Factorising Quadratic Expressions
Factorising quadratic expressions means rewriting a quadratic such as x² + 5x + 6 as a product of two brackets, the reverse of expanding. This Edexcel IGCSE Maths guide covers the three cases you are examined on: taking out a common factor from a two-term expression, the difference of two squares, and finding the factor pair for a three-term quadratic x² + bx + c. Work through the examples, then use the randomly generated, auto-marked practice rooms below to factorise as many quadratics as you like, checking each answer by expanding the brackets back out.
What does factorising a quadratic mean?
A quadratic expression has an \(x^2\) term, for example \(x^2 + 5x + 6\). Factorising it means writing it as a product of brackets that multiply back to give the original. It is the exact reverse of expanding:
Expand: \((x+2)(x+3) = x^2+5x+6\)
Factorise: \(x^2+5x+6 = (x+2)(x+3)\)
There are three cases in the Edexcel IGCSE course: a two-term expression with a common factor, the difference of two squares, and a three-term quadratic \(x^2+bx+c\) that needs a factor pair. Always check a factorisation by expanding the brackets back out.
A picture of the factor pair
The factor pair for \(x^2+5x+6\) is the same as the side lengths of a rectangle whose area is \(x^2+5x+6\). The two sides are \((x+2)\) and \((x+3)\), and the four pieces inside add up to the quadratic.
Sides \((x+2)\) and \((x+3)\); areas \(x^2 + 3x + 2x + 6 = x^2 + 5x + 6\).
Worked examples
💡 Example 1: common factor
Factorise \(x^2 + 6x\).
What is happening?
Both terms contain \(x\). Take it out in front of a bracket and write what is left inside. Because this is an expression, the next line starts with an equals sign on a new line, not a one-line equation.
💡 Example 2: difference of two squares
Factorise \(x^2 - 49\).
What is happening?
Each term is a square and they are subtracted, so use \(x^2-a^2=(x-a)(x+a)\) with \(a=7\).
💡 Example 3: last sign plus
Factorise \(x^2 + 7x + 12\).
How to find the two numbers
Find two numbers that multiply to \(12\) (the last term) and add to \(7\) (the middle number). List the factor pairs of \(12\):
- \(1\) and \(12\): add to \(13\)
- \(2\) and \(6\): add to \(8\)
- \(3\) and \(4\): add to \(7\)
The last sign is plus and the middle is plus, so both numbers are positive: \((x+3)(x+4)\).
💡 Example 4: last sign minus
Factorise \(x^2 - 2x - 15\).
How to find the two numbers
The last sign is minus, so the two numbers have opposite signs. They must multiply to \(-15\) and add to \(-2\). Try the factor pairs of \(15\) with one sign each:
- \(+1\) and \(-15\): add to \(-14\)
- \(+3\) and \(-5\): add to \(-2\)
The pair \(+3\) and \(-5\) works, giving \((x-5)(x+3)\).
💡 Example 5: take out the common factor first
Factorise \(2x^2 - 18\).
What is happening?
Both terms share a factor of \(2\), so take it out first. That leaves \(x^2-9\), which is a difference of two squares, so factorise it as \((x-3)(x+3)\). Always factorise completely: the final answer keeps the \(2\) out in front.
where \(m+n=b\) and \(mn=c\). A sum of squares \(x^2+a^2\) does not factorise over integers.
🔑 Key Points
- Always take out a common factor first, then factorise what is left.
- For \(x^2+bx+c\), find two numbers that multiply to \(c\) and add to \(b\).
- Last sign plus: both brackets share the sign of the middle term. Last sign minus: the signs differ.
- Difference of two squares: \(x^2-a^2=(x-a)(x+a)\).
- Check every answer by expanding the brackets back out.
⚠️ Common Pitfalls
- Missing the common factor: \(x^2+6x\) is \(x(x+6)\), not \((x+2)(x+3)\).
- Sign slips: picking numbers with the wrong signs for the factor pair.
- Trying to factorise \(x^2+a^2\). A sum of two squares will not factorise.
- Not factorising fully: \(2(x^2+5x)\) still has a factor of \(x\); the complete answer is \(2x(x+5)\).