Factorisation: To rewrite an expression using brackets
Understanding the Core Concept: From Expansion to Factorisation
Think of factorisation as "packing" an expression. When you expand, you distribute a factor across terms inside brackets. For example, $3(x + 4)$ expands to $3x + 12$. Factorisation is the opposite journey: looking at $3x + 12$ and asking, "What is common to both terms that I can 'factor out' and place outside brackets?" The number $3$ is common, so we rewrite it as $3(x + 4)$.
The general goal is to find the Highest Common Factor (HCF)[1]. For the expression $8y^2 + 12y$, the HCF of the coefficients $8$ and $12$ is $4$, and both terms contain at least one $y$. Therefore, the HCF is $4y$. Dividing each term by $4y$ gives $2y$ and $3$, resulting in the factorised form: $4y(2y + 3)$.
Key Methods of Factorisation for Different Expressions
The method used depends on the structure of the expression. Here are the primary techniques, progressing in complexity.
1. Factorising by Taking Out a Common Factor: This is the first and most important step to check for any expression. Look for numbers, variables, or combinations that appear in every term.
- Example: $6a^2b - 9ab^2$. The common factors are $3$, $a$, and $b$. So, HCF = $3ab$. Factorised: $3ab(2a - 3b)$.
- Example: $5(x+2) + y(x+2)$. Here, the common factor is the binomial $(x+2)$ itself. Factorised: $(x+2)(5 + y)$.
2. Factorising by Grouping: Used when an expression has four or more terms and no immediate common factor for all. We group terms that do share a factor.
- Example: $ax + ay + bx + by$. Group as $(ax + ay) + (bx + by)$. Factor each group: $a(x+y) + b(x+y)$. Now $(x+y)$ is common: $(x+y)(a + b)$.
3. Factorising Quadratics (Expressions of the form $ax^2 + bx + c$): This is a major topic. We look for two numbers that multiply to give $ac$ and add to give $b$.
- Example: Factorise $x^2 + 5x + 6$. We need two numbers that multiply to $6$ and add to $5$. They are $2$ and $3$. So: $x^2 + 5x + 6 = (x + 2)(x + 3)$.
- Example: Factorise $2x^2 - 7x + 3$. Here, $ac = 2 * 3 = 6$. We need two numbers that multiply to $6$ and add to $-7$. They are $-6$ and $-1$. Rewrite the middle term: $2x^2 - 6x - x + 3$. Now group: $(2x^2 - 6x) + (-x + 3) = 2x(x-3) -1(x-3) = (x-3)(2x-1)$.
4. Factorising Using Special Algebraic Identities: Recognizing these patterns speeds up the process immensely.
| Identity Name | Expanded Form | Factorised Form | Example |
|---|---|---|---|
| Difference of Two Squares | $a^2 - b^2$ | $(a - b)(a + b)$ | $x^2 - 9 = (x - 3)(x + 3)$ |
| Perfect Square Trinomial | $a^2 + 2ab + b^2$ | $(a + b)^2$ | $x^2 + 10x + 25 = (x + 5)^2$ |
| Perfect Square Trinomial | $a^2 - 2ab + b^2$ | $(a - b)^2$ | $4x^2 - 12x + 9 = (2x - 3)^2$ |
| Sum/Difference of Two Cubes[2] | $a^3 + b^3$ | $(a + b)(a^2 - ab + b^2)$ | $8 + y^3 = (2 + y)(4 - 2y + y^2)$ |
Applying Factorisation: Solving Equations and Simplifying Fractions
Factorisation is not an end in itself; it is a powerful tool for solving problems. Two of its most important applications are solving quadratic equations and simplifying algebraic fractions.
Solving Quadratic Equations by Factorisation: The principle is simple: if the product of two factors is zero, then at least one of the factors must be zero. This is called the Zero Product Property.
- Ensure the equation is in the standard form: $ax^2 + bx + c = 0$.
- Factorise the quadratic expression completely.
- Set each factor equal to zero and solve the resulting simple equations.
Step 1: Rearrange: $x^2 - 5x - 14 = 0$.
Step 2: Factorise: Find two numbers that multiply to $-14$ and add to $-5$. They are $-7$ and $+2$. So, $(x - 7)(x + 2) = 0$.
Step 3: Set each factor to zero: $x - 7 = 0$ gives $x = 7$. $x + 2 = 0$ gives $x = -2$.
The solutions are $x = 7$ and $x = -2$.
Simplifying Algebraic Fractions: Just like simplifying $\frac{12}{18}$ to $\frac{2}{3}$ by dividing numerator and denominator by their HCF (6), we simplify algebraic fractions by factorising first.
- Example: Simplify $\frac{2x^2 + 6x}{x^2 - 9}$.
Factor numerator: $2x(x + 3)$. Factor denominator using difference of squares: $(x - 3)(x + 3)$.
The fraction becomes $\frac{2x(x + 3)}{(x - 3)(x + 3)}$. Cancel the common factor $(x + 3)$ to get $\frac{2x}{x - 3}$, provided $x \neq -3$.
Important Questions About Factorisation
Factorisation is a foundational skill. While the quadratic formula can solve any quadratic equation, factorisation is often faster and more elegant for simple equations. More importantly, it is essential for simplifying expressions, finding common denominators in algebra, and solving higher-degree polynomial equations. It also helps in understanding the properties of graphs, such as finding the x-intercepts (roots) of a parabola.
First, double-check all common methods: taking out a common factor, grouping, and checking for special identities. Some quadratic expressions, like $x^2 + 4 = 0$, have no real factors (they are "prime" or "irreducible" over the real numbers). In such cases, you may use the quadratic formula, which will yield complex number solutions. Always ensure the expression is written in its simplest form before concluding it doesn't factorise.
Q3: How do I know which factorisation method to use first?
Follow a systematic checklist:
- Always look for a common factor first. This is the most important step and can simplify the expression dramatically.
- Count the number of terms.
- 2 terms: Check for difference of squares or sum/difference of cubes.
- 3 terms: Likely a quadratic. Try the method for factoring $ax^2 + bx + c$ or check for a perfect square trinomial.
- 4 or more terms: Try factorising by grouping.
- If you get stuck, rearrange the terms or consider if the expression is already fully factorised.
Footnote
[1] HCF (Highest Common Factor): Also known as GCF (Greatest Common Factor). The largest number or expression that is a factor of two or more numbers or terms. For example, the HCF of 12 and 18 is 6.
[2] Sum/Difference of Two Cubes: These are specific algebraic identities for expressions where two perfect cube terms are added or subtracted. The corresponding factorization formulas are given in the table above.