Remainder and Factor Theorems

Hello! Welcome back to our exploration of polynomials.

In our last lesson, you learned the mechanical processes of polynomial long division and synthetic division. We ended on a key observation: when a polynomial division results in a remainder of 0, it means the divisor is a factor of the dividend. For example, you saw that $(x^{3} - 7 x - 6) \div (x + 2)$ gives a remainder of 0, meaning $(x + 2)$ is a factor.

Today, we're going to formalise this idea and explore its powerful consequences. We will introduce two of the most useful theorems in introductory algebra: the Remainder Theorem and the Factor Theorem. These theorems provide an elegant shortcut, allowing us to find remainders and test for factors simply by substituting a value into a polynomial, completely bypassing the need for long division.

By the end of this lesson, you will be able to apply the Remainder and Factor Theorems to analyse polynomial functions, including finding unknown coefficients within them.

1. The Remainder Theorem: A Shortcut to the Remainder

Let's start with a fundamental question: Is there a way to find the remainder of a polynomial division without actually performing the division? The answer is yes, and the method is called the Remainder Theorem.

The Remainder Theorem states:

When a polynomial $f (x)$ is divided by a linear expression $(x - a)$ , the remainder is equal to the value of the polynomial at $x = a$ . In other words, the remainder is $f (a)$ .

Why does this work? Remember from the last lesson that any division can be expressed as:

f (x) = (x - a) \cdot Q (x) + R

where $f (x)$ is the dividend, $(x - a)$ is the divisor, $Q (x)$ is the quotient, and $R$ is the remainder.

If we substitute $x = a$ into this equation, watch what happens to the divisor term:

f (a) = (a - a) \cdot Q (a) + R

f (a) = (0) \cdot Q (a) + R

f (a) = R

The entire quotient part of the expression vanishes, leaving us with just the remainder!

Let's see this powerful idea explained with a concrete example.

IGCSE - Remainder Theorem and Factor Theorem

This video from the channel 'Y=mx+c' provides an excellent explanation of the logic behind the Remainder Theorem and demonstrates how to apply it.

Watch from 8:34 to 12:39. The presenter clearly explains that the goal is to make the divisor part of the equation zero. Pay attention to how he finds the remainder of a division by substituting the root of the divisor into the polynomial.

So, to find the remainder when a polynomial $f (x)$ is divided by $(x - a)$ , you simply need to calculate $f (a)$ . If the divisor is $(x + a)$ , you would calculate $f (- a)$ , since $(x + a)$ is zero when $x = - a$ .

2. The Factor Theorem: A Special Case of the Remainder Theorem

Now, what if the remainder, $f (a)$ , is equal to 0? This tells us that the polynomial divides perfectly, with nothing left over. This is the entire basis of the Factor Theorem.

The Factor Theorem states:

A linear expression $(x - a)$ is a factor of a polynomial $f (x)$ if and only if $f (a) = 0$ .

This is a direct consequence of the Remainder Theorem. If $f (a) = 0$ , it means the remainder is 0, which by definition means $(x - a)$ is a factor. Your preference for the term "root" is very relevant here: if $f (a) = 0$ , then $a$ is a root of the polynomial, and $(x - a)$ is a factor.

Let's watch a quick video that illustrates this with simple examples.

The Factor Theorem and The Remainder Theorem

The channel 'Maths Genie' offers a very clear and direct demonstration of how to use the Factor Theorem to check if a given linear expression is a factor of a polynomial.

Watch from 3:42 to 5:37. Notice how in both examples, showing that an expression is a factor simply requires substituting the corresponding root and verifying that the result is zero.

The image below shows both theorems in action on a single polynomial. Part (a) uses the idea of the Remainder Theorem (though implemented via synthetic division) to find $P (5)$ , and part (b) uses the Factor Theorem to show $(x - 2)$ is a factor because $P (2) = 0$ .

Polynomials: The Remainder and Factor Theorems Example — Caption: This image shows two applications for the polynomial $P(x) = x^3 - x^2 + x - 6$. On the left, the Remainder Theorem is used to find that the remainder upon division by $(x-5)$ is $P(5)=99$. On the right, the Factor Theorem is used to confirm that $(x-2)$ is a factor because $P(2)=0$.

Test your understanding!

Consider the polynomial $P (x) = x^{3} - x^{2} - 14 x + 24$ .

What is the remainder when $P (x)$ is divided by $(x - 1)$ ? Use the Remainder Theorem.
Use the Factor Theorem to determine if $(x - 2)$ is a factor of $P (x)$ .

Show answer

To find the remainder when dividing by $(x - 1)$ , we need to calculate $P (1)$ :
$P (1) = (1)^{3} - (1)^{2} - 14 (1) + 24 = 1 - 1 - 14 + 24 = 10$
The remainder is 10.
To check if $(x - 2)$ is a factor, we need to calculate $P (2)$ :
$P (2) = (2)^{3} - (2)^{2} - 14 (2) + 24 = 8 - 4 - 28 + 24 = 0$
Since $P (2) = 0$ , $(x - 2)$ is a factor of $P (x)$ .

3. Applying the Theorems to Solve Problems

The Remainder and Factor theorems are not just for finding remainders; they are powerful tools for solving more complex algebraic problems. A very common application is finding unknown coefficients in a polynomial when given information about its factors or remainders.

This type of problem requires you to use the theorems to set up a system of equations—usually simultaneous linear equations, which you've solved before.

IGCSE - Remainder Theorem and Factor Theorem

Let's watch a detailed walkthrough of an exam-style question. This problem asks to find two unknown coefficients, 'A' and 'B', in a cubic polynomial using clues about its factors and remainders.

Watch the segment from 17:05 to 22:36. Follow the strategy carefully: Translate the statement "is divisible by x-3" into the equation f(3) = 0 using the Factor Theorem. Translate the statement "leave the remainder of 20 when divided by x+1" into the equation f(-1) = 20 using the Remainder Theorem. See how these two equations are used to solve for A and B.

This example beautifully demonstrates the practical power of these theorems. By translating verbal statements about factors and remainders into simple algebraic equations, we can solve for unknowns that would be very difficult to find otherwise.

The video continues with a second part (from 22:36 onwards) where, having found the full polynomial, the presenter uses the known factor $(x - 3)$ and polynomial division to break the cubic down into a quadratic. This allows them to find the remaining roots of the cubic equation. This is a perfect illustration of the general strategy for factorising higher-degree polynomials, which we will focus on in our next lesson.

Conclusion

In this lesson, you've moved beyond the mechanics of polynomial division to understand the powerful theory that underpins it.

Key Takeaways:

The Remainder Theorem provides a shortcut to find the remainder of a division. To find the remainder when $f (x)$ is divided by $(x - a)$ , you simply calculate $f (a)$ .
The Factor Theorem is a special case of the Remainder Theorem. $(x - a)$ is a factor of $f (x)$ if and only if $f (a) = 0$ . This connects the roots of a polynomial to its factors.
These theorems can be used to set up equations to solve for unknown coefficients within a polynomial.

Aspect	Remainder Theorem	Factor Theorem
Definition	The remainder of $f (x) \div (x - c)$ is $f (c)$ .	$(x - c)$ is a factor of $f (x)$ if $f (c) = 0$ .
Application	Finds remainders and evaluates polynomials.	Finds factors and identifies roots.
Relationship	The Factor Theorem is the special case where the remainder, $f (c)$ , is zero.	General case for any linear divisor.

Preview of the Next Lesson:

So far, we have used the Factor Theorem to test if a given expression like $(x - 2)$ is a factor. But what if you aren't given any factors to test? How do you find a starting point to begin factorising a cubic or quartic polynomial from scratch?

In our next lesson, we will learn the Rational Root Theorem. This theorem provides a systematic method for generating a list of all possible rational roots of a polynomial. We can then use the Factor Theorem to test these candidates, find a factor, and begin breaking the polynomial down.

Recommended resources

1. The Remainder Theorem: A Shortcut to the Remainder

2. The Factor Theorem: A Special Case of the Remainder Theorem

3. Applying the Theorems to Solve Problems

Conclusion