Factorisation by regrouping terms

12.2.2 Factorisation by regrouping terms

Description

Quick Overview

This section teaches the method of factorisation by regrouping terms in algebraic expressions, highlighting the process through specific examples.

Standard

Factorisation by regrouping involves rearranging terms in an expression to reveal common factors, enabling the expression to be rewritten as a product of simpler expressions. The section illustrates this method with examples such as 2xy + 2y + 3x + 3, emphasizing the importance of recognizing groupable terms.

Detailed

Detailed Summary

In this section, we explore the method of factorisation by regrouping terms, which is particularly useful in algebraic expressions where no single common factor is apparent. The process begins by identifying terms that can be grouped based on shared factors. For example, in the expression 2xy + 2y + 3x + 3, we can observe that the first two terms share 2y as a common factor, while the last two terms share 3. This allows us to factor each group individually:

  • 2xy + 2y = 2y(x + 1)
  • 3x + 3 = 3(x + 1)

This results in the expression being rewritten as (x + 1)(2y + 3). The significance of this method lies in its utility for expressions where obvious single factors do not exist, enabling a clearer pathway to factorisation. Furthermore, the technique encourages a deeper understanding of the relationship between terms and the common factors they may contain. Through multiple examples and explanations, students can develop skills to recognise how to rearrange terms effectively for factorisation.

Example : Factorise 5xy - 10y + 15 - 5x.

Solution:
Step 1
Check if there is a common factor among all terms. There is none.
Step 2
Think of grouping. Notice that the first two terms have a common factor of 5y:

5xy - 10y = 5y(x - 2)

What about the last two terms? Observe them. If you change their order to -5x + 15, it will come out:

-5x + 15 = -5(x - 3)

Step 3
Putting (a) and (b) together,

5xy - 10y + 15 - 5x = 5y(x - 2) - 5(x - 3) = 5y(x - 2) - 5(x - 3)

The factors of (5xy - 10y + 15 - 5x) are (5y - 5)(x - 2)

Key Concepts

  • Factorisation by Regrouping: A method to factor expressions by reorganizing them into groups that reveal common factors.

  • Identifying Common Factors: Look for similarities between terms to group potential factors.

  • Rearrangement Technique: The technique involves rearranging terms to facilitate easy grouping.

Memory Aids

🎡 Rhymes Time

  • To group and to factor is quite a fun game, rearranging the terms will lead you to fame!

πŸ“– Fascinating Stories

  • Once there was a wise math wizard who loved to solve puzzles. One day he faced a tricky expression, but he rearranged the numbers, grouping them cleverly until they revealed hidden treasuresβ€”common factors.

🧠 Other Memory Gems

  • RAP - Rearrange, Analyze, and Factor.

🎯 Super Acronyms

GFG - Group Factors Gradually.

Examples

  • Example: Factorising 2xy + 2y + 3x + 3 results in (x + 1)(2y + 3).

  • Example: Factorising 6xy - 4y + 6 - 9x leads to (3y - 2)(2x - 1).

Glossary of Terms

  • Term: Regrouping

    Definition:

    The process of rearranging terms in an expression to form manageable groups that reveal common factors.

  • Term: Common Factor

    Definition:

    A number or term that divides two or more numbers or terms evenly.

  • Term: Factorisation

    Definition:

    The process of breaking down an expression into its constituent factors.

  • Term: Irreducible Factors

    Definition:

    Factors that cannot be further broken down into simpler components.

Similar Question

Factorise 8ab - 4a + 12 - 4b.

Solution:
Step 1
Check if there is a common factor among all terms. There is none.
Step 2
Think of grouping. Notice that the first two terms have a common factor of 4:

4(2ab - a)

What about the last two terms? Observe them. If you change their order to -4b + 12, it will come out:

-4b + 12 = -4(b - 3)

Step 3
Putting (a) and (b) together,

8ab - 4a + 12 - 4b = 4(2ab - a) - 4(b - 3) = 4(2ab - a - b + 3)

The factors of (8ab - 4a + 12 - 4b) are (4)(2ab - a - b + 3)