Ever wondered why some equations break down like LEGO bricks, letting you pull them apart piece by piece?
In simple words, a quadratic equation is an expression where the highest power of x is 2. By factorisation we rewrite it as a product of two simpler brackets, then set each bracket to zero to find x.
How to solve a quadratic equation by factorisation
When the question says “solve quadratic equation by factorisation”, it expects you to turn the equation into something that looks like (ax + b)(cx + d)=0. The word “factor” means to split a number or expression into parts that multiply together, just like splitting a pizza into slices.
Step‑by‑step method
- Bring everything to one side so the equation reads quadratic = 0.
- Look for two numbers that multiply to give the constant term (the number without x) and add to give the coefficient of x.
- Write the quadratic as a product of two linear factors using those numbers.
- Apply the Zero Product Property: if AB = 0, then either A = 0 or B = 0.
- Solve each simple equation for x.
graph TD
A[Start: Write equation as ax^2+bx+c=0] --> B[Find two numbers that multiply to a*c and add to b]
B --> C[Rewrite middle term using those numbers]
C --> D[Factor by grouping into (px+q)(rx+s)]
D --> E[Set each factor = 0]
E --> F[Solve for x]
F --> G[Answer]
Worked example 1
Solve x^2 - 5x + 6 = 0.
- It’s already in the form ax^2+bx+c=0, so no rearranging needed.
- We need two numbers that multiply to 6 (the constant) and add to -5 (the coefficient of x). Those numbers are -2 and -3 because (-2)×(-3)=6 and (-2)+(-3)=-5.
- Rewrite the middle term: x^2 - 2x - 3x + 6 = 0.
- Group: (x^2 - 2x) + (-3x + 6) = 0.
- Factor each group: x(x - 2) - 3(x - 2) = 0.
- Now we have a common factor (x‑2): (x - 3)(x - 2) = 0.
- Set each factor to zero: x - 3 = 0 → x = 3; x - 2 = 0 → x = 2.
Answer: x = 2 or x = 3.
Worked example 2 (with a leading coefficient)
Solve 2x^2 + 7x + 3 = 0.
- Multiply the leading coefficient (2) with the constant (3) → 6.
- Find two numbers that multiply to 6 and add to 7. Those are 1 and 6.
- Rewrite 7x as 1x + 6x: 2x^2 + 1x + 6x + 3 = 0.
- Group: (2x^2 + 1x) + (6x + 3) = 0.
- Factor each group: x(2x + 1) + 3(2x + 1) = 0.
- Common factor (2x + 1): (x + 3)(2x + 1) = 0.
- Set each to zero: x + 3 = 0 → x = -3; 2x + 1 = 0 → x = -½.
Answer: x = -3 or x = -½.
Quick reference table
| Step | What you do | Why it works |
|---|---|---|
| 1 | Move everything to one side | Creates the standard form ax^2+bx+c=0 |
| 2 | Find pair of numbers | Turns the middle term into two parts that can be grouped |
| 3 | Rewrite and group | Allows common factor to appear |
| 4 | Factor out common term | Gives product of two brackets |
| 5 | Apply zero‑product rule | If a product is zero, at least one factor is zero |
Common mistakes to avoid
- Forgetting to set the equation to zero before factoring.
- Choosing the wrong pair of numbers – always double‑check both product and sum.
- Mixing up signs; remember that a negative sign flips both product and sum.
- Leaving a common factor hidden, e.g., forgetting to factor out a greatest common divisor first.
📝 Likely Exam Questions
- Solve x^2 + 4x - 12 = 0 by factorisation.
Answer: (x + 6)(x - 2)=0 → x = -6 or x = 2. - Factorise 3x^2 - 14x + 8 = 0 and state the solutions.
Answer: (3x - 2)(x - 4)=0 → x = 2/3 or x = 4. - Explain why the Zero Product Property is essential in solving quadratic equations by factorisation.
Answer: It lets us split a product equal to zero into separate equations, each giving a possible value of x. - Given that one root of 2x^2 - kx + 8 = 0 is 2, find k using factorisation.
Answer: Substitute x=2: 2·4 - 2k + 8 = 0 → 8 - 2k + 8 =0 → 16 = 2k → k = 8.
#ICSE#Class 10#Mathematics#Quadratic Equations#Factorisation
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