krit.club logo

Factorisation - Factorisation by Regrouping Terms

Grade 8ICSE

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

πŸ”‘Concepts

β€’

Regrouping is a factorisation technique used when the terms of an algebraic expression do not have a common factor across all of them, but can be arranged into smaller groups that share common factors. Visually, you look at the expression as segments, like two distinct blocks of terms.

β€’

The most common grouping for Grade 8 is the 2-2 split, where an expression with four terms is divided into two pairs. For example, in the expression ax+ay+bx+byax + ay + bx + by, the first pair is (ax+ay)(ax + ay) and the second pair is (bx+by)(bx + by).

β€’

A successful regrouping must result in a Common Binomial Factor. After factoring out the Highest Common Factor (HCF) from each group, the terms inside the parentheses must be identical. If you see a(x+y)+b(xβˆ’y)a(x + y) + b(x - y), the binomials do not match, and you must try a different grouping.

β€’

Order of terms matters: Sometimes terms are given in a scrambled order. If grouping the first two terms doesn't work, try rearranging the terms. For instance, in x2βˆ’bc+abβˆ’xcx^2 - bc + ab - xc, rearranging it to x2+abβˆ’bcβˆ’xcx^2 + ab - bc - xc or x2βˆ’xc+abβˆ’bcx^2 - xc + ab - bc may make the common factors more obvious.

β€’

Handling Negative Signs: If the third term in a four-term expression is negative, you must be careful when factoring out a negative sign from the second group. Visually, imagine pulling a minus sign out like a plug; this changes the signs of all terms remaining inside that set of brackets.

β€’

The final factored form is always written as the product of the common binomial factor and the sum/difference of the remaining monomial factors. It looks like this: (CommonΒ Binomial)Γ—(TermΒ 1+TermΒ 2)(\text{Common Binomial}) \times (\text{Term 1} + \text{Term 2}).

πŸ“Formulae

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

axβˆ’ay+bxβˆ’by=a(xβˆ’y)+b(xβˆ’y)=(a+b)(xβˆ’y)ax - ay + bx - by = a(x - y) + b(x - y) = (a + b)(x - y)

ax+ayβˆ’bxβˆ’by=a(x+y)βˆ’b(x+y)=(aβˆ’b)(x+y)ax + ay - bx - by = a(x + y) - b(x + y) = (a - b)(x + y)

πŸ’‘Examples

Problem 1:

Factorise the expression: 15xyβˆ’6x+5yβˆ’215xy - 6x + 5y - 2

Solution:

Step 1: Group the terms into two pairs: (15xyβˆ’6x)+(5yβˆ’2)(15xy - 6x) + (5y - 2). \ Step 2: Find the HCF for each group. For the first group, the HCF of 15xy15xy and 6x6x is 3x3x. For the second group, the only common factor is 11. \ Step 3: Factor out the HCFs: 3x(5yβˆ’2)+1(5yβˆ’2)3x(5y - 2) + 1(5y - 2). \ Step 4: Identify the common binomial (5yβˆ’2)(5y - 2) and factor it out: (5yβˆ’2)(3x+1)(5y - 2)(3x + 1).

Explanation:

We first paired terms with common numerical and literal coefficients. Since the second pair (5yβˆ’2)(5y - 2) was already identical to the binomial factor of the first pair, we factored out 11 to maintain the structure for the final binomial product.

Problem 2:

Factorise: zβˆ’7+7xyβˆ’xyzz - 7 + 7xy - xyz

Solution:

Step 1: Rearrange the terms to group terms with zz together and terms without zz together: (zβˆ’xyz)+(7xyβˆ’7)(z - xyz) + (7xy - 7). \ Step 2: Factor out zz from the first pair and 77 from the second pair: z(1βˆ’xy)+7(xyβˆ’1)z(1 - xy) + 7(xy - 1). \ Step 3: Notice that (1βˆ’xy)(1 - xy) and (xyβˆ’1)(xy - 1) are opposites. Factor out βˆ’1-1 from the second group to make them identical: z(1βˆ’xy)βˆ’7(1βˆ’xy)z(1 - xy) - 7(1 - xy). \ Step 4: Factor out the common binomial (1βˆ’xy)(1 - xy): (1βˆ’xy)(zβˆ’7)(1 - xy)(z - 7).

Explanation:

This problem requires rearranging terms first. A key step is recognizing that (xyβˆ’1)(xy - 1) is the negative of (1βˆ’xy)(1 - xy). By changing the sign of the outside factor from +7+7 to βˆ’7-7, we align the binomial factors to complete the factorisation.