Factorising
Breaking down expressions into their simplest factors
๐ Introduction
Factorising is the reverse process of expanding brackets. It involves expressing a mathematical expression as a product of its factors. This skill is essential for simplifying expressions, solving equations, and working with algebraic fractions.
๐ฏ Learning Objectives
By the end of this lesson, you will be able to:
- Identify and extract common factors from expressions
- Factorise quadratic expressions of the form $ax^2 + bx + c$
- Recognise and factorise the difference of two squares
- Apply factorising techniques to more complex expressions
๐ Key Concepts
Taking Out Common Factors
The simplest form of factorising involves identifying and extracting common factors from each term in an expression.
๐ Definition
Common Factor: A term that divides exactly into every term of an expression.
To factorise by taking out common factors:
- Identify the highest common factor (HCF) of all terms
- Express each term as the HCF multiplied by what remains
- Write the expression as the HCF multiplied by a bracket containing the remaining terms
Factorising Quadratics
A quadratic expression is of the form $ax^2 + bx + c$, where $a$, $b$, and $c$ are constants, and $a \neq 0$.
๐ Definition
Factorising Quadratics: Expressing a quadratic expression as a product of two linear factors.
For quadratics in the form $x^2 + bx + c$, we need to find two numbers that:
- Multiply to give $c$
- Sum to give $b$
Difference of Two Squares
๐ Definition
Difference of Two Squares: An expression in the form $a^2 - b^2$, which can always be factorised as $(a+b)(a-b)$.
๐ก Pro Tip
Always look to take out common factors first before attempting other factorising methods. This can often simplify the problem significantly.
๐ Worked Examples
๐ Example 1: Taking Out Common Factors
Question: Factorise $6x^3 + 9x^2 - 12x$.
Solution:
Identify the highest common factor (HCF) of all terms.
The HCF of $6x^3$, $9x^2$, and $-12x$ is $3x$ because:
- The HCF of 6, 9, and 12 is 3
- $x$ appears in all terms, with the lowest power being $x^1$
Express each term as the HCF multiplied by what remains.
Write the expression as the HCF multiplied by a bracket containing the remaining terms.
We have factorised the expression by taking out the common factor $3x$.
๐ Example 2: Factorising a Quadratic Expression
Question: Factorise $x^2 + 7x + 12$.
Solution:
We need to find two numbers that multiply to give 12 and add to give 7.
Let's list the factors of 12: (1, 12), (2, 6), (3, 4)
Check which pair adds to 7:
- $1 + 12 = 13$ (too large)
- $2 + 6 = 8$ (too large)
- $3 + 4 = 7$ (correct!)
The numbers we need are 3 and 4. So we can factorise the expression as:
Check: Expanding $(x+3)(x+4)$ gives $x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ โ
๐ Example 3: Difference of Two Squares
Question: Factorise $16x^2 - 25$.
Solution:
Identify this as a difference of two squares by expressing each term as a square.
Apply the difference of two squares formula: $a^2 - b^2 = (a+b)(a-b)$
๐ Example 4: Factorising More Complex Quadratics
Question: Factorise $2x^2 + 5x - 3$.
Solution:
For a quadratic in the form $ax^2 + bx + c$ where $a \neq 1$, we need to find two numbers $p$ and $q$ such that:
- $p \times q = a \times c = 2 \times (-3) = -6$
- $p + q = b = 5$
Find two numbers that multiply to give -6 and add to give 5.
The possible pairs of factors of -6 are: (-1, 6), (1, -6), (-2, 3), (2, -3)
Check which pair adds to 5:
- $-1 + 6 = 5$ (correct!)
- $1 + (-6) = -5$ (not right)
- $-2 + 3 = 1$ (not right)
- $2 + (-3) = -1$ (not right)
So our numbers are -1 and 6.
Rewrite the middle term using these numbers:
Now group the terms:
Factor out common terms from each group:
Check: Expanding $(2x - 1)(x + 3)$ gives $2x^2 + 6x - x - 3 = 2x^2 + 5x - 3$ โ
โ ๏ธ Common Mistakes
๐ซ Mistake 1: Missing Common Factors
What students often do wrong: Jumping straight to factorising quadratics without first checking for common factors.
Why it's wrong: This can make the problem much harder than necessary and often leads to errors.
How to avoid it: Always check for common factors first before applying other factorisation methods.
๐ซ Mistake 2: Incorrect Signs When Factorising
What students often do wrong: Getting the signs wrong in the brackets when factorising expressions like $x^2 - 5x + 6$.
Why it's wrong: The signs must be chosen so that when you expand the brackets, you get the original expression back.
How to avoid it: Be systematic when finding factors. For the constant term, find factors that multiply to give $c$. For the middle term, these factors must add to give $b$. Always check your answer by expanding the brackets.
๐ช Practice Problems
Try these problems to test your understanding:
๐ฏ Practice Question 1
Factorise completely:
(a) $4x^2y + 8xy^2$
(b) $x^2 - 9$
(c) $x^2 + 8x + 15$
๐ Show Solution
(a) $4x^2y + 8xy^2 = 4xy(x + 2y)$
(b) $x^2 - 9 = x^2 - 3^2 = (x+3)(x-3)$
(c) $x^2 + 8x + 15 = (x+3)(x+5)$
๐ฏ Practice Question 2
Factorise the following expressions:
(a) $3x^2 - 12$
(b) $x^2 - 5x - 14$
(c) $2x^2 + 7x - 4$
๐ Show Solution
(a) $3x^2 - 12 = 3(x^2 - 4) = 3(x+2)(x-2)$
(b) $x^2 - 5x - 14 = (x-7)(x+2)$
(c) $2x^2 + 7x - 4 = (2x-1)(x+4)$
๐ Summary
๐ฏ Key Takeaways
- Always look for common factors first when factorising any expression
- For quadratic expressions of the form $x^2 + bx + c$, find numbers that multiply to give $c$ and add to give $b$
- Remember the difference of two squares formula: $a^2 - b^2 = (a+b)(a-b)$
- For quadratics with a coefficient of $x^2$ (i.e., $ax^2 + bx + c$ where $a \neq 1$), the factorisation process is more complex and may require the grouping method
๐ What's Next?
Now that you understand factorising, you're ready to learn about negative and fractional indices, which will expand your algebraic toolbox and allow you to work with more complex expressions.