The Remainder Theorem
Aligned to A-APR.B.2 — Common Core State Standards for Mathematics.
What this lesson teaches
For a polynomial p(x), the remainder when dividing by (x − a) equals p(a). So evaluating the polynomial at a quickly gives that remainder, and if p(a) = 0 then (x − a) is a factor.
Worked examples
Example 1
Let p(x) = x^2 − 5x + 6 and divide by (x − 2). Compute p(2) = 4 − 10 + 6 = 0, so the remainder is 0 and (x − 2) is a factor of p(x).
Example 2
Let p(x) = 2x³ + x² − 8x + 5 and divide by (x + 2). The theorem says the remainder is p(−2) = 2(−8) + 4 + 16 + 5 = −16 + 25 = 9. Since the remainder is not zero, (x + 2) is not a factor — one evaluation settles it, with no division at all.
Example 3
Common mistake: using the theorem with a divisor that is not linear. Dividing by (x² − 1) leaves a remainder that can be an entire linear expression, like 3x + 2, so there is no single number p(c) to evaluate. The Remainder Theorem applies only to divisors of the form (x − c). Factor (x² − 1) into (x − 1)(x + 1) and test each linear piece separately.
Practice questions
- Find the remainder when p(x) = x^2 + 3x − 4 is divided by (x − 1).
- Is (x + 2) a factor of p(x) = x^2 + x − 2? Use p(−2).
- Find p(3) for p(x) = x^3 − 2x^2 + x − 5 to get the remainder on division by (x − 3).
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 del Residuo
Para un polinomio p(x), el residuo al dividir entre (x − a) es igual a p(a). Así, evaluar el polinomio en a da rápidamente ese residuo, y si p(a) = 0 entonces (x − a) es un factor.
Ejemplo 1: Sea p(x) = x^2 − 5x + 6 y divide entre (x − 2). Calcula p(2) = 4 − 10 + 6 = 0, así que el residuo es 0 y (x − 2) es un factor de p(x).
Ejemplo 2: Sea p(x) = 2x³ + x² − 8x + 5 y divide entre (x + 2). El teorema dice que el residuo es p(−2) = 2(−8) + 4 + 16 + 5 = −16 + 25 = 9. Como el residuo no es cero, (x + 2) no es factor — una sola evaluación lo resuelve, sin dividir nada.
Ejemplo 3: Error común: usar el teorema con un divisor que no es lineal. Dividir entre (x² − 1) deja un residuo que puede ser toda una expresión lineal, como 3x + 2, así que no hay un solo número p(c) que evaluar. El teorema del residuo se aplica solo a divisores de la forma (x − c). Factoriza (x² − 1) como (x − 1)(x + 1) y prueba cada parte lineal por separado.
More Algebra lessons
Teach this lesson today
An AI tutor that guides with questions instead of just giving answers — all K–12 core subjects, every child in the family.
Start free — no card required