The Remainder Theorem
Aligned to A-APR.B.2 — Common Core State Standards for Mathematics.
What this lesson teaches
When a polynomial p(x) is divided by (x − a), the remainder equals p(a). So you can find the remainder just by evaluating the polynomial at a, without doing full division.
Worked examples
Example 1
Find the remainder when p(x) = x^3 − 2x^2 + 5x − 1 is divided by (x − 2). By the theorem, the remainder is p(2) = 8 − 8 + 10 − 1 = 9. So dividing leaves a remainder of 9.
Example 2
Use the theorem as a factor test. Is (x + 3) a factor of p(x) = x³ + 2x² − 5x − 6? The factor (x + 3) means c = −3, so evaluate p(−3) = −27 + 18 + 15 − 6 = 0. The remainder is 0, so yes, (x + 3) is a factor — no long division needed to find out.
Example 3
Common mistake: using the wrong sign for c. Dividing by (x + 4) means c = −4, because x + 4 = x − (−4). Evaluating p(4) answers a different question entirely. Write the divisor in the form (x − c) first, then read c off directly — for (x + 4) that gives c = −4, and for (x − 4) it gives c = 4.
Practice questions
- Find the remainder when p(x) = x^2 + 3x − 4 is divided by (x − 1).
- Use the theorem to find the remainder of x^3 − 8 divided by (x + 2).
- Is (x − 3) a factor of x^2 − 9? Explain using p(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
Cuando un polinomio p(x) se divide entre (x − a), el residuo es igual a p(a). Así puedes encontrar el residuo solo evaluando el polinomio en a, sin hacer la división completa.
Ejemplo 1: Encuentra el residuo cuando p(x) = x^3 − 2x^2 + 5x − 1 se divide entre (x − 2). Por el teorema, el residuo es p(2) = 8 − 8 + 10 − 1 = 9. Así que dividir deja un residuo de 9.
Ejemplo 2: Usa el teorema como prueba de factor. ¿Es (x + 3) un factor de p(x) = x³ + 2x² − 5x − 6? El factor (x + 3) significa c = −3, así que evalúa p(−3) = −27 + 18 + 15 − 6 = 0. El residuo es 0, así que sí, (x + 3) es factor — sin necesidad de hacer la división larga.
Ejemplo 3: Error común: usar el signo equivocado para c. Dividir entre (x + 4) significa c = −4, porque x + 4 = x − (−4). Evaluar p(4) responde a una pregunta totalmente distinta. Escribe primero el divisor en la forma (x − c) y lee c directamente — para (x + 4) eso da c = −4, y para (x − 4) da c = 4.
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