Remainder Theorem
Lesson Objectives
By the end of this lesson, students should be able to:
- State the Remainder Theorem.
- Apply the Remainder Theorem to evaluate remainders.
- Determine if a number is a root of a polynomial using substitution.
- Relate Remainder Theorem to Factor Theorem conceptually.
Introduction
Imagine you have a machine that divides polynomials and instantly tells you the remainder. That’s exactly what the Remainder Theorem does! Instead of doing long division, you can simply substitute a number into the polynomial. Let’s learn how!
Core Lesson Content
The Remainder Theorem states that: If a polynomial \(f(x)\) is divided by \(x - a\), the remainder is \(f(a)\).
Worked Example
Example 1: Find the remainder when \(f(x) = x^2 - 4x + 3\) is divided by \(x - 1\).
\(f(1) = (1)^2 - 4(1) + 3\)
\(= 1 - 4 + 3\)
\(= 0\)
The remainder is 0.
\(f(1) = (1)^2 - 4(1) + 3\)
\(= 1 - 4 + 3\)
\(= 0\)
The remainder is 0.
Example 2: Find the remainder when \(f(x) = 2x^3 + 3x^2 - x + 1\) is divided by \(x + 2\).
Use \(f(-2)\) because \(x + 2 = x - (-2)\)
\(f(-2) = 2(-2)^3 + 3(-2)^2 - (-2) + 1\)
\(= 2(-8) + 3(4) + 2 + 1\)
\(= -16 + 12 + 2 + 1 = -1\)
The remainder is -1.
Use \(f(-2)\) because \(x + 2 = x - (-2)\)
\(f(-2) = 2(-2)^3 + 3(-2)^2 - (-2) + 1\)
\(= 2(-8) + 3(4) + 2 + 1\)
\(= -16 + 12 + 2 + 1 = -1\)
The remainder is -1.
Example 3: If \(f(x) = x^3 - 2x^2 + x - 5\), find the remainder when divided by \(x - 3\).
\(f(3) = (3)^3 - 2(3)^2 + 3 - 5\)
\(= 27 - 18 + 3 - 5 = 7\)
The remainder is 7.
\(f(3) = (3)^3 - 2(3)^2 + 3 - 5\)
\(= 27 - 18 + 3 - 5 = 7\)
The remainder is 7.
Example 4: Determine if \(x + 1\) is a factor of \(x^3 + 3x^2 + 3x + 1\).
Use \(f(-1)\):
\(f(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1\)
\(= -1 + 3 - 3 + 1 = 0\)
Since the remainder is 0, \(x + 1\) is a factor.
Use \(f(-1)\):
\(f(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1\)
\(= -1 + 3 - 3 + 1 = 0\)
Since the remainder is 0, \(x + 1\) is a factor.
Example 5: A polynomial \(f(x) = 5x^4 - 3x^2 + x - 2\) is divided by \(x - 1\). Find the remainder.
\(f(1) = 5(1)^4 - 3(1)^2 + 1 - 2\)
\(= 5 - 3 + 1 - 2 = 1\)
The remainder is 1.
\(f(1) = 5(1)^4 - 3(1)^2 + 1 - 2\)
\(= 5 - 3 + 1 - 2 = 1\)
The remainder is 1.
Exercises
- \(f(x) = x^2 - 2x + 1\), find the remainder when divided by \(x - 1\).
- \(f(x) = x^3 + 2x^2 + x + 1\), find the remainder when divided by \(x + 1\).
- \(f(x) = 3x^3 - x^2 + 4x - 7\), find the remainder when divided by \(x - 2\).
- [WAEC] \(f(x) = x^3 - 6x + 7\); find the remainder when divided by \(x - 1\). (Past Question)
- \(f(x) = x^4 - 16\); what is the remainder when divided by \(x + 2\)?
- [NABTEC] \(f(x) = x^3 + 4x^2 + 5x + 2\); find the remainder when divided by \(x - 3\). (Past Question)
- \(f(x) = 2x^3 - x^2 + 3x - 1\), find the remainder when divided by \(x - 0.5\).
- [WAEC] \(f(x) = x^3 - 9x + 8\); find the remainder when divided by \(x + 2\). (Past Question)
- \(f(x) = x^2 - 5x + 6\); determine if \(x - 3\) is a factor.
- [JAMB] \(f(x) = 4x^3 + x - 6\); find the remainder when divided by \(x - 2\). (Past Question)
Conclusion / Recap
The Remainder Theorem is a fast way to evaluate the remainder of a polynomial division. If \(f(a) = 0\), then \(x - a\) is a factor. In the next lesson, we’ll build on this by exploring the Factor Theorem.
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