How to Solve Quadratic Equations by Factorisation
Solving quadratic equations by factorisation is frequently tested in GCSE and IGCSE Maths exams. This page shows how to rearrange quadratics into standard form, factorise accurately, and solve each factor to find all solutions. Worked examples and instant-feedback practice questions help reinforce correct method and exam technique.
What is a quadratic equation?
A quadratic equation is an equation involving \(x^2\). Quadratics can appear as three terms (\(ax^2+bx+c\)) or just two terms (\(ax^2+bx\)). Solving quadratic equations by factorisation works for both monic and non-monic forms: with two terms you can pull out a common factor; with three terms, use the sign rule and the table tennis method below. Always move everything to the left so it equals zero before you begin.
Step-by-step (general)
- Rearrange into standard form with \(=0\).
- Look at the sign of \(c\): it fixes the bracket signs before you hunt for any numbers.
- Factorise: take out a common factor for two terms; table tennis the factor pairs for three terms.
- Set each factor equal to zero.
- Solve for \(x\).
Step 1: The sign rule
For example, in \(2x^2+11x+12=0\): \(a=2\), \(b=11\) and \(c=12\).
Before hunting for numbers, look at the sign of \(c\). It tells you the bracket signs immediately: if \(c\) is positive, both brackets take the same sign as \(b\); if \(c\) is negative, the brackets always have one positive and one negative.
| Signs in the quadratic | Bracket signs | Example |
|---|---|---|
| \(c\) positive, \(b\) positive | Both positive: ( + )( + ) | \(x^2+7x+10=(x+2)(x+5)\) |
| \(c\) positive, \(b\) negative | Both negative: ( − )( − ) | \(x^2-7x+10=(x-2)(x-5)\) |
| \(c\) negative | One of each: ( + )( − ) | \(x^2+3x-10=(x+5)(x-2)\) |
When \(c\) is negative, the larger product takes the sign of \(b\): in \(x^2+3x-10\), the bigger number (5) is positive because \(b\) is \(+3\).
Step 2: The table tennis method
For non-monic quadratics (where \(a\) is not 1), write the factor pairs of \(a\) and \(c\) in two columns and multiply along the diagonals, like a ball bouncing across a table tennis net. The two diagonal products must combine to give \(b\). If they do not, bounce again with a different pair or order. When they do, read the brackets across the rows.
Factors of 2 on the left, factors of 12 on the right. The sign rule says both brackets are positive.
The bounce is the whole point: if the diagonal products miss \(b\), swap the right-hand pair round or pick a new factor pair and try again. Each attempt takes seconds, so a few bounces is normal.
Worked Examples
💡 Example 1: Monic (3 terms)
Solve \(x^2-5x+6=0\)
\(x^2-5x+6=0\)
\((x-2)(x-3)=0\)
\(x-2=0\) or \(x-3=0\)
\(x=2\) or \(x=3\)
What's happening?
Sign rule: \(c\) is \(+6\) and \(b\) is \(-5\), so both brackets are negative.
Now find the pair of 6 that sums to 5: that is 2 and 3.
💡 Example 2: Non-monic (3 terms)
Solve \(2x^2+5x-12=0\)
\(2x^2+5x-12=0\)
\((2x-3)(x+4)=0\)
\(2x-3=0\) or \(x+4=0\)
\(x=\tfrac{3}{2}\) or \(x=-4\)
What's happening?
Sign rule: \(c\) is \(-12\), so one bracket of each sign.
Table tennis the pairs of 2 and 12: the diagonals 8 and 3 must differ by 5, with the larger one positive to match \(+5x\).
💡 Example 3: Two terms (common factor)
Solve \(x^2+7x=0\)
\(x^2+7x=0\)
\(x(x+7)=0\)
\(x=0\) or \(x+7=0\)
\(x=0\) or \(x=-7\)
What's happening?
With two terms, pull out the common factor of \(x\) first.
Do not divide both sides by \(x\): that throws away the solution \(x=0\).
💡 Example 4: Rearrange first
Solve \(x^2=5x\)
\(x^2=5x\)
\(x^2-5x=0\)
\(x(x-5)=0\)
\(x=0\) or \(x=5\)
What's happening?
Move everything to the left before factorising.
Dividing both sides by \(x\) would give only \(x=5\) and lose \(x=0\).
🔑 Key Points
- Rearrange to standard form \(ax^2+bx+c=0\).
- Sign rule first: \(c\) positive means both brackets take \(b\)'s sign; \(c\) negative means one of each.
- Table tennis: diagonals multiply, rows make the brackets.
- Set each factor equal to zero, then solve. A repeated factor such as \((x-3)^2=0\) gives a single repeated solution.
⚠️ Common Pitfalls
- Forgetting \(=0\): always move all terms to the left before factorising.
- Reading the brackets along the diagonals: diagonals are only for multiplying.
- Ignoring the sign rule and hunting numbers with the wrong signs.
- Solving one factor only: you must solve both brackets.
- Dividing both sides by \(x\): this loses the solution \(x=0\).
FAQ: Solving quadratic equations by factorisation
When can I use factorisation? Use it when the quadratic can be written as a product of two linear brackets, e.g. \((ax+p)(bx+q)=0\). If it does not factorise nicely, try completing the square or the quadratic formula.
What is the table tennis method? A quick way to factorise non-monic quadratics: write the factor pairs of \(a\) and \(c\) in two columns, multiply along the diagonals, and bounce between pairs until the diagonal products combine to give \(b\). The brackets are then read across the rows.
What if there are only two terms? You are still solving quadratic equations by factorisation: take out the common factor first, e.g. \(x^2+7x=x(x+7)\).
Do I always get two solutions? Often yes, but a repeated factor such as \((x-3)^2=0\) gives a single (repeated) solution \(x=3\).