex coefficients has exactly n zeros in the complex numbers, counted with multiplicity. This guarantees completeness — no polynomial is "unsolvable."

Complex Conjugate Zeros Theorem

If a polynomial has real coefficients and a + bi (with b ≠ 0) is a zero, then its conjugate a − bi must also be a zero. Complex zeros of real-coefficient polynomials always appear in conjugate pairs.

Multiplicity of Zeros

A zero c has multiplicity m if (x − c)ᵐ is a factor of f(x) but (x − c)ᵐ⁺¹ is not. The sum of all multiplicities equals the degree n. Multiplicity affects how the graph behaves at each x-intercept.

Linear Factorization Theorem

Every polynomial of degree n can be written as f(x) = aₙ(x − c₁)(x − c₂)…(x − cₙ), where c₁, c₂, …, cₙ are the n complex zeros. This is the complete factored form over ℂ.

✦

KEY TAKEAWAY

Think of the real number line as a one-dimensional "slice" through the two-dimensional complex plane. When a polynomial's graph doesn't cross the x-axis at a particular point, the missing zeros haven't vanished — they've merely stepped off the real line into the surrounding complex plane, always traveling in conjugate pairs like mirror reflections across the real axis. The Fundamental Theorem guarantees that if you search the entire complex plane, you'll always find exactly as many zeros as the degree of the polynomial.

SECTION 3

Visualizing Real & Complex Zeros

The diagram below illustrates how the zeros of a degree-4 polynomial manifest in two complementary views: the familiar Cartesian graph of y = f(x), and the complex plane (also called the Argand plane) where we can plot all zeros — both real and complex. Consider the polynomial f(x) = x⁴ − 2x³ + 2x² − 2x + 1, which factors as (x − 1)²(x² + 1). It has a real zero at x = 1 with multiplicity 2, and two complex zeros at x = i and x = −i. On the Cartesian graph, the curve touches the x-axis at x = 1 and bounces back without crossing, while the complex zeros are invisible. In the complex plane, however, all four zeros are visible as distinct points.

Left: The Cartesian graph of f(x) = x⁴ − 2x³ + 2x² − 2x + 1 shows only the real zero at x = 1, where the curve touches but does not cross the x-axis (multiplicity 2). Right: The Argand diagram reveals all four zeros — the real zero at 1 (cyan) and the conjugate pair i and −i (pink), symmetric about the real axis.

Notice several critical features in the diagram. On the Cartesian graph, the curve's behavior at the real zero reflects its multiplicity: because x = 1 has even multiplicity, the graph touches and rebounds from the x-axis rather than passing through it. A zero with odd multiplicity would instead cross the axis. Meanwhile, the complex plane view reveals the hidden structure: the zeros i and −i form a conjugate pair reflected symmetrically across the real axis. This symmetry is not coincidental — it is mandated by the Complex Conjugate Zeros Theorem whenever the polynomial's coefficients are real.

SECTION 4

Mathematical Framework

The algebraic machinery for working with polynomial zeros rests on several interconnected results. We begin with the general form and the factorization theorems, then connect these to the behavior of complex conjugate pairs and their corresponding irreducible quadratic factors.

LINEAR FACTORIZATION OVER ℂ

f(x) = aₙ(x − c₁)(x − c₂)(x − c₃) ⋯ (x − cₙ)

where aₙ is the leading coefficient and c₁, c₂, …, cₙ are the n complex zeros (counted with multiplicity). Each factor (x − cₖ) is linear in x.

CONJUGATE PAIR → IRREDUCIBLE QUADRATIC

(x − (a + bi))(x − (a − bi)) = x² − 2ax + (a² + b²)

When a conjugate pair a ± bi (with b ≠ 0) appears, multiplying the two linear factors yields a quadratic with real coefficients and a negative discriminant (4a² − 4(a² + b²) = −4b² < 0). This quadratic is irreducible over ℝ — it cannot be factored further using only real numbers.

COMPLETE REAL FACTORIZATION

f(x) = aₙ × (linear factors from real zeros) × (irreducible quadratic factors from conjugate pairs)

Every polynomial with real coefficients can be factored into a product of linear factors (from real zeros) and irreducible quadratic factors (from complex conjugate pairs). This is the most refined factorization possible without leaving ℝ.

A crucial consequence is a parity constraint on the number of complex zeros. Since complex zeros of real-coefficient polynomials come in conjugate pairs, the number of non-real complex zeros must be even. For a polynomial of odd degree, there must be at least one real zero because you cannot account for all n zeros using only pairs. Conversely, a polynomial of even degree could theoretically have all non-real zeros (e.g., x² + 1 has zeros ±i and no real zeros). The relationship between the degree, the number of real zeros, and the number of complex conjugate pairs is summarized by: n = (number of real zeros, counting multiplicity) + 2 × (number of conjugate pairs).

💡

Degree-Parity Rule

An odd-degree polynomial with real coefficients must have at least one real zero. An even-degree polynomial may have zero real zeros. In both cases, the total number of complex (non-real) zeros is always even.

SECTION 5

Classifying Zeros by Degree

Understanding the possible distributions of real and complex zeros for polynomials of various degrees helps build intuition for graph analysis and factorization. The table below classifies the possible zero configurations for polynomials of degrees 2 through 5 with real coefficients. Each row shows a distinct scenario for how the n zeros might partition into real zeros and conjugate pairs.

Degree	Real Zeros	Complex Conjugate Pairs	Example Polynomial
2	2	0	x² − 5x + 6 = (x − 2)(x − 3)
2	0	1	x² + 4 → zeros ±2i
3	3	0	x³ − 6x² + 11x − 6 = (x−1)(x−2)(x−3)
3	1	1	x³ + x = x(x² + 1) → zeros 0, ±i
4	4	0	(x−1)(x−2)(x−3)(x−4)
4	2	1	(x−1)(x−2)(x²+1)
4	0	2	(x² + 1)(x² + 4) → zeros ±i, ±2i
5	5	0	(x−1)(x−2)(x−3)(x−4)(x−5)
5	1	2	x(x²+1)(x²+4) → zeros 0, ±i, ±2i

Possible zero configurations for polynomials of degrees 2–5 with real coefficients

Four polynomials illustrating different zero distributions. Top-left: a cubic with 3 real zeros. Top-right: a cubic with 1 real zero and 1 conjugate pair. Bottom-left: a quartic with 4 real zeros. Bottom-right: a quartic with 0 real zeros and 2 conjugate pairs — the graph never crosses the x-axis.

The bottom-right panel is particularly instructive: the quartic k(x) = x⁴ + 5x² + 4 = (x² + 1)(x² + 4) has no real zeros whatsoever. Its graph stays entirely above the x-axis because both irreducible quadratic factors are always positive for real x. Yet in the complex plane, all four zeros are clearly visible: ±i and ±2i. This demonstrates why the complex plane is essential for a complete accounting of polynomial zeros.

SECTION 6

Worked Example: Finding All Zeros

Let us find all zeros of f(x) = x⁴ − 6x³ + 17x² − 28x + 20, given that 1 + 2i is a zero. We will use the Complex Conjugate Zeros Theorem to identify a second zero, then factor systematically.

Find all zeros of f(x) = x⁴ − 6x³ + 17x² − 28x + 20

1
Step 1 — Apply the Conjugate Zeros Theorem

Since f(x) has real coefficients and 1 + 2i is a zero, the conjugate 1 − 2i must also be a zero. We now know two of the four zeros.

Zeros so far: 1 + 2i, 1 − 2i
2
Step 2 — Build the quadratic factor from the conjugate pair

Multiply the corresponding linear factors: (x − (1 + 2i))(x − (1 − 2i)). Expand using the difference-of-squares pattern: ((x − 1) − 2i)((x − 1) + 2i) = (x − 1)² − (2i)² = (x − 1)² + 4 = x² − 2x + 1 + 4 = x² − 2x + 5.

Quadratic factor: x² − 2x + 5
3
Step 3 — Polynomial long division

Divide f(x) = x⁴ − 6x³ + 17x² − 28x + 20 by x² − 2x + 5. First: x⁴ ÷ x² = x². Multiply: x²(x² − 2x + 5) = x⁴ − 2x³ + 5x². Subtract: (−6x³ + 17x²) − (−2x³ + 5x²) = −4x³ + 12x². Bring down: −4x³ + 12x² − 28x + 20. Next: −4x³ ÷ x² = −4x. Multiply: −4x(x² − 2x + 5) = −4x³ + 8x² − 20x. Subtract: (12x² − 28x) − (8x² − 20x) = 4x² − 8x. Bring down: 4x² − 8x + 20. Next: 4x² ÷ x² = 4. Multiply: 4(x² − 2x + 5) = 4x² − 8x + 20. Subtract: 0. The quotient is x² − 4x + 4.

f(x) = (x² − 2x + 5)(x² − 4x + 4)
4
Step 4 — Factor the remaining quadratic

The quotient x² − 4x + 4 is a perfect square: x² − 4x + 4 = (x − 2)². So x = 2 is a real zero with multiplicity 2.

Remaining zeros: x = 2 (multiplicity 2)
5
Step 5 — State the complete factorization and all zeros

The complete factorization over ℝ is f(x) = (x² − 2x + 5)(x − 2)². Over ℂ, this becomes f(x) = (x − (1 + 2i))(x − (1 − 2i))(x − 2)². The four zeros are: 1 + 2i, 1 − 2i, and 2 (with multiplicity 2). We can verify: 2 real zeros (counting multiplicity) + 2 complex zeros = 4 = degree.

All zeros: x = 1 + 2i, x = 1 − 2i, x = 2 (mult. 2)

⚠

Note on the Worked Example

The polynomial used in this worked example is f(x) = x⁴ − 6x³ + 17x² − 28x + 20, not the initially stated polynomial from Step 3. This was corrected to ensure 1 + 2i is truly a zero. On the AP exam, always verify a given zero by direct substitution before proceeding.

SECTION 7

Strategies, Strengths & Common Pitfalls

Working with complex zeros involves several powerful techniques, but each has limitations and common error traps. The table below compares the major zero-finding strategies you should have in your toolkit, along with their appropriate contexts and potential pitfalls.

Strategy	When to Use	Common Pitfall
Conjugate Zeros Theorem	Given one complex zero of a real-coefficient polynomial; immediately double it with the conjugate	Applying the theorem when coefficients are not all real — it does not hold for complex-coefficient polynomials
Rational Root Theorem	Integer coefficients; test p/q candidates to find rational real zeros before tackling the reduced polynomial	Assuming all real zeros are rational — irrational zeros (like √2) won't appear among p/q candidates
Synthetic/Long Division	After finding a zero (real or complex), reduce the degree; repeat until a quadratic remains	Arithmetic errors during division; always verify by multiplying quotient × divisor + remainder = original
Quadratic Formula	Final step: apply to the remaining degree-2 factor; complex zeros emerge when discriminant < 0	Forgetting the ± on √(b² − 4ac), or miscalculating the discriminant sign, leading to wrong zero type
Descartes' Rule of Signs	Narrow down possible numbers of positive/negative real zeros before testing candidates	The rule gives upper bounds only — sign changes = max possible real zeros of that sign (reduced by 2s)

Comparison of zero-finding strategies for polynomials with real coefficients

✦

KEY TAKEAWAY

Finding polynomial zeros is like solving a jigsaw puzzle where the Fundamental Theorem tells you exactly how many pieces to expect. The Conjugate Zeros Theorem, Rational Root Theorem, and Descartes' Rule of Signs function as edge-sorting strategies — each narrows the search space differently. The most efficient approach combines multiple strategies: use Descartes' Rule to estimate the distribution, the Rational Root Theorem to test for easy wins, and the Conjugate Zeros Theorem to exploit any given complex zero. Ultimately, every polynomial reduces to a sequence of quadratic and linear factors.

SECTION 8

Connections to Advanced Theory

The ideas you've learned in this lesson form the bridge between precalculus algebra and several deep areas of higher mathematics. The Fundamental Theorem of Algebra, while stated simply, requires tools from complex analysis or topology to prove rigorously. In a first course in abstract algebra, you'll encounter the concept of algebraic closure — the property that makes ℂ special among number fields. The real numbers ℝ are not algebraically closed (x² + 1 has no real root), but ℂ is: every non-constant polynomial with complex coefficients splits completely into linear factors over ℂ.

AP Precalculus Concept	Advanced Extension
Complex zeros come in conjugate pairs	Galois theory: the conjugation map z → z̄ is the nontrivial automorphism of ℂ/ℝ, and conjugate pairing reflects the structure of the Galois group
Multiplicity affects graph behavior	Calculus: a zero of multiplicity m means the first m − 1 derivatives also vanish there; Taylor expansion begins with the (x − c)ᵐ term
Factor Theorem: f(c) = 0 ⟹ (x − c) \| f(x)	Ring theory: polynomial rings over fields are principal ideal domains; the Factor Theorem is a special case of the division algorithm in R[x]
Degree n ⟹ exactly n complex zeros	Complex analysis: proven via Liouville's theorem or the argument principle; connects polynomial algebra to the topology of the complex plane

How AP Precalculus concepts connect to college-level mathematics

For the AP Precalculus exam, you don't need these advanced perspectives, but knowing they exist reveals why this material matters. The techniques you practice here — identifying zero structure, applying the conjugate pairs theorem, and performing complete factorizations — are the operational skills that underpin everything from signal processing (where polynomial roots govern filter behavior) to quantum mechanics (where characteristic polynomials determine energy eigenvalues). Mastering the fundamentals now sets a strong foundation for wherever your mathematical journey leads.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A polynomial f(x) of degree 5 with real coefficients has exactly three distinct real zeros. Which of the following could be the complete list of zeros of f?

A. −1, 0, 2, 3i
B. −1, 0, 2, 3i, −3i
C. −1, 0, 2, 1 + i, 2 − i
D. −1, 0, 2, i, 2i
E. −1, 0, 2, 1 + i, 1 − i, 3

Show answer (B)

The correct answer is B. A degree-5 polynomial has exactly 5 zeros (counting multiplicity), and complex zeros of real-coefficient polynomials must appear in conjugate pairs. Choice A has only 4 zeros listed (3i appears without its conjugate −3i), so the total count is wrong and the conjugate rule is violated. Choice B has 5 zeros with 3 real (−1, 0, 2) and one conjugate pair (3i, −3i), satisfying all requirements. Choice C lists 5 zeros but 1 + i and 2 − i are NOT conjugates of each other; the conjugate of 1 + i is 1 − i and the conjugate of 2 − i is 2 + i, so two complex zeros are missing. Choice D has 5 zeros but i appears without −i and 2i appears without −2i — two unpaired complex numbers violating the conjugate theorem. Choice E lists 6 zeros, exceeding the degree of the polynomial.

Question 2: BASIC CALCULATION

If 2 − 3i is a zero of a polynomial with real coefficients, what is the corresponding irreducible quadratic factor?

A. x² − 4x + 13
B. x² + 4x + 13
C. x² − 4x + 5
D. x² − 4x − 5
E. x² + 4x + 5

Show answer (A)

The correct answer is A. By the Conjugate Zeros Theorem, 2 + 3i is also a zero. The irreducible quadratic factor is (x − (2 − 3i))(x − (2 + 3i)) = ((x − 2) + 3i)((x − 2) − 3i) = (x − 2)² − (3i)² = (x − 2)² + 9 = x² − 4x + 4 + 9 = x² − 4x + 13. Choice B has +4x instead of −4x, which would correspond to zeros with real part −2, not 2. Choice C gives x² − 4x + 5, which would factor as (x − (2 − i))(x − (2 + i)), corresponding to the zero 2 − i not 2 − 3i. Choice D has a negative constant term, meaning the discriminant would be positive (16 + 20 = 36 > 0), so it would factor over the reals and is not irreducible. Choice E corresponds to zeros with real part −2 and imaginary part ±1.

Question 3: INTERMEDIATE

A degree-4 polynomial f(x) with real coefficients satisfies f(1) = 0, f(−1) = 4, and has the complex zero −1 + i. The leading coefficient is 1. Which of the following is f(x)?

A. (x − 1)(x² + 2x + 2)
B. (x − 1)²(x² + 2x + 2)
C. (x − 1)(x + 1)(x² + 2x + 2)
D. (x − 1)(x² + 2x + 2)(x² − 2x + 2)
E. (x − 1)²(x² − 2x + 2)

Show answer (B)

The correct answer is B. The complex zero −1 + i forces its conjugate −1 − i to also be a zero (Conjugate Zeros Theorem). The quadratic factor from this pair is (x − (−1 + i))(x − (−1 − i)) = (x + 1 − i)(x + 1 + i) = (x + 1)² + 1 = x² + 2x + 2. This accounts for 2 of the 4 zeros. We know x = 1 is a zero of multiplicity 2, contributing factor (x − 1)². Together, (x − 1)²(x² + 2x + 2) is degree 4 with leading coefficient 1. We verify f(0) = (−1)²(2) = 2, but we also check f(−1) = (−1 − 1)²(1 − 2 + 2) = 4 · 1 = 4, confirming the condition f(−1) = 4. Choice A is degree 3. Choice C gives f(−1) = (−2)(0)(1) = 0, not 4. Choice D is degree 5. Choice E uses x² − 2x + 2, which has zeros 1 ± i, not −1 ± i.

Question 4: APPLIED

5 pts

An engineer models a vibration system with the characteristic polynomial p(x) = x⁴ + 2x³ + 10x² + 18x + 9. (a) Use the Rational Root Theorem to identify any rational zeros of p(x). (b) Factor p(x) completely over the reals. (c) Find all complex zeros of p(x) and state each zero's multiplicity. (d) Explain what the presence of complex zeros tells the engineer about the physical system's behavior.

Show answer

Point 1 (Rational Root Theorem): Possible rational roots are ±1, ±3, ±9 (factors of 9 over factors of 1). Testing x = −1: (−1)⁴ + 2(−1)³ + 10(−1)² + 18(−1) + 9 = 1 − 2 + 10 − 18 + 9 = 0. So x = −1 is a zero. Testing x = −3: 81 − 54 + 90 − 54 + 9 = 72 ≠ 0. No other rational roots work. Point 2 (First factoring step): Divide p(x) by (x + 1) using synthetic division: coefficients 1, 2, 10, 18, 9. Bring down 1; multiply by −1: −1; add to 2: 1; multiply by −1: −1; add to 10: 9; multiply by −1: −9; add to 18: 9; multiply by −1: −9; add to 9: 0. Quotient: x³ + x² + 9x + 9. Factor by grouping: x²(x + 1) + 9(x + 1) = (x + 1)(x² + 9). So p(x) = (x + 1)²(x² + 9). Point 3 (Complete factorization over ℝ): p(x) = (x + 1)²(x² + 9). The quadratic x² + 9 has discriminant 0 − 36 = −36 < 0, so it is irreducible over ℝ. This is the complete real factorization. Point 4 (All complex zeros): x = −1 with multiplicity 2 (from (x + 1)²); x = 3i and x = −3i each with multiplicity 1 (from x² + 9 = (x − 3i)(x + 3i)). Total zeros counting multiplicity: 2 + 1 + 1 = 4 = degree. Point 5 (Physical interpretation): The complex zeros 3i and −3i are purely imaginary, corresponding to undamped oscillatory modes in the vibration system. In control theory and mechanical engineering, purely imaginary roots of a characteristic polynomial indicate sustained oscillation at a natural frequency (here proportional to 3). The real zero at −1 (with multiplicity 2) indicates an exponentially decaying component. The engineer would conclude the system exhibits both a transient decay and a persistent oscillation.

Question 5: CRITICAL THINKING

4 pts

A student claims: "A polynomial of degree 6 with real coefficients that has no real zeros must have exactly 3 pairs of complex conjugate zeros." (a) Explain whether this claim is always true, and justify your reasoning using the Fundamental Theorem of Algebra and the Conjugate Zeros Theorem. (b) Construct a specific degree-6 polynomial with real coefficients and no real zeros. (c) Could such a polynomial be written as a product of exactly two irreducible quadratic factors? Explain.

Show answer

Point 1 (Evaluating the claim): The claim is true. By the Fundamental Theorem of Algebra, a degree-6 polynomial has exactly 6 zeros in ℂ counting multiplicity. If there are no real zeros, all 6 zeros are non-real complex numbers. By the Conjugate Zeros Theorem (applicable since coefficients are real), non-real zeros come in conjugate pairs. Therefore 6 non-real zeros form exactly 3 conjugate pairs. Point 2 (Constructing an example): One such polynomial is f(x) = (x² + 1)(x² + 4)(x² + 9) = x⁶ + 14x⁴ + 49x² + 36. Each quadratic factor x² + k (k > 0) has discriminant −4k < 0 and thus no real zeros. The 6 complex zeros are ±i, ±2i, ±3i, forming 3 conjugate pairs. Point 3 (Two irreducible quadratic factors?): No, this is impossible. Two irreducible quadratic factors would give a polynomial of degree 2 + 2 = 4, not degree 6. To have degree 6 with no real zeros, you need at minimum 3 irreducible quadratic factors (since linear factors introduce real zeros). You could also have one irreducible quadratic with multiplicity 3, such as (x² + 1)³, which is still degree 6 with no real zeros. Point 4 (Multiplicity consideration): The student's claim holds even when zeros have multiplicity greater than 1. For example, (x² + 1)³ has zeros i and −i each with multiplicity 3. There are still 3 "pairs" in the sense that 6 non-real zeros = 3 conjugate pairs (though in this case it's the same pair repeated 3 times). The key insight is that 6 = 2 × 3, and the factor of 2 comes from the pairing requirement.

SUMMARY

Lesson Summary

Summary

The Fundamental Theorem of Algebra guarantees that every polynomial of degree n has exactly n zeros in the complex numbers when counted with multiplicity. For polynomials with real coefficients, the Complex Conjugate Zeros Theorem ensures that non-real zeros appear in conjugate pairs a + bi and a − bi. Each such pair produces an irreducible quadratic factor x² − 2ax + (a² + b²) with real coefficients and a negative discriminant.

When solving for all zeros of a polynomial, combine the Rational Root Theorem, synthetic division, the Conjugate Zeros Theorem, and the quadratic formula to systematically reduce and factor. Remember that polynomials of odd degree always have at least one real zero, while even-degree polynomials may have none. The Linear Factorization Theorem allows any polynomial to be written as a product of n linear factors over ℂ, providing the most complete factored form possible.