Remainder Theorem

Learn how to find the remainder when dividing a polynomial by a linear factor without performing long division.

Advanced25 minLesson

Definition

The Remainder Theorem states that when a polynomial f(x)f(x) is divided by (x−c)(x - c), the remainder equals f(c)f(c).
In other words:
f(x)=(x−c)⋅q(x)+rf(x) = (x - c) \cdot q(x) + r
where q(x)q(x) is the quotient and r=f(c)r = f(c) is the remainder.
Key insight: To find the remainder, simply substitute cc into the polynomial - no division needed!
Example: To find the remainder when f(x)=x3−4x+2f(x) = x^3 - 4x + 2 is divided by (x−2)(x - 2):
f(2)=23−4(2)+2=8−8+2=2f(2) = 2^3 - 4(2) + 2 = 8 - 8 + 2 = 2
The remainder is 22.

Try it now

According to the Remainder Theorem, when f(x)=x2+3x−4f(x) = x^2 + 3x - 4 is divided by (x−1)(x - 1), what is the remainder?

Worked Examples

Find the remainder when f(x)=x3+2x2−5x+3f(x) = x^3 + 2x^2 - 5x + 3 is divided by (x−1)(x - 1).

1

Identify the value of cc

Divisor is (x−1)(x - 1), so c=1c = 1

2

Substitute cc into f(x)f(x)

f(1)=13+2(1)2−5(1)+3f(1) = 1^3 + 2(1)^2 - 5(1) + 3 → f(1)=1+2−5+3f(1) = 1 + 2 - 5 + 3

3

Calculate

1+2−5+3=11 + 2 - 5 + 3 = 1 → Remainder = 11

Common Mistakes

Using the wrong sign for cc when the divisor is (x+a)(x + a)

Why it's wrong: The theorem uses (x−c)(x - c), so (x+2)(x + 2) means c=−2c = -2, not c=2c = 2.

Correct: Rewrite (x+a)(x + a) as (x−(−a))(x - (-a)) to correctly identify c=−ac = -a.

Forgetting to include the constant term when evaluating

Why it's wrong: Every term matters! The constant term is part of the polynomial.

Correct: Always write out all terms: f(c)=c3+2c2−5c+3f(c) = c^3 + 2c^2 - 5c + 3, including the +3+3.

Confusing the remainder with the quotient

Why it's wrong: The Remainder Theorem only gives the remainder, not the quotient polynomial.

Correct: Use synthetic division if you need both the quotient and remainder.

Interactive Visual

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y = x
Slope (m)1
Y-Intercept (b)0
b
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11
22

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Practice Problems

18 problems
Problem 1 of 18
Easy

According to the Remainder Theorem, when f(x)=x2+3x−4f(x) = x^2 + 3x - 4 is divided by (x−1)(x - 1), what is the remainder?

Why It Matters

The Remainder Theorem is a powerful shortcut that saves time and reduces errors:
  • Quick evaluation: Find f(c)f(c) by substitution instead of lengthy polynomial division
  • Root testing: If f(c)=0f(c) = 0, then (x−c)(x - c) is a factor (this is the Factor Theorem!)
  • Graphing: Find y-values at specific x-coordinates quickly
  • Engineering: Used in control systems and signal processing
  • Computer science: Polynomial error-correction codes rely on remainder calculations
Instead of performing synthetic or long division every time, the Remainder Theorem lets you evaluate in seconds!

Real World Applications

Polynomial Root Testing

The Remainder Theorem is fundamental for testing potential roots of polynomial equations.

Example:

To check if x=2x = 2 is a root of x3−5x2+8x−4=0x^3 - 5x^2 + 8x - 4 = 0, calculate f(2)f(2). If f(2)=0f(2) = 0, then x=2x = 2 is a root.

1Try It Yourself

You need to test if x=4x = 4 is a root of f(x)=x3−7x2+14x−8f(x) = x^3 - 7x^2 + 14x - 8.

Is x=4x = 4 a root?

Step 1: Write the mathematical expression

Calculate f(4)f(4):

Error Detection in Data Transmission

Computer scientists use polynomial remainder calculations in cyclic redundancy checks (CRC) to detect transmission errors.

Example:

When you download a file, your computer uses polynomial division to verify data integrity - any non-zero remainder indicates corruption.

2Try It Yourself

A simple polynomial check: verify that x2+3x+2x^2 + 3x + 2 is divisible by (x+1)(x + 1).

What is the remainder when dividing by (x+1)(x + 1)?

Step 1: Write the mathematical expression

Calculate f(−1)f(-1):

Key Takeaways

  • 1The Remainder Theorem: when f(x)f(x) is divided by (x−c)(x - c), the remainder is f(c)f(c)
  • 2For (x+a)(x + a), substitute c=−ac = -a (watch the sign!)
  • 3If f(c)=0f(c) = 0, then (x−c)(x - c) is a factor (Factor Theorem)
  • 4This method is faster than long division for finding remainders
  • 5The theorem can be used to find unknown coefficients in polynomials

Frequently Asked Questions

The Factor Theorem is a special case of the Remainder Theorem. The Remainder Theorem says f(c)f(c) equals the remainder. The Factor Theorem adds: if that remainder is zero, then (x−c)(x - c) is a factor.
The Factor Theorem is a special case of the Remainder Theorem. The Remainder Theorem says f(c)f(c) equals the remainder. The Factor Theorem adds: if that remainder is zero, then (x−c)(x - c) is a factor.
Not directly. The Remainder Theorem requires divisors of the form (x−c)(x - c). For (2x−3)(2x - 3), you would rewrite it as 2(x−32)2(x - \frac{3}{2}) and use c=32c = \frac{3}{2}, but the result needs adjustment.
Use synthetic division when you need the quotient polynomial, not just the remainder. The Remainder Theorem only gives you the remainder value.

Glossary

Remainder
The value left over after division; in polynomial division by (x−c)(x - c), it equals f(c)f(c)
Quotient
The result of division; when f(x)f(x) is divided by (x−c)(x - c), the quotient is a polynomial of degree one less
Factor
A polynomial that divides evenly into another polynomial (remainder = 0)
Root (or Zero)
A value cc where f(c)=0f(c) = 0; also where the graph crosses the x-axis

Formula Card

Remainder Theorem

If f(x)÷(x−c), then remainder =f(c)\text{If } f(x) \div (x - c), \text{ then remainder } = f(c)

The remainder when dividing polynomial f(x) by (x-c) equals f(c)

Polynomial Division Identity

f(x)=(x−c)⋅q(x)+rf(x) = (x - c) \cdot q(x) + r

Any polynomial can be written as divisor times quotient plus remainder

Factor Theorem (Special Case)

f(c)=0⇔(x−c) is a factor of f(x)f(c) = 0 \Leftrightarrow (x - c) \text{ is a factor of } f(x)

If f(c)=0, then (x-c) divides f(x) evenly with no remainder

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