Why bother with factorisation?
Imagine you have a puzzle where the pieces fit together perfectly only when you spot the hidden pairs. That's what factorisation does for a quadratic equation – it splits the problem into two simple pieces you can solve instantly.
💡 In Simple Words: A quadratic equation is a math sentence that looks like ax²+bx+c=0. By factorising, we rewrite it as (mx+n)(px+q)=0, then each little bracket can be set to zero to find the answers.
What is a quadratic equation?
A quadratic equation is any equation where the highest power of the variable (usually x) is 2. The standard form is ax² + bx + c = 0, where a, b, and c are numbers and a isn’t zero.
When does factorisation work?
Factorisation is like using a key that fits the lock. It works best when the quadratic can be broken into whole‑number factors – that is, when a, b, and c are nice integers and the equation factors without leaving fractions.
Step‑by‑step method
Follow these five easy steps, and you’ll be solving quadratics by factorisation in no time.
1. Write the equation in standard form
Make sure every term is on the left side and the right side is just 0. If you see something like x² = 5x + 6, bring everything left: x² – 5x – 6 = 0.
2. Identify a, b, c
Spot the numbers in front of x², x, and the constant term. In x² – 5x – 6 = 0, a = 1, b = –5, c = –6.
3. Factorise
Look for two numbers that multiply to a·c (that's a times c) and add up to b. For our example, a·c = 1·(–6) = –6. The pair –6 and +1 works because –6 + 1 = –5.
Rewrite the middle term using those two numbers: x² – 6x + x – 6 = 0. Then group: (x² – 6x) + (x – 6) = 0.
Factor each group: x(x – 6) + 1(x – 6) = 0.
Now pull out the common binomial (x – 6): (x + 1)(x – 6) = 0.
4. Set each factor to zero
When a product equals zero, at least one factor must be zero. So, x + 1 = 0 or x – 6 = 0.
5. Solve for x and check
From x + 1 = 0, we get x = –1. From x – 6 = 0, we get x = 6. Plug both back into the original equation to be safe – they both work.
Another worked example
Solve 2x² + 7x – 3 = 0 by factorisation.
- Step 1: Equation is already in standard form.
- Step 2: a = 2, b = 7, c = –3.
- Step 3: Find two numbers that multiply to a·c = 2·(–3) = –6 and add to b = 7. The pair 9 and –2 works because 9 × (–2) = –18 (oops, not –6). Try again. The correct pair is 9 and –2 for a·c = –6? Wait, 9 × (–2) = –18, not –6. Let's instead use the “splitting the middle term” trick: we need numbers that multiply to a·c = –6 and add to 7. Those numbers are 9 and –2? No, 9 + (–2) = 7, but 9 × (–2) = –18. So we made a mistake. The right pair is 9 and –2 only works for a·c = –18. Since we can't find integers, we try a different approach: factor out the common factor first if possible. Here, there is none. So factorisation with integers fails – we would need to use the quadratic formula. This shows why factorisation only works when the right pair exists.
Lesson: always check the product a·c first. If you can’t find a pair, move on to another method.
Quick summary table
| Step | What to do | Why |
|---|---|---|
| 1 | Write in standard form | All terms on one side makes factoring easier |
| 2 | Find a·c and look for two numbers | These numbers help split the middle term |
| 3 | Rewrite and group | Turns a single quadratic into a product of two binomials |
| 4 | Set each factor to zero | Zero product rule gives possible x values |
| 5 | Check solutions | Ensures no accidental mistakes |
Common pitfalls and how to avoid them
- Skipping the standard form: If any term sits on the right, you’ll miss the correct a·c product.
- Wrong pair of numbers: Double‑check both the sum (b) and the product (a·c). If one fails, try another pair.
- Forgetting to factor out the greatest common factor (GCF): Sometimes a simple number like 2 can be taken out first, making the rest easier.
- Not checking the answer: Plugging back catches sign errors quickly.
📝 Likely Exam Questions
- Solve x² – 4x – 12 = 0 by factorisation.
Answer: (x – 6)(x + 2)=0 → x=6 or x=–2. - Factorise 3x² – 14x + 8 = 0 and find the roots.
Answer: a·c=24, numbers –12 and –2 give –14 sum. 3x² –12x –2x +8 =0 → 3x(x –4) –2(x –4)=0 → (3x –2)(x –4)=0 → x=2/3 or x=4. - Explain why the equation 2x² + 5x + 3 = 0 cannot be solved by simple factorisation.
Answer: a·c=6, no pair of integers multiplies to 6 and adds to 5, so integer factorisation fails; use quadratic formula instead.