Why factorising quadratics matters
Imagine you have a puzzle where the pieces fit together perfectly only when you split the picture into two halves. In algebra, many quadratic equations become easy to solve once you break them into two simpler "pieces" – that’s factorisation.
💡 In Simple Words: A quadratic equation is a math sentence like ax²+bx+c=0. By factorising, you rewrite it as a product of two brackets, like (x‑p)(x‑q)=0. Then each bracket can be set to zero, giving the answers instantly.
What is a quadratic equation?
A quadratic equation is any equation where the highest power of the variable (usually x) is 2. The general shape looks like ax² + bx + c = 0, where a, b and c are numbers and a isn’t zero. Think of it like a hill – the graph of a quadratic makes a smooth “U” or upside‑down “U”.
Factorisation: the idea in plain language
Factorisation means writing a number or expression as a product of simpler factors. For quadratics, we want to express ax²+bx+c as (mx+n)(px+q). The magic comes from the zero product property – if the product of two things is zero, at least one of them must be zero. It’s just like if a river stops flowing, either the left dam or the right dam must be closed.
Step‑by‑step method
Here’s the recipe most students use when the equation can be factorised without a calculator:
Worked example 1: Simple quadratic
Solve x² – 5x + 6 = 0.
- Step 1: It is already in standard form (ax²+bx+c=0).
- Step 2: Look for two numbers that multiply to c (6) and add to b (‑5). Those numbers are ‑2 and ‑3 because (‑2)·(‑3)=6 and (‑2)+(‑3)=‑5.
- Step 3: Write the quadratic as (x‑2)(x‑3)=0. You can check by expanding: (x‑2)(x‑3)=x²‑5x+6.
- Step 4: Apply zero product property: set each bracket to zero.
- x‑2=0 → x=2
- x‑3=0 → x=3
- Answer: x = 2 or x = 3.
Worked example 2: Quadratic with a leading coefficient
Solve 2x² + 7x + 3 = 0.
- Step 1: Standard form – already there.
- Step 2: Multiply a (2) and c (3) → 6. Find two numbers that multiply to 6 and add to b (7). Those are 1 and 6.
- Step 3: Split the middle term: 2x² + 1x + 6x + 3 = 0.
- Step 4: Group: (2x² + 1x) + (6x + 3) = 0. Factor each group: x(2x+1) + 3(2x+1) = 0.
- Step 5: Factor out the common binomial (2x+1): (2x+1)(x+3) = 0.
- Step 6: Zero product property:
- 2x+1=0 → x = -½
- x+3=0 → x = -3
- Answer: x = -½ or x = -3.
Quick checklist before you start
- Make sure the equation is in the form ax²+bx+c=0. Move everything to one side if needed.
- Identify a·c (product of the first and last coefficients).
- Find two numbers that multiply to a·c and add to b.
- Split the middle term using those numbers, then factor by grouping.
- Write the final brackets and set each to zero.
Comparison of common solving methods
| Method | When it works best | Steps needed | Typical exam time |
|---|---|---|---|
| Factorisation | Coefficients are small and the quadratic easily splits into integer factors. | Find two numbers, split, group, factor, apply zero product. | 5‑7 minutes |
| Quadratic formula | Any quadratic, especially when factorisation is messy. | Plug a, b, c into x = [-b ± √(b²‑4ac)]/(2a). | 7‑10 minutes |
| Completing the square | When you need to derive the vertex form or solve equations with fractions. | Move c, add (b/2a)², rewrite as (x+… )² = …, take square root. | 8‑12 minutes |
📝 Likely Exam Questions
- Solve by factorisation: x² – 4x – 12 = 0.
Model answer: Numbers –6 and +2 → (x‑6)(x+2)=0 → x=6 or x=‑2. - Factorise and find roots: 3x² – 14x + 8 = 0.
Model answer: a·c = 24. Numbers –12 and –2 give –14. Split: 3x²‑12x‑2x+8 = 0 → 3x(x‑4)‑2(x‑4)=0 → (3x‑2)(x‑4)=0 → x=2/3 or x=4. - Given that (x‑5) is a factor of 2x² + kx – 15, find k and the other root.
Model answer: Substitute x=5 → 2·25 + 5k –15 =0 → 50 +5k –15=0 → 5k =‑35 → k=‑7. Polynomial becomes 2x²‑7x‑15 = (2x+3)(x‑5). Other root = –3/2. - Explain why the zero product property works for quadratics.
Model answer: If a product AB = 0, the only way a multiplication can give zero is if A=0 or B=0 (or both). Since factorised quadratics are written as a product of two linear terms, setting each term to zero yields all possible solutions. - Solve: 4x² – 12x + 9 = 0 by factorisation.
Model answer: Recognise perfect square: (2x‑3)²=0 → 2x‑3=0 → x=3/2.