How To Factorise Algebra | Unpacking Expressions

Algebraic factorisation allows us to break down complex expressions into their fundamental building blocks, much like disassembling a machine to understand its individual components. This process is essential for simplifying equations, solving for unknown variables, and gaining deeper insights into mathematical relationships.

Understanding Factorisation: The Core Idea

Factorisation is the reverse operation of expansion. When we expand an expression, we multiply factors together to remove parentheses; factorisation involves converting an expression into a product of its factors. For example, expanding `3(x + 2)` yields `3x + 6`, while factorising `3x + 6` returns `3(x + 2)`.

The goal is to express a polynomial as a product of two or more polynomials of lower degree. This skill is foundational in algebra, appearing in topics from solving quadratic equations to simplifying rational expressions. It helps us find the “roots” or x-intercepts of polynomial functions, where the expression equals zero.

Identifying Common Factors: The First Step

The most straightforward method of factorisation involves finding the Greatest Common Factor (GCF) among all terms in an expression. This GCF can be a number, a variable, or a combination of both.

To apply this method, identify the largest number that divides into all coefficients and the lowest power of any common variable present in every term. Once the GCF is determined, it is written outside a set of parentheses, and each original term is divided by the GCF to find the terms inside the parentheses.

• Example: Factorise `6x²y + 9xy² – 3xy`
  1. Identify the GCF of the coefficients (6, 9, -3), which is 3.
  2. Identify the GCF of the variables (`x²y`, `xy²`, `xy`). The lowest power of `x` is `x`, and the lowest power of `y` is `y`. So, the variable GCF is `xy`.
  3. The overall GCF is `3xy`.
  4. Divide each term by `3xy`:
     • `6x²y / 3xy = 2x`
     • `9xy² / 3xy = 3y`
     • `-3xy / 3xy = -1`
  5. The factorised expression is `3xy(2x + 3y – 1)`.

Factoring Special Forms: Difference of Two Squares

Certain algebraic expressions follow specific patterns that allow for rapid factorisation. The “difference of two squares” is a prominent example, defined by the pattern `a² – b² = (a – b)(a + b)`. This pattern only applies when two perfect square terms are separated by a subtraction sign.

Recognising perfect squares is key here. Numbers like 1, 4, 9, 16, 25, etc., are perfect squares, as are variables raised to an even power, like `x²`, `y⁴`, `z⁶`. The square root of `a²` is `a`, and the square root of `b²` is `b`.

• Example: Factorise `49x² – 81y²`
  1. Identify `a²` as `49x²`, so `a = √(49x²) = 7x`.
  2. Identify `b²` as `81y²`, so `b = √(81y²) = 9y`.
  3. Apply the formula `(a – b)(a + b)`: `(7x – 9y)(7x + 9y)`.

This method simplifies expressions quickly and is often used in conjunction with other factorisation techniques. For a deeper understanding of algebraic concepts, resources like Khan Academy offer extensive lessons and practice exercises.

Mastering Quadratic Trinomials: x² + bx + c

Quadratic trinomials are expressions of the form `ax² + bx + c`. When `a = 1`, the trinomial simplifies to `x² + bx + c`. To factorise these, we look for two numbers that satisfy two conditions:

• Their product equals the constant term `c`.
• Their sum equals the coefficient of the `x` term, `b`.

Once these two numbers, let’s call them `p` and `q`, are found, the trinomial can be factorised into `(x + p)(x + q)`. This method relies on systematically testing pairs of factors for `c`.

• Example: Factorise `x² + 7x + 10`
  1. Identify `b = 7` and `c = 10`.
  2. List factor pairs of `c` (10):
     • (1, 10) -> Sum = 11
     • (2, 5) -> Sum = 7
  3. The pair (2, 5) satisfies both conditions (product is 10, sum is 7).
  4. The factorised expression is `(x + 2)(x + 5)`.

Dealing with Negative Terms

When `c` is positive and `b` is negative, both `p` and `q` must be negative. When `c` is negative, one of `p` or `q` must be positive and the other negative. The sign of `b` then helps determine which number is larger in magnitude.

• Example: Factorise `x² – 3x – 18`
  1. Identify `b = -3` and `c = -18`.
  2. List factor pairs of `c` (-18) and their sums:
     • (1, -18) -> Sum = -17
     • (-1, 18) -> Sum = 17
     • (2, -9) -> Sum = -7
     • (-2, 9) -> Sum = 7
     • (3, -6) -> Sum = -3
     • (-3, 6) -> Sum = 3
  3. The pair (3, -6) satisfies both conditions.
  4. The factorised expression is `(x + 3)(x – 6)`.

Advanced Quadratic Factorisation: ax² + bx + c

When the leading coefficient `a` is not 1, factorising `ax² + bx + c` requires a slightly more involved process. Two common methods are the “Grouping Method” and “Trial and Error.” The grouping method is often more systematic.

The Grouping Method

This method transforms the trinomial into a four-term expression, which can then be factorised by grouping.

1. Multiply `a` and `c`. Let this product be `P = a c`.
2. Find two numbers, `p` and `q`, that multiply to `P` and add to `b`.
3. Rewrite the middle term `bx` as `px + qx`. This creates a four-term expression: `ax² + px + qx + c`.
4. Factorise by grouping the first two terms and the last two terms separately.
5. Identify the common binomial factor and factor it out.

• Example: Factorise `2x² + 11x + 12`
  1. `a = 2`, `b = 11`, `c = 12`. Calculate `P = a c = 2 12 = 24`.
  2. Find two numbers that multiply to 24 and add to 11. These numbers are 3 and 8.
  3. Rewrite `11x` as `3x + 8x`: `2x² + 3x + 8x + 12`.
  4. Group terms: `(2x² + 3x) + (8x + 12)`.
  5. Factor out the GCF from each group: `x(2x + 3) + 4(2x + 3)`.
  6. Factor out the common binomial `(2x + 3)`: `(2x + 3)(x + 4)`.

Here is a summary of common factorisation patterns:

Pattern Name	Form	Factorised Form
Common Factor	`ax + ay`	`a(x + y)`
Difference of Two Squares	`a² – b²`	`(a – b)(a + b)`
Perfect Square Trinomial (Sum)	`a² + 2ab + b²`	`(a + b)²`
Perfect Square Trinomial (Difference)	`a² – 2ab + b²`	`(a – b)²`

Factorising by Grouping: Four-Term Expressions

Factorisation by grouping is not limited to quadratic trinomials with `a ≠ 1`. It is a general technique applicable to any polynomial with four or more terms where pairs of terms share common factors. The core idea is to find common factors within smaller groups of terms, leading to a common binomial factor across the entire expression.

This method is particularly effective when dealing with cubic polynomials or other higher-degree polynomials that do not fit standard patterns. The order of terms can sometimes be rearranged to facilitate grouping, though this is not always necessary.

1. Arrange the terms, if necessary, so that the first two terms share a common factor and the last two terms share a common factor.
2. Factor out the GCF from the first pair of terms.
3. Factor out the GCF from the second pair of terms. Ensure that the remaining binomial factor is identical in both groups.
4. Factor out the common binomial factor.

• Example: Factorise `x³ + 2x² + 5x + 10`
  1. Group the terms: `(x³ + 2x²) + (5x + 10)`.
  2. Factor out GCF from the first group: `x²(x + 2)`.
  3. Factor out GCF from the second group: `5(x + 2)`.
  4. Notice the common binomial factor `(x + 2)`.
  5. Factor out `(x + 2)`: `(x + 2)(x² + 5)`.

Sometimes, after the first round of grouping, the resulting factors might be factorable further, such as a difference of two squares. Always check for further factorisation opportunities.

Educational policy and curriculum development often emphasize these core algebraic skills. The Department of Education outlines standards that include proficiency in algebraic manipulation.

The Sum and Difference of Cubes: A Deeper Look

Beyond squares, factorisation extends to cubes. There are specific formulas for the sum and difference of two cubes. These are less frequently encountered than quadratic factorisation but are important for a complete understanding of polynomial factorisation.

• Sum of Two Cubes: `a³ + b³ = (a + b)(a² – ab + b²)`
• Difference of Two Cubes: `a³ – b³ = (a – b)(a² + ab + b²)`

Identifying perfect cubes is the initial step, similar to identifying perfect squares. Numbers like 1, 8, 27, 64, 125 are perfect cubes, as are variables raised to powers that are multiples of three, such as `x³`, `y⁶`, `z⁹`.

• Example (Sum of Cubes): Factorise `8x³ + 27`
  1. Identify `a³` as `8x³`, so `a = ³√(8x³) = 2x`.
  2. Identify `b³` as `27`, so `b = ³√(27) = 3`.
  3. Apply the sum of cubes formula: `(2x + 3)((2x)² – (2x)(3) + 3²)`.
  4. Simplify: `(2x + 3)(4x² – 6x + 9)`.

• Example (Difference of Cubes): Factorise `y⁶ – 64`
  1. Identify `a³` as `y⁶`, so `a = ³√(y⁶) = y²`.
  2. Identify `b³` as `64`, so `b = ³√(64) = 4`.
  3. Apply the difference of cubes formula: `(y² – 4)((y²)² + (y²)(4) + 4²)`.
  4. Simplify: `(y² – 4)(y⁴ + 4y² + 16)`.
  5. Notice that `(y² – 4)` is a difference of two squares and can be factorised further: `(y – 2)(y + 2)`.
  6. The fully factorised expression is `(y – 2)(y + 2)(y⁴ + 4y² + 16)`.

Here is a quick reference for trinomial factorisation strategies:

Trinomial Type	Strategy	Key Steps
`x² + bx + c`	Find two numbers	Multiply to `c`, Add to `b`
`ax² + bx + c` (`a ≠ 1`)	Grouping Method	Multiply `ac`, find factors that add to `b`, rewrite `bx`, then group.
Perfect Square Trinomial	Recognise pattern	`a² ± 2ab + b² = (a ± b)²`

A Systematic Approach to Factorisation

When faced with an expression to factorise, a systematic approach helps ensure no method is overlooked and the expression is fully factorised. This involves a sequence of checks, starting with the simplest and progressing to more complex techniques.

1. Look for a Greatest Common Factor (GCF) first: Always extract any common factors from all terms. This simplifies the remaining expression, making subsequent steps easier.
2. Count the number of terms:
   • Two terms: Check for Difference of Two Squares (`a² – b²`), Sum of Two Cubes (`a³ + b³`), or Difference of Two Cubes (`a³ – b³`).
   • Three terms (trinomials):
     • If it’s `x² + bx + c`, find two numbers that multiply to `c` and add to `b`.
     • If it’s `ax² + bx + c` (`a ≠ 1`), use the grouping method or trial and error.
     • Check if it’s a Perfect Square Trinomial (`a² ± 2ab + b²`).
   • Four terms: Try factorisation by grouping.
3. Check for further factorisation: After applying a method, examine each resulting factor to see if it can be factorised again. For example, a difference of two squares might appear within a larger factor.
4. Verify by expansion: Multiply the factors back together to ensure they return the original expression. This step confirms the accuracy of your factorisation.

Following these steps provides a reliable framework for successfully factorising a wide range of algebraic expressions. Each technique builds upon the others, forming a cohesive set of tools for algebraic manipulation.