Fundamental Theorem of Algebra: A Basic Example

Theorem Statement

Every non-constant polynomial with complex coefficients has at least one complex root.

P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + ... + a₁z + a₀, (n ≥ 1, aₙ ≠ 0)

⇒ ∃ r ∈ ℂ such that P(r) = 0.

Concrete Example

Polynomial Without Real Roots

Consider the simple quadratic equation:

P(x) = x² + 9

Step 1: Check for Real Roots

Setting P(x) = 0:

x² + 9 = 0
x² = -9

No real number squared gives -9. In ℝ, this polynomial has no roots.

Step 2: Apply the Theorem

The Fundamental Theorem guarantees at least one complex root exists.

x² = -9 ⇒ x = ±√(-9) = ±3i

where i = √(-1) is the imaginary unit.

Step 3: Complete Factorization

Since degree = 2, there must be exactly 2 complex roots (counting multiplicity):

P(x) = x² + 9 = (x - 3i)(x + 3i)

Key Consequence Illustrated

The polynomial of degree 2 has exactly 2 complex roots: 3i and -3i.

These are complex conjugates, which always occur in pairs for polynomials with real coefficients.

Important Insight

Without complex numbers, the equation x² + 9 = 0 appears to have no solution.

By extending to complex numbers (ℂ), we get a complete solution and factorization.

This demonstrates why ℂ is called an algebraically closed field — polynomial equations always have solutions.

Visual Interpretation

The graph of y = x² + 9 never crosses the x-axis (no real roots), but in the complex plane, it intersects the plane z = 0 at two points: (0 + 3i) and (0 - 3i).

Summary: The Fundamental Theorem assures us that working in the complex numbers is sufficient to solve any polynomial equation completely. This example shows how what's impossible in reals becomes possible and complete in complexes.

Mathematical Notation: ℝ = real numbers, ℂ = complex numbers, i = imaginary unit (i² = -1).