The factor theorem Revision Notes for AQA GCSE Further Maths

The Factor Theorem

What is the factor theorem?

The factor theorem is a powerful tool that creates a direct connection between the factors of a polynomial and its roots. Understanding this theorem will help you solve polynomial equations more efficiently and factorise complex expressions.

A polynomial is an algebraic expression made up of terms with different powers of a variable. For example, $x^3 + 2x^2 - x - 2$ is a cubic polynomial (degree 3), while $x^2 - 5x - 6$ is a quadratic polynomial (degree 2). The degree of a polynomial is determined by its highest power.

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The factor theorem is one of the most practical tools in algebra because it allows you to work backwards from roots to find factors, or forwards from factors to find roots. This two-way relationship makes it incredibly versatile for solving complex polynomial problems.

The key relationship

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The Factor Theorem states a fundamental relationship in two parts:

Part 1: If $(x - a)$ is a factor of polynomial $f(x)$ , then $f(a) = 0$

Part 2: Conversely, if $f(a) = 0$ , then $(x - a)$ is a factor of $f(x)$

This means that when you substitute a value into a polynomial and get zero, you've found both a root of the equation and identified a linear factor.

This two-way relationship makes the factor theorem incredibly useful for both finding factors and solving polynomial equations. The beauty of this theorem lies in its simplicity - if you can make a polynomial equal zero by substituting a value, you've automatically found a factor.

Finding factors using substitution

To use the factor theorem effectively, you need to test different values by substituting them into the polynomial. When $f(a) = 0$ , you know that $(x - a)$ is a factor.

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Worked Example: Finding factors of $x^3 + 2x^2 - x - 2$

Given $f(x) = x^3 + 2x^2 - x - 2$ , we need to find values that make $f(x) = 0$ .

Testing $x = 1$ : $f(1) = 1^3 + 2(1)^2 - 1 - 2 = 1 + 2 - 1 - 2 = 0$

Since $f(1) = 0$ , we know that (x - 1) is a factor.

Testing $x = -1$ : $f(-1) = (-1)^3 + 2(-1)^2 - (-1) - 2 = -1 + 2 + 1 - 2 = 0$

Since $f(-1) = 0$ , we know that (x + 1) is a factor.

Testing $x = 2$ : $f(2) = 2^3 + 2(2)^2 - 2 - 2 = 8 + 8 - 2 - 2 = 12$

Since $f(2) ≠ 0$ , $(x - 2)$ is not a factor.

Testing $x = -2$ : $f(-2) = (-2)^3 + 2(-2)^2 - (-2) - 2 = -8 + 8 + 2 - 2 = 0$

Since $f(-2) = 0$ , we know that (x + 2) is a factor.

Therefore, $f(x) = (x - 1)(x + 1)(x + 2)$ , and the roots are $x = 1$ , $x = -1$ , and $x = -2$ .

Systematic approach to finding factors

When looking for factors, it's helpful to test values systematically. For a polynomial with integer coefficients, good values to try first are the factors of the constant term (the term without $x$ ).

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Worked Example: Working with $x^3 + 3x^2 - x - 3$

For this polynomial, the constant term is $-3$ , so we should test the factors of 3: $±1, ±3$ .

Testing $x = -1$ : $f(-1) = (-1)^3 + 3(-1)^2 - (-1) - 3 = -1 + 3 + 1 - 3 = 0$

Since $f(-1) = 0$ , we know (x + 1) is a factor.

The factors of the constant term $(-3)$ give us the values $±1, ±3$ to test. From our calculation, we found that testing $x = 1$ and $x = -3$ would also yield roots, giving us the complete factorization.

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Strategy for systematic testing:

List all positive and negative factors of the constant term
Test these values in order of increasing absolute value
Stop when you've found enough factors (remember a polynomial of degree $n$ has at most $n$ linear factors)

Using long division after finding one factor

Sometimes you may only find one linear factor initially. In such cases, you can use polynomial long division to find the remaining factors.

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Worked Example: Factorizing $x^3 - x^2 - 3x - 1$

If we find that $(x + 1)$ is a factor, we can divide the original polynomial by this factor to find what remains.

The long division process involves:

Dividing the first term of the polynomial by the first term of the factor
Multiplying the entire factor by this result
Subtracting from the original polynomial
Repeating until complete

Through this process, we discover that: $x^3 - x^2 - 3x - 1 = (x + 1)(x^2 - 2x - 1)$

The remaining quadratic factor $x^2 - 2x - 1$ can then be solved using the quadratic formula if needed.

Working with different types of factors

Finding quadratic factors

When given that one linear factor exists, you can find other factors systematically. The process involves using the relationship between the original polynomial and its known factors.

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Worked Example: Given $(x + 2)$ is a factor of $x^3 - 5x - 2$

Since we know $(x + 2)$ is a factor, we can write: $x^3 - 5x - 2 = (x + 2)(x^2 + bx + c)$

By expanding the right side and comparing coefficients, or by using polynomial division, we can find the values of $b$ and $c$ to complete the factorization.

This systematic approach ensures we don't miss any factors and helps verify our results.

Extension to fractional roots

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Extended Factor Theorem: If $f\left(\frac{b}{a}\right) = 0$ , then $(axe - b)$ is a factor of $f(x)$ .

This extension is particularly useful for polynomials with leading coefficients other than 1.

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Worked Example: Working with fractional factors

Consider the polynomial $3x^4 + 4x^3 - 16x^2 - 7x + 10$ .

If we suspect that $(3x - 2)$ might be a factor, we test $x = \frac{2}{3}$ :

$f\left(\frac{2}{3}\right)$ involves substituting $\frac{2}{3}$ for every $x$ in the polynomial. If this equals zero, then (3x - 2) is indeed a factor.

This technique expands our ability to find factors beyond simple integer values.

Key techniques summary

Here are the essential methods you need to master when working with the factor theorem:

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Essential Techniques:

Test systematic values: Start with factors of the constant term, including both positive and negative values.
Use the factor theorem: If $f(a) = 0$ , then $(x - a)$ is a factor.
Apply long division: When you find one factor, use polynomial long division to find the remaining factors.
Check your work: Always verify your factors by expanding them back to the original polynomial.
Consider fractional roots: For polynomials with leading coefficients other than 1, test fractional values.

Common exam techniques

Understanding how to apply the factor theorem in exam situations is crucial for success:

When asked to "show that (x - a) is a factor," substitute $x = a$ into the polynomial and demonstrate that the result is zero.
If asked to "find all factors," use the factor theorem systematically and apply long division as needed.
For "solve f(x) = 0," find all the factors first, then set each factor equal to zero.

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Common Mistake to Avoid: Don't forget to test negative values! Many students only test positive factors of the constant term and miss important roots. Always test both positive and negative values systematically.

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Key Points to Remember:

The factor theorem creates a direct link between factors and roots: if $(x - a)$ is a factor, then $x = a$ is a root, and vice versa.
Always test factors of the constant term first when looking for integer roots.
Use polynomial long division when you've found one factor but need to find the remaining ones.
The theorem works for all polynomial degrees, making it a versatile tool for solving complex equations.
Double-check your work by expanding your factors back to the original polynomial.
For polynomials with leading coefficients other than 1, consider testing fractional values using the extended form of the theorem.

The factor theorem (AQA GCSE Further Maths): Revision Notes