Why bother with factorising quadratics?
Ever seen a ball bounce back to the same spot? That back‑and‑forth motion is a lot like a quadratic equation returning to zero. If you can split the equation into easy pieces, the bounce becomes predictable – and the exam gets easier.
💡 In Simple Words: A quadratic equation is a math sentence with an x² term. By factorising, we rewrite it as a product of two simple brackets. Then we set each bracket to zero and solve for x. It’s like breaking a big LEGO wall into two smaller walls you can knock down one at a time.
What does ‘factorisation’ mean?
Factorisation (or factoring) is the process of expressing a number or expression as a product of its building blocks. For a quadratic, those building blocks are linear factors – expressions that look like (x + a). Think of it as turning a tangled rope into two neat strands.
Step‑by‑step method to solve a quadratic by factorisation
Follow these four moves and you’ll always land on the right answer.
1. Write the equation in standard form
Standard form means ax² + bx + c = 0. The right‑hand side must be zero, just like a balanced scale.
2. Look for two numbers that multiply to a·c and add to b
Here a, b, c are the coefficients (the numbers in front of x², x, and the constant). Those two numbers are the secret keys that let you split the middle term.
3. Split the middle term and factor by grouping
Rewrite bx as the sum of the two numbers you found, then pull out common factors from each pair.
4. Set each bracket to zero and solve
When a product equals zero, at least one factor must be zero – that’s the zero‑product property. Solve each simple linear equation for x.
Worked example 1
Problem: Solve 2x² + 7x + 3 = 0 by factorisation.
- Standard form is already there.
- a·c = 2 × 3 = 6. We need two numbers that multiply to 6 and add to 7. Those are 6 and 1.
- Rewrite 7x as 6x + x:
2x² + 6x + x + 3 = 0
Group: (2x² + 6x) + (x + 3) = 0
Factor each group: 2x(x + 3) + 1(x + 3) = 0
Now factor out the common binomial (x + 3): (2x + 1)(x + 3) = 0
- Set each factor to zero:
2x + 1 = 0 → x = -½
x + 3 = 0 → x = -3
- Check: Plug -½ and -3 back into the original equation – both work.
Worked example 2 (when a = 1)
Problem: Solve x² - 5x + 6 = 0.
- a·c = 1 × 6 = 6. Numbers that multiply to 6 and add to -5 are -2 and -3.
- Rewrite: x² - 2x - 3x + 6 = 0
- Group: (x² - 2x) + (-3x + 6) = 0
- Factor: x(x - 2) -3(x - 2) = 0
- Common binomial: (x - 2)(x - 3) = 0
- Solutions: x = 2 or x = 3
Quick checklist – what to remember
| Step | What to do |
|---|---|
| 1 | Bring everything to one side → ax²+bx+c=0 |
| 2 | Find two numbers that multiply to a·c and add to b |
| 3 | Split bx, factor by grouping |
| 4 | Use zero‑product property → set each factor = 0 |
| 5 | Check answers in the original equation |
Common pitfalls and how to avoid them
- Forgetting to move the constant term to the left side. It’s like leaving a loose screw in a chair – the whole thing can wobble.
- Choosing the wrong pair of numbers. Double‑check by multiplying and adding.
- Skipping the check step. A quick substitution catches sign errors.
📝 Likely Exam Questions
- Solve 3x² - 14x + 8 = 0 by factorisation.
Answer: (3x - 2)(x - 4)=0 → x=2/3 or x=4 - Factorise x² + 4x + 4 and state the solutions of the equation x² + 4x + 4 = 0.
Answer: (x + 2)²=0 → x = -2 - A quadratic equation has roots -1 and 5. Write the equation in standard form and solve it by factorisation.
Answer: (x + 1)(x - 5)=0 → x² - 4x - 5=0; solving gives x = -1, 5 - Explain why the zero‑product property works for quadratics.
Answer: If a product of two numbers equals zero, at least one factor must be zero; this lets us split a factored quadratic into two simple equations. - Solve 2x² - x - 3 = 0. Show each step.
Answer: a·c = -6, numbers 2 and -3 → 2x² + 2x - 3x - 3 = 0 → 2x(x + 1) -3(x + 1)=0 → (2x - 3)(x + 1)=0 → x=3/2 or x=-1