Remainder Theorem Revision Notes for Grade 12 NSC Matric Mathematics

Remainder Theorem

What is the remainder theorem?

The remainder theorem is a powerful tool that helps us find the remainder when a polynomial is divided by a linear expression without actually performing long division.

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Definition: When a polynomial p(x) is divided by a linear expression (cx - d), the remainder equals p(d/c).

In mathematical terms: R = p(d/c)

This means you can find the remainder by substituting the value x = d/c directly into the original polynomial.

How the remainder theorem works

When we divide any polynomial p(x) by a linear expression in the form (cx - d), we can write this division as:

$p(x) = (cx - d) \times Q(x) + R$

Where:

Q(x) is the quotient (the result of division)
R is the remainder (a constant value)

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The remainder theorem tells us that instead of performing the full division, we can simply:

Set the linear divisor equal to zero: $cx - d = 0$
Solve for x: $x = \frac{d}{c}$
Substitute this value into the original polynomial: $R = p(\frac{d}{c})$

Key steps for applying the remainder theorem

The remainder theorem can be applied systematically by following a clear process:

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Step-by-Step Process:

Step 1: Identify the linear divisor in the form $(cx - d)$
Step 2: Find the value to substitute by solving $cx - d = 0$ , giving $x = \frac{d}{c}$
Step 3: Substitute $x = \frac{d}{c}$ into the original polynomial $p(x)$
Step 4: Calculate the result - this is your remainder

Worked example 1: Finding remainders

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Worked Example: Finding Remainders

Question: Use the remainder theorem to determine the remainder when $p(x) = 3x^3 + 5x^2 - x + 1$ is divided by the following linear polynomials:

$x + 2$
$2x - 1$
$x + m$

Solution:

Part 1: Division by $x + 2$

Rewrite as: $x - (-2)$ , so $d = -2$ and $c = 1$
Substitute $x = -2$ into $p(x)$ :
$p(-2) = 3(-2)^3 + 5(-2)^2 - (-2) + 1$
$p(-2) = 3(-8) + 5(4) + 2 + 1$
$p(-2) = -24 + 20 + 3 = -1$
Therefore, R = -1

Part 2: Division by $2x - 1$

Rewrite as: $2x - 1$ , so $d = 1$ and $c = 2$
Substitute $x = \frac{1}{2}$ into $p(x)$ :
$p(\frac{1}{2}) = 3(\frac{1}{2})^3 + 5(\frac{1}{2})^2 - (\frac{1}{2}) + 1$
$p(\frac{1}{2}) = 3(\frac{1}{8}) + 5(\frac{1}{4}) - \frac{1}{2} + 1$
$p(\frac{1}{2}) = \frac{3}{8} + \frac{5}{4} - \frac{1}{2} + 1$
$p(\frac{1}{2}) = \frac{3}{8} + \frac{10}{8} - \frac{4}{8} + \frac{8}{8} = \frac{17}{8}$
Therefore, R = 17/8

Part 3: Division by $x + m$

Rewrite as: $x - (-m)$ , so substitute $x = -m$
$p(-m) = 3(-m)^3 + 5(-m)^2 - (-m) + 1$
Therefore, R = -3m³ + 5m² + m + 1

Worked example 2: Using the remainder theorem to solve for unknowns

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Worked Example: Finding Unknown Coefficients

Question: Given that $f(x) = 2x^3 + x^2 + kx + 5$ divided by $2x - 3$ gives a remainder of $9\frac{1}{2}$ , use the remainder theorem to determine the value of k.

Solution:

Step 1: Apply the remainder theorem

From the divisor $2x - 3$ , we get $x = \frac{3}{2}$
We know that $f(\frac{3}{2}) = 9\frac{1}{2}$

Step 2: Set up the equation

$f(x) = 2x^3 + x^2 + kx + 5$
$f(\frac{3}{2}) = 2(\frac{3}{2})^3 + (\frac{3}{2})^2 + k(\frac{3}{2}) + 5$
$f(\frac{3}{2}) = 2(\frac{27}{8}) + (\frac{9}{4}) + k(\frac{3}{2}) + 5$
$f(\frac{3}{2}) = \frac{27}{4} + \frac{9}{4} + \frac{3k}{2} + 5$

Step 3: Solve for k

$\frac{19}{2} = \frac{27}{4} + \frac{9}{4} + \frac{3k}{2} + 5$
$\frac{19}{2} = \frac{36}{4} + \frac{3k}{2} + 5$
$\frac{19}{2} = 9 + \frac{3k}{2} + 5$
$\frac{19}{2} = 14 + \frac{3k}{2}$
$\frac{19}{2} - 14 = \frac{3k}{2}$
$-\frac{9}{2} = \frac{3k}{2}$
Therefore, k = -3

Exam tips

Understanding the remainder theorem is essential for exam success, but there are several key points to remember when applying it.

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Essential Exam Guidelines:

Always check your divisor is in the correct form $(cx - d)$ before applying the theorem
Be careful with negative signs when substituting values
Remember that the remainder is always a constant (not a polynomial)
When the remainder is zero, the divisor is a factor of the polynomial
Show all substitution steps clearly for full marks

Remember!

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Key Points to Remember:

The remainder theorem states: When $p(x)$ is divided by $(cx - d)$ , the remainder $R = p(\frac{d}{c})$
To apply the theorem: Set $cx - d = 0$ , solve for $x$ , then substitute into $p(x)$
The remainder is always a constant value, never a polynomial expression
If the remainder is zero, then $(cx - d)$ is a factor of $p(x)$
This method saves time compared to long polynomial division in exams

Remainder Theorem (Grade 12 NSC Matric Mathematics): Revision Notes