Class 8 - Factorisation by Grouping

Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication. We have seen several examples of factoring already. However, for this article, you should be especially familiar with taking common factors using the distributive property. For example:6x2+4x=2x(3x+2)

Table of Contents

• What is Factorisation by Grouping?
• Solved Examples On Factorisation by Grouping
  
  

What is Factorisation by Grouping?

Factorisation by grouping is a method where the terms of an algebraic expression are rearranged into groups, each group is factorised separately, and then a common factor is extracted from the groups

ac+ad+bc+bd=a(c+d)+b(c+d)=(a+b)(c+d)

Read more: Important Questions on Factorization - Class 8

Solved Examples On Factorisation by Grouping

Example 1: Basic grouping

Problem: Factorise: ax + ay + bx + by

Solution:

Step 1: Group the terms: (ax + ay) + (bx + by)

Step 2: Factor each group: a(x + y) + b(x + y)

Step 3: Take out the common binomial factor (x + y)

(x + y)(a + b)

Answer: ax + ay + bx + by = (x + y)(a + b)

Example 2: Grouping with subtraction

Problem: Factorise: 6xy - 4y + 6 - 9x

Solution:

Step 1: Rearrange terms: 6xy - 9x - 4y + 6

Step 2: Group: (6xy - 9x) + (- 4y + 6)

Step 3: Factor each group

3x(2y - 3) + (-2)(2y - 3)

= 3x(2y - 3) - 2(2y - 3)

Step 4: Common binomial factor

(2y - 3)(3x - 2)

Answer: 6xy - 4y + 6 - 9x = (2y - 3)(3x - 2)

Example 3: Factoring out negative from a group

Problem: Factorise: x² + xz - xy - yz

Solution:

Step 1: Group

(x² + xz) + (-xy - yz)

Step 2: Factor each group

x(x + z) + (-y)(x + z)

= x(x + z) - y(x + z)

Step 3: Common binomial factor

(x + z)(x - y)

(x + z)(x - y) = x² - xy + xz - yz.

Answer: x² + xz - xy - yz = (x + z)(x - y)

Example 4: Grouping with numerical coefficients

Problem: Factorise: 2a² + 6ab + a + 3b

Solution:

Step 1: Group

(2a² + 6ab) + (a + 3b)

Step 2: Factor each group

2a(a + 3b) + 1(a + 3b)

Step 3: Common binomial factor

(a + 3b)(2a + 1)

Answer: 2a² + 6ab + a + 3b = (a + 3b)(2a + 1)

Example 5: All terms with same variable

Problem: Factorise: x³ + x² + x + 1

Solution:

Step 1: Group: (x³ + x²) + (x + 1)

Step 2: Factor each group: x²(x + 1) + 1(x + 1)

Step 3: Common binomial factor: (x + 1)(x² + 1)

(x + 1)(x² + 1) = x³ + x + x² + 1 = x³ + x² + x + 1.

Answer: x³ + x² + x + 1 = (x + 1)(x² + 1)

Example 6: Grouping with squared terms

Problem: Factorise: a²b + ab² + a + b

Solution:

Step 1: Group: (a²b + ab²) + (a + b)

Step 2: Factor each group: ab(a + b) + 1(a + b)

Step 3: Common binomial factor: (a + b)(ab + 1)

Answer: a²b + ab² + a + b = (a + b)(ab + 1)