class 8 math case study factorisation worksheet

Class 8 Math Case Study Factorisation Worksheet | Download PDF

Class 8 Math Case Study Factorisation Worksheet

A class 8 math case study factorisation worksheet is designed to help students apply algebraic formulas through real-world mathematical problems. These case studies improve logical reasoning and build strong conceptual understanding. Moreover, they strengthen preparation for school-level assessments.

Benefits of Practicing Factorisation Worksheets

Practicing factorisation worksheets enables students to simplify algebraic expressions easily. It also supports learning for Case Study math questions for class 9 and math case study questions class 9. Additionally, consistent practice helps boost speed and confidence in solving problems.

Developing Skills Through Case Studies

Students should regularly practice math case study questions from worksheets and NCERT books. Furthermore, repeated exercises help in mastering factorisation techniques efficiently.

Case Study 2: Optimising Classroom Flooring with Algebraic Factorisation

A Class 8 mathematics club at Riverside Academy has been asked to design and cost a new floor covering for their multipurpose classroom. The head teacher wants a central rectangular area covered with a special anti-slip tile and a surrounding corridor with cheaper tiles. The students model the central rectangle’s length by \((2a+5)\) metres and its width by \((a-1)\) metres, where \(a\) is a positive integer chosen based on the classroom modular units. They must determine the total number of special tiles required (area in square metres), simplify algebraic expressions arising during budgeting, and investigate how small adjustments to the length and width (by adding or removing modular units) change the area and hence the cost.

During the project the students apply factorisation by common factors and grouping, and the standard identities

\((a+b)^2=a^2+2ab+b^2,\qquad (a-b)^2=a^2-2ab+b^2,\qquad (a+b)(a-b)=a^2-b^2,\)

to simplify expansions, compute differences of areas, and factor quadratic expressions that appear in supplier quotations. They also check algebraic manipulations by substituting sample integer values for \(a\) to verify practical feasibility. The exercise stresses careful symbolic manipulation (no unjustified cancellations), strategic grouping to factor polynomials, and using identities to convert between expanded and factorised forms for quick mental checks during on-site measurements.

1. The exact algebraic expression for the area (in square metres) of the central special-tiled region is:
Solution:

Area of a rectangle \(=\) length \(\times\) width. Thus

\(\text{Area}=(2a+5)(a-1)=2a^2+5a-2a-5=2a^2+3a-5.\)
2. A supplier listed two square tile samples with side lengths \((2a+5)\) and \((a-1)\). Using the identity \((x+y)(x-y)=x^2-y^2\), the difference of the squares of their side-lengths simplifies to:
Solution:

Let \(x=2a+5\) and \(y=a-1\). Then

\(x^2-y^2=(x+y)(x-y)=(2a+5+a-1)\big((2a+5)-(a-1)\big)=(3a+4)(a+6).\)

Option (b) gives the correct factorised product; note option (c) is a valid expanded form but is not factored as the identity suggests. Expanding our factorised result also yields

\((3a+4)(a+6)=3a^2+18a+4a+24=3a^2+22a+24,\)

which matches \(x^2-y^2\) after expansion.

3. Factorise the quadratic \(2a^2+3a-5\) (which represents the tiled area) into two linear factors.
Solution:

We check which pair expands to \(2a^2+3a-5\). Compute

\((2a+5)(a-1)=2a^2+5a-2a-5=2a^2+3a-5.\)

Hence \(2a^2+3a-5=(2a+5)(a-1)\), which is option (b).

4. Use factorisation by grouping to factor the polynomial \(a^2+7a+10\) that appears when students combine two sub-areas. The correct factorisation is:
Solution:

We seek two numbers whose product is \(10\) and sum is \(7\): \(2\) and \(5\). Thus

\(a^2+7a+10=a^2+2a+5a+10=a(a+2)+5(a+2)=(a+5)(a+2).\)

Option (a) matches; note order may vary, \((a+2)(a+5)\) is equivalent.

5. The students propose modifying the central area by increasing the length by \(2\) metres and decreasing the width by \(1\) metre (i.e. new length \(=2a+7\), new width \(=a-2\)). What is the algebraic change in area (new area minus original area)?
Solution:

Original area \(A_{\text{orig}}=(2a+5)(a-1)=2a^2+3a-5\). New area

\(A_{\text{new}}=(2a+7)(a-2)=2a^2+7a-4a-14=2a^2+3a-14.\)

Therefore the change is

\(A_{\text{new}}-A_{\text{orig}}=(2a^2+3a-14)-(2a^2+3a-5)=-9.\)

Thus the area decreases by \(9\) square metres, independently of \(a\), so option (b) is correct.

Review your answers and solutions below:

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