Factorising Quadratics: Every Method You Need

Why Factorising Quadratics Is a Must-Know Skill

Quadratic expressions and equations appear throughout the IGCSE Maths syllabus. Whether you are solving equations, sketching graphs, finding turning points, or working with algebraic fractions, the ability to factorise a quadratic quickly and accurately is essential.

Factorising means breaking a quadratic expression into a product of two brackets. The reverse of expanding, factorising turns ax² + bx + c into something like (px + q)(rx + s). Once in this form, solving the equation becomes trivial — just set each bracket to zero.

There are several methods for factorising quadratics, and the best students know when to use each one. This guide covers all of them.

Method 1: Common Factor First

Before attempting any other method, always check for a common factor. If every term in the quadratic shares a common factor, take it out first.

For example:

• 3x² + 12x = 3x(x + 4)
• 2x² − 8x + 6 = 2(x² − 4x + 3) = 2(x − 1)(x − 3)

Taking out the common factor simplifies the expression inside the brackets and makes subsequent factorisation much easier. Skipping this step is one of the most frequent mistakes students make.

Method 2: Simple Factorisation (a = 1)

When the coefficient of x² is 1, the quadratic has the form x² + bx + c. You need to find two numbers that:

• Multiply to give c (the constant term)
• Add to give b (the coefficient of x)

Worked Example

Factorise x² + 7x + 12.

• You need two numbers that multiply to 12 and add to 7
• The pairs that multiply to 12 are: (1, 12), (2, 6), (3, 4)
• The pair that adds to 7 is (3, 4)
• Therefore: x² + 7x + 12 = (x + 3)(x + 4)

Handling Negatives

When c is negative, one of your numbers must be positive and the other negative:

• x² + 2x − 15: numbers multiply to −15 and add to 2 → (5) and (−3) → (x + 5)(x − 3)
• x² − x − 6: numbers multiply to −6 and add to −1 → (−3) and (2) → (x − 3)(x + 2)

When c is positive but b is negative, both numbers are negative:

• x² − 9x + 20: numbers multiply to 20 and add to −9 → (−4) and (−5) → (x − 4)(x − 5)

Tip: Write out all factor pairs of c systematically. This prevents you from missing the correct pair.

Method 3: Factorising When a ≠ 1

When the coefficient of x² is not 1, the quadratic has the form ax² + bx + c. This requires a different approach. The most reliable method for IGCSE students is splitting the middle term.

The Splitting Method

1. Multiply a × c
2. Find two numbers that multiply to ac and add to b
3. Rewrite the middle term (bx) as the sum of two terms using these numbers
4. Factor by grouping

Worked Example

Factorise 6x² + 11x + 3.

• Step 1: a × c = 6 × 3 = 18
• Step 2: Find two numbers that multiply to 18 and add to 11 → 9 and 2
• Step 3: Rewrite: 6x² + 9x + 2x + 3
• Step 4: Group: 3x(2x + 3) + 1(2x + 3) = (3x + 1)(2x + 3)

Another Example

Factorise 4x² − 8x − 5.

• Step 1: a × c = 4 × (−5) = −20
• Step 2: Numbers that multiply to −20 and add to −8 → −10 and 2
• Step 3: Rewrite: 4x² − 10x + 2x − 5
• Step 4: Group: 2x(2x − 5) + 1(2x − 5) = (2x + 1)(2x − 5)

Common mistake: When grouping, make sure both brackets are identical. If they are not, check your splitting — you may have the wrong number pair.

Method 4: Difference of Two Squares

This is a special case that is easy to spot and quick to factorise:

a² − b² = (a + b)(a − b)

Look for expressions where both terms are perfect squares and they are separated by a minus sign:

• x² − 25 = (x + 5)(x − 5)
• 4x² − 9 = (2x + 3)(2x − 3)
• 16x² − 1 = (4x + 1)(4x − 1)

Warning: The sum of two squares (a² + b²) cannot be factorised using real numbers. Only the difference works.

Method 5: The Quadratic Formula

When factorising by inspection or splitting is difficult — or when the question specifically asks you to solve rather than factorise — use the quadratic formula:

x = (−b ± √(b² − 4ac)) / 2a

This formula always works for any quadratic equation ax² + bx + c = 0, whether or not the expression factorises neatly.

Worked Example

Solve 2x² − 5x − 3 = 0 using the quadratic formula.

• a = 2, b = −5, c = −3
• b² − 4ac = 25 − 4(2)(−3) = 25 + 24 = 49
• x = (5 ± √49) / 4 = (5 ± 7) / 4
• x = 12/4 = 3 or x = −2/4 = −0.5

The Discriminant

The expression under the square root, b² − 4ac, is called the discriminant. It tells you about the nature of the roots:

• b² − 4ac > 0: Two distinct real roots (the quadratic crosses the x-axis twice)
• b² − 4ac = 0: One repeated root (the quadratic touches the x-axis at one point)
• b² − 4ac < 0: No real roots (the quadratic does not cross the x-axis)

IGCSE questions sometimes ask you to find the value of a constant that makes the discriminant equal to zero. This is a common “show that” or “find k” style question.

Method 6: Completing the Square

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. While not strictly factorising, it is closely related and frequently tested:

For x² + bx + c:

1. Halve the coefficient of x: p = b/2
2. Write (x + b/2)²
3. Subtract the extra term: (x + b/2)² − (b/2)² + c

Worked Example

Complete the square for x² + 6x + 2.

• Half of 6 is 3
• (x + 3)² = x² + 6x + 9
• So x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7

This form reveals that the minimum value of the expression is −7, occurring when x = −3.

Choosing the Right Method

• Always start by checking for common factors
• If a = 1, use simple inspection (Method 2)
• If a ≠ 1, use the splitting method (Method 3)
• If you recognise a difference of two squares, use Method 4
• If the question says “solve” and factorising looks difficult, use the formula (Method 5)
• If the question asks for minimum/maximum values or the form (x + p)² + q, use completing the square (Method 6)

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