The Fundamental Theorem of Algebra
Aligned to N-CN.C.9 — Common Core State Standards for Mathematics.
What this lesson teaches
This is a (+) advanced standard. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n roots among the complex numbers, counting multiplicity. A quadratic (degree 2) therefore always has two roots.
Worked examples
Example 1
Consider x^2 + 2x + 5, which has degree 2. Solving gives x = (−2 ± √(4 − 20))/2 = (−2 ± 4i)/2 = −1 ± 2i — exactly two complex roots, just as the theorem predicts.
Example 2
Consider p(x) = x³ − 1, of degree 3. One root is obvious: x = 1. Factor it out: x³ − 1 = (x − 1)(x² + x + 1). The quadratic gives x = (−1 ± √(1 − 4))/2 = (−1 ± i√3)/2. Three roots for a cubic, exactly as the theorem guarantees — one real and two complex conjugates.
Example 3
Common mistake: counting distinct roots and concluding the theorem failed. p(x) = (x − 3)² has degree 2 but only one visible root, x = 3. The theorem counts roots WITH MULTIPLICITY: here x = 3 appears twice, so the count is still 2. Similarly x⁴ = 0 has the single root 0 with multiplicity 4. Add up the exponents on the factors, and the total always equals the degree.
Practice questions
- How many roots does a degree-5 polynomial have among the complex numbers?
- Show that x^2 + 1 has exactly two roots.
- Explain why a quadratic with a negative discriminant still has two roots.
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
El Teorema Fundamental del Álgebra
Este es un estándar avanzado (+). El Teorema Fundamental del Álgebra establece que todo polinomio de grado n tiene exactamente n raíces entre los números complejos, contando la multiplicidad. Por lo tanto, una cuadrática (grado 2) siempre tiene dos raíces.
Ejemplo 1: Considera x^2 + 2x + 5, que tiene grado 2. Al resolver se obtiene x = (−2 ± √(4 − 20))/2 = (−2 ± 4i)/2 = −1 ± 2i — exactamente dos raíces complejas, tal como predice el teorema.
Ejemplo 2: Considera p(x) = x³ − 1, de grado 3. Una raíz es evidente: x = 1. Factorízala: x³ − 1 = (x − 1)(x² + x + 1). La cuadrática da x = (−1 ± √(1 − 4))/2 = (−1 ± i√3)/2. Tres raíces para una cúbica, exactamente como garantiza el teorema — una real y dos complejas conjugadas.
Ejemplo 3: Error común: contar raíces distintas y concluir que el teorema falló. p(x) = (x − 3)² tiene grado 2 pero solo una raíz visible, x = 3. El teorema cuenta las raíces CON MULTIPLICIDAD: aquí x = 3 aparece dos veces, así que el conteo sigue siendo 2. Igual, x⁴ = 0 tiene la única raíz 0 con multiplicidad 4. Suma los exponentes de los factores y el total siempre iguala al grado.
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