1. Introduction
Solving a quadratic equation by factorisation means expressing the quadratic expression in the form:
\( (x - p)(x - q) = 0 \)
Once the quadratic is written as a product of two linear factors, we can apply the zero product principle, which states:
If \(AB = 0\), then either \(A = 0\) or \(B = 0\).
This method is efficient when the quadratic expression splits nicely into factors.
2. Steps for Solving by Factorisation
The factorisation method involves the following steps:
- Write the equation in standard form \(ax^2 + bx + c = 0\).
- Multiply \(a\) and \(c\) (first and last coefficients).
- Find two numbers whose product = ac and sum = b.
- Split the middle term using these two numbers.
- Factor by grouping.
- Apply the zero product principle to get roots.
2.1. Example of Steps
Example: Solve \(x^2 + 5x + 6 = 0\) by factorisation.
- Here, \(a = 1, b = 5, c = 6\).
- Product \(ac = 6\).
- Find two numbers whose product is 6 and sum is 5 → 2 and 3.
- Split middle term: \(x^2 + 2x + 3x + 6 = 0\).
- Group terms: \(x(x + 2) + 3(x + 2) = 0\).
- Factor: \((x + 2)(x + 3) = 0\).
Thus, the solutions are:
\(x = -2\) and \(x = -3\)
3. Special Cases to Recognise
Some quadratics can be factored quickly if they fit common patterns.
3.1. Perfect Square Trinomials
Expressions of the form:
- \(a^2 + 2ab + b^2 = (a + b)^2\)
- \(a^2 - 2ab + b^2 = (a - b)^2\)
Example: \(x^2 + 6x + 9 = (x + 3)^2\)
3.2. Difference of Squares
\(a^2 - b^2 = (a - b)(a + b)\)
Example: \(x^2 - 16 = (x - 4)(x + 4)\)
3.3. Common Factors Before Factorisation
Always check for a common factor first.
Example: \(2x^2 + 4x = 2x(x + 2)\)
4. Worked Examples
Practice seeing different ways to factorise.
4.1. Example 1
Solve: \(x^2 - 7x + 12 = 0\)
Numbers with product 12 and sum -7 → -3, -4
Factor: \((x - 3)(x - 4) = 0\)
Solutions: \(x = 3, x = 4\)
4.2. Example 2
Solve: \(2x^2 + 7x + 3 = 0\)
Product = 6, sum = 7 → 6 and 1
Split: \(2x^2 + 6x + x + 3\)
Group: \(2x(x + 3) + 1(x + 3)\)
Factor: \((x + 3)(2x + 1) = 0\)
Solutions: \(x = -3, x = -\dfrac{1}{2}\)
4.3. Example 3
Solve: \(3x^2 - 12 = 0\)
Factor out common term:
\(3(x^2 - 4) = 0\)
Difference of squares:
\(3(x - 2)(x + 2) = 0\)
Solutions: \(x = 2, x = -2\)
5. Zero Product Principle
This principle is key to solving by factorisation. It states:
If \((A)(B) = 0\), then either \(A = 0\) or \(B = 0\).
Once a quadratic is written as \((x - p)(x - q) = 0\), immediately the solutions are:
\(x = p\) and \(x = q\)
6. Common Mistakes
- Choosing incorrect pair of numbers while splitting the middle term.
- Forgetting to factor out common terms first.
- Incorrect signs during grouping.
- Stopping after factorising but forgetting to set each factor equal to zero.
7. Quick Practice
Factorise and solve the following:
- \(x^2 + 8x + 15 = 0\)
- \(2x^2 - 5x + 3 = 0\)
- \(4x^2 - 25 = 0\)
- \(x^2 - x - 20 = 0\)
8. Summary
- Factorisation involves expressing a quadratic as a product of two linear factors.
- Find two numbers whose product is \(ac\) and sum is \(b\).
- Use the zero product principle to get solutions.
- Recognise special cases like perfect squares and difference of squares.