Remainder Theorem

Understand the Remainder Theorem, its statement, how to apply it, and solve examples using a simple student-friendly approach.

1. Introduction

The Remainder Theorem is a useful shortcut to find the remainder when a polynomial is divided by a linear divisor of the form \(x - a\). Instead of performing long division, we can simply evaluate the polynomial at \(x = a\).

This theorem saves time and helps in solving many algebraic problems efficiently.

2. Statement of the Remainder Theorem

If a polynomial \(p(x)\) is divided by \(x - a\), then the remainder obtained is simply \(p(a)\).

In short:

Remainder = value of the polynomial at \(x = a\)

Example: Remainder when \(p(x) = x^2 - 3x + 2\) is divided by \(x - 1\) is:

\(p(1) = 1 - 3 + 2 = 0\)

3. How to Use the Remainder Theorem

To find the remainder when a polynomial is divided by \(x - a\), follow these steps:

  1. Identify the value of a from the divisor \(x - a\).
  2. Substitute this value into the polynomial.
  3. Simplify to get the result. This result is the remainder.

3.1. Example of Steps

Find the remainder when \(p(x) = 2x^2 + 5x - 1\) is divided by \(x - 2\).

Here, \(a = 2\).

Substitute into \(p(x)\):

\(p(2) = 2(2)^2 + 5(2) - 1 = 8 + 10 - 1 = 17\)

So, remainder = 17.

4. Worked Examples

Let’s apply the Remainder Theorem in different problems.

4.1. Example 1

Find the remainder when \(p(x) = x^2 + 4x + 4\) is divided by \(x - 2\).

\(p(2) = 2^2 + 4(2) + 4 = 4 + 8 + 4 = 16\)

Remainder = 16

4.2. Example 2

Find the remainder when \(p(x) = 3x^3 - 2x + 1\) is divided by \(x + 1\).

Divisor = \(x + 1 = x - (-1)\), so \(a = -1\).

\(p(-1) = 3(-1)^3 - 2(-1) + 1 = -3 + 2 + 1 = 0\)

Remainder = 0

4.3. Example 3

Find the remainder when \(p(x) = 5x^2 - 7x + 3\) is divided by \(x - 4\).

\(p(4) = 5(4)^2 - 7(4) + 3 = 80 - 28 + 3 = 55\)

Remainder = 55

5. Why the Remainder Theorem Works (Simple Explanation)

Any polynomial \(p(x)\) can be written in division form as:

\(p(x) = (x - a)Q(x) + R\)

If we substitute \(x = a\), the term \((x - a)Q(x)\) becomes 0.

So, \(p(a) = R\).

That is why evaluating \(p(a)\) gives the remainder directly.

6. Common Mistakes

  • Using the wrong value of \(a\) from the divisor. For \(x + 3\), \(a = -3\).
  • Incorrect substitution or sign errors.
  • Thinking this theorem works for divisors that are not of the form \(x - a\).
  • Forgetting to simplify exponents first.

7. Quick Practice

Use the Remainder Theorem to find the remainder:

  1. \(p(x) = x^3 + 1\), divided by \(x - 1\)
  2. \(p(x) = 2x^2 - 5x + 3\), divided by \(x + 2\)
  3. \(p(x) = x^3 - 4x\), divided by \(x - 2\)
  4. \(p(x) = 6x^2 + x - 1\), divided by \(x - 3\)

8. Summary

  • Remainder Theorem gives a quick way to find the remainder when dividing by \(x - a\).
  • The remainder is simply \(p(a)\).
  • Always identify \(a\) correctly before substitution.
  • This theorem helps in factorisation and leads to the Factor Theorem.