Polynomial Remainder Theorem

TipLearning Objectives

By the end of this lesson, you’ll be able to:

  • Evaluate polynomials efficiently using the Remainder Theorem.
  • Use \(f(c)\) to determine the remainder when dividing by \((x - c)\).
  • Recognize when a number indicates a factor (remainder 0).

Key Ideas

The Remainder Theorem gives a shortcut to polynomial division.

If a polynomial \(f(x)\) is divided by \((x - c)\), then:

\[ \text{Remainder} = f(c) \]

This avoids long or synthetic division and works for any polynomial.

Common Problem Types

1. Finding a Remainder

Compute \(f(c)\) directly.

2. Testing Factors

If \(f(c) = 0\), then \((x - c)\) is a factor of the polynomial.

3. Evaluating Large Polynomials

Use the theorem to evaluate without substituting carelessly.

4. Linking to Synthetic Division

Synthetic division’s final value matches \(f(c)\).

Strategies

  • Substitute carefully: evaluate each term at \(x = c\).
  • Keep signs clear—negative \(c\) values cause common mistakes.
  • If \(f(c) = 0\), note that the divisor is a factor (Factor Theorem).
  • For polynomials with many terms, compute step-by-step to avoid arithmetic errors.
  • Use synthetic division whenever you want a quotient along with the remainder.

Worked Examples

Example 1 — Find a Remainder

Find the remainder when
\[ f(x) = 3x^4 - x^3 + 2x - 5 \]
is divided by \((x - 2)\).

Solution:
Evaluate \(f(2)\):

\[ \begin{split} f(2) &= 3(2^4) - (2^3) + 2(2) - 5 \\ &= 3(16) - 8 + 4 - 5 \\ &= 39 \end{split} \]

So the remainder is 39.

Example 2 — Checking a Factor

Is \((x - 3)\) a factor of \[ x^3 - 7x + 6? \]

Solution:
Compute \(f(3)\):

\[ \begin{split} f(3) &= 3^3 - 7(3) + 6 \\ &= 27 - 21 + 6 \\ &= 12 \end{split} \]

Since \(f(3) \neq 0\), \((x - 3)\) is not a factor.

WarningCommon Mistakes
  • Plugging in \(-c\) instead of \(c\) for a divisor of \((x - c)\).
  • Forgetting to evaluate every term of the polynomial.
  • Confusing the Remainder Theorem with the Factor Theorem.

Practice Problems

  1. Remainder of \((x^3 - 4x^2 + x - 6)\) divided by \((x - 1)\).
  2. Evaluate \(f(-2)\) for \(f(x) = 2x^3 - x + 5\).
  3. Is \((x + 1)\) a factor of \(x^3 + 3x^2 + 3x + 1\)?
  4. Remainder of \(5x^4 + x\) divided by \((x - 3)\).
  5. Evaluate \(f(4)\) for \(f(x) = x^3 - 2x + 1\).

1.
\[ f(1) = 1 - 4 + 1 - 6 = -8 \]

2.
\[ f(-2) = 2(-8) - (-2) + 5 = -16 + 2 + 5 = -9 \]

3.
Check \(f(-1)\):
\[ (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0 \]
Yes, remainder 0 → it is a factor.

4.
\[ f(3) = 5(3^4) + 3 = 5(81) + 3 = 408 \]

5.
\[ f(4) = 64 - 8 + 1 = 57 \]

Summary

  • Dividing by \((x - c)\) gives remainder \(f(c)\).
  • If \(f(c) = 0\), then \((x - c)\) is a factor.
  • Evaluating \(f(c)\) is usually faster than long or synthetic division.
  • The theorem works for any polynomial degree.
  • Synthetic division’s final value matches the remainder from the theorem.
  • For divisors of \((x - c)\), plug in \(c\), not \(-c\).
  • Check for factors by testing \(f(c) = 0\).
  • Evaluate polynomials step-by-step to avoid arithmetic slips.
  • Use synthetic division when you need both quotient and remainder.