Introduction to Factorisation and the Factor Theorem

Welcome to one of the most powerful toolkits in Algebra! In this chapter, we are going to learn how to break down complex algebraic expressions into simpler "building blocks." Think of it like taking a finished Lego model and figuring out exactly which individual bricks were used to build it.

By the end of these notes, you will be able to factorise expressions, use the Factor Theorem to hunt for hidden factors, and even perform Algebraic Division on cubic expressions. These skills are essential for solving higher-level equations in your Pearson Edexcel International GCSE (Specification B) exams.

Note: This chapter focuses on cubic expressions and division. For a refresher on basic quadratic factorisation, see the chapter on "Linear, quadratic and cubic equations."

1. Simple Factorisation

Factorising is the inverse of expanding brackets. It means writing an expression as a product of its factors.

Common Factors

The simplest way to factorise is to look for the Highest Common Factor (HCF) in every term of the expression.

Example: Factorise \(3x^2 + 12x\)

1. Look at the numbers: The HCF of \(3\) and \(12\) is \(3\).
2. Look at the variables: The HCF of \(x^2\) and \(x\) is \(x\).
3. Put the HCF outside a bracket: \(3x(x + 4)\).

Quick Review: Difference of Two Squares

Keep an eye out for this special pattern: \(a^2 - b^2 = (a - b)(a + b)\).
For example, \(x^2 - 16\) becomes \((x - 4)(x + 4)\).

Key Takeaway: Always look for a common factor first before trying more complex methods!

2. The Factor Theorem

When we deal with cubic expressions (expressions where the highest power is \(x^3\)), we can't always see the factors easily. This is where the Factor Theorem becomes our best friend.

What is the Factor Theorem?

If we have a polynomial function \(f(x)\):
If \(f(k) = 0\), then \((x - k)\) is a factor of \(f(x)\).

This also works for more complex linear factors like \((ax - b)\):
If \(f(\frac{b}{a}) = 0\), then \((ax - b)\) is a factor of \(f(x)\).

How to use it (Step-by-Step):

1. Pick a small integer value for \(x\) (usually starting with \(1, -1, 2,\) or \(-2\)).
2. Substitute this value into the expression.
3. If the result is \(0\), you have found a factor!
4. If the result is not \(0\), try a different number.

Example: Show that \((x - 2)\) is a factor of \(f(x) = x^3 - 7x + 6\).

Substitute \(x = 2\) into the function:
\(f(2) = (2)^3 - 7(2) + 6\)
\(f(2) = 8 - 14 + 6\)
\(f(2) = 0\)
Since \(f(2) = 0\), the Factor Theorem tells us that \((x - 2)\) is a factor.

Did you know? The values you test are usually factors of the constant term (the number at the end). In the example above, \(6\) is the constant, so we test factors of \(6\) like \(\pm 1, \pm 2, \pm 3\).

3. Algebraic Division

Once you have found one linear factor (like \(x - 2\)) of a cubic expression, you need to find the remaining factors. We do this using Algebraic Long Division. This process is very similar to the long division you learned in primary school, just with more \(x\)'s!

Step-by-Step Process:

Let’s divide \(x^3 + 2x^2 - 5x - 6\) by \((x + 1)\).

Step 1: Divide
Divide the first term of the cubic (\(x^3\)) by the first term of the divisor (\(x\)).
\(x^3 \div x = x^2\). Write \(x^2\) on top.

Step 2: Multiply
Multiply \(x^2\) by the whole divisor \((x + 1)\).
\(x^2(x + 1) = x^3 + x^2\). Write this underneath the cubic.

Step 3: Subtract
Subtract \((x^3 + x^2)\) from the original expression.
\((x^3 + 2x^2) - (x^3 + x^2) = x^2\). Bring down the next term (\(-5x\)).

Step 4: Repeat
Now divide your new term (\(x^2\)) by \(x\), which gives \(+x\). Repeat the multiply and subtract steps until you reach the end. Since \((x + 1)\) is a factor, your final remainder should be \(0\).

Key Takeaway: Algebraic division turns a cubic expression into a linear factor multiplied by a quadratic expression.

4. Factorising Cubics Completely

The exam often asks you to "Factorise completely" or "Solve \(f(x) = 0\)" for a cubic. Here is the master plan:

1. Find one factor: Use the Factor Theorem by trying \(x = \pm 1, \pm 2\), etc.
2. Divide: Use Algebraic Division to divide the cubic by your factor. This gives you a quadratic.
3. Factorise the quadratic: Use standard quadratic methods (brackets or the formula) to break down the result from Step 2.
4. Write the final answer: List all three factors together.

Example: Factorise \(x^3 - 6x^2 + 11x - 6\) completely.

- Try \(x = 1\): \(1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0\). So \((x - 1)\) is a factor.
- Divide \(x^3 - 6x^2 + 11x - 6\) by \((x - 1)\) to get the quadratic \(x^2 - 5x + 6\).
- Factorise the quadratic: \(x^2 - 5x + 6 = (x - 2)(x - 3)\).
- Final Answer: \((x - 1)(x - 2)(x - 3)\).

5. Common Pitfalls to Avoid

Don't worry if this seems tricky at first! Most students make mistakes in these areas:

  • Negative Signs: When subtracting during algebraic division, be extremely careful with negatives. Subtracting a negative is the same as adding! \( -5x - (-x) = -4x \).
  • Missing Powers: If an expression is missing a power (e.g., \(x^3 - 7x + 6\) has no \(x^2\) term), write it as \(0x^2\) during division to keep your columns straight.
  • The "Sign Flip": If \(f(2) = 0\), the factor is \((x - 2)\). If \(f(-3) = 0\), the factor is \((x + 3)\). Don't mix up the signs!

Summary Box:
- Factor Theorem: \(f(k) = 0 \implies (x - k)\) is a factor.
- Division: Used to reduce a cubic to a quadratic.
- Solving: A cubic usually has three linear factors.