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Factorisation - Factorisation by Regrouping Terms

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Factorisation is the process of rewriting an algebraic expression as the product of its constituent factors. For example, the expression 2x+42x + 4 is factorised into 2(x+2)2(x + 2), where 22 and (x+2)(x + 2) are factors.

Regrouping terms is a technique used when all terms in an expression do not share a single common factor, but subsets of terms do. It involves rearranging and clustering terms into groups (usually pairs) to facilitate factorisation.

Visualizing the Grouping: When looking at a four-term expression like ax+ay+bx+byax + ay + bx + by, imagine it as two distinct blocks: (ax+ay)(ax + ay) and (bx+by)(bx + by). This mental separation helps identify the common factor within each 'block' independently.

Application of the Distributive Law in Reverse: Regrouping relies on the identity a(b+c)=ab+aca(b + c) = ab + ac. By reversing this, we transform the sum of terms ab+acab + ac back into the product a(b+c)a(b + c).

Identifying the 'Twin Parentheses': After the first stage of grouping, a successful regrouping will result in two terms sharing an identical binomial factor, such as p(x+y)+q(x+y)p(x + y) + q(x + y). Visualising these identical brackets as a single common entity allows you to pull them out as a single factor.

Managing Signs and Negative Factors: Careful attention must be paid to signs when grouping. For example, in the expression axaybx+byax - ay - bx + by, when you group the last two terms as (bxby)-(bx - by), the signs inside the bracket 'flip' to ensure the binomial factor (xy)(x - y) matches the first group.

Final Product Verification: The end result of factorisation by regrouping should always be a product of two or more algebraic expressions. You can verify the result by multiplying the factors back together to see if you reach the original expression.

📐Formulae

ab+ac=a(b+c)ab + ac = a(b + c)

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

axay+bxby=a(xy)+b(xy)=(a+b)(xy)ax - ay + bx - by = a(x - y) + b(x - y) = (a + b)(x - y)

a(x+y)b(x+y)=(ab)(x+y)a(x + y) - b(x + y) = (a - b)(x + y)

💡Examples

Problem 1:

Factorise the expression: 15xy6x+5y215xy - 6x + 5y - 2

Solution:

Step 1: Group the first two terms and the last two terms: (15xy6x)+(5y2)(15xy - 6x) + (5y - 2). \ Step 2: Identify the Highest Common Factor (HCF) for each group. For (15xy6x)(15xy - 6x), the HCF is 3x3x. For (5y2)(5y - 2), the HCF is 11. \ Step 3: Factor out the HCF from each group: 3x(5y2)+1(5y2)3x(5y - 2) + 1(5y - 2). \ Step 4: Notice the common binomial factor (5y2)(5y - 2). Factor it out from the whole expression: (5y2)(3x+1)(5y - 2)(3x + 1).

Explanation:

We grouped the terms such that each pair had a common factor. By factoring out 3x3x from the first pair, we created a binomial (5y2)(5y - 2) that matched the second pair exactly, allowing for a final step of common factor extraction.

Problem 2:

Factorise the expression: z7+7xyxyzz - 7 + 7xy - xyz

Solution:

Step 1: Rearrange the terms to group those with common variables: (zxyz)+(7xy7)(z - xyz) + (7xy - 7). \ Step 2: Factor out the HCF from the first group: z(1xy)z(1 - xy). \ Step 3: Factor out the HCF from the second group: 7(xy1)7(xy - 1). \ Step 4: Notice that (1xy)(1 - xy) and (xy1)(xy - 1) are opposites. To make them match, factor out a 1-1 from the second group: z(1xy)7(1xy)z(1 - xy) - 7(1 - xy). \ Step 5: Factor out the common binomial (1xy)(1 - xy): (1xy)(z7)(1 - xy)(z - 7).

Explanation:

This example demonstrates the importance of term rearrangement and sign management. By factoring out 7-7 instead of +7+7, we ensured the binomial factors were identical, which is essential for the regrouping method.