Class 8 Math Case Study on Factorisation and Identities | Worksheets & Solutions

Class 8 Math Case Study on Factorisation and Identities

A class 8 math case study on factorisation and identities helps students connect algebraic formulas to real-life applications. It focuses on simplification techniques and math case study questions that improve logical and analytical thinking. Moreover, it strengthens the foundation for higher-level algebra.

Understanding Factorisation and Identities

Practicing factorisation and identities helps students identify algebraic patterns and solve expressions easily. It prepares them for Case Study math questions for class 9 and math case study questions class 9. Additionally, solving worksheets enhances clarity and speed during exams.

Practice and Application

Students should practice math case study questions regularly using NCERT examples. Furthermore, solving case-based problems boosts confidence and deepens conceptual understanding.

Case Study 4: Calculating Materials for a School Auditorium Partition

The school intends to partition a large rectangular section of the auditorium to create a rehearsal space and a storage bay using removable panels. The partitioned rectangle’s length (parallel to the stage) is modelled by \((3a+2)\) metres and its width (perpendicular to the stage) by \((a-2)\) metres, where \(a\) is a positive integer representing the number of standard panel modules available. Inside this rectangle the team will remove a smaller rectangular storage footprint of dimensions \((a+1)\) by \((a-2)\) metres to be used as a locked store; the remaining region will be covered with acoustic panels. The project requires the students to compute total area, the net panel-covered area after removing storage, use identities to compute differences between squared dimensions when ordering square acoustic mats, and to re-evaluate material requirements when modular changes occur (for example, if the length is decreased by one module and the width increased by three modules). The work emphasises choosing the appropriate factorisation strategy (common factor extraction, grouping) and applying standard identities such as \((x+y)^2\), \((x-y)^2\), and \((x+y)(x-y)\) to simplify algebraic expressions arising in budgeting and ordering. Students must also perform sanity checks by substituting integer values of \(a\) to confirm algebraic results correspond to feasible quantities of panels.

1. The algebraic expression for the total area (in square metres) of the partitioned rectangle is:
Solution:

Area \(=\) length \(\times\) width. Thus

\(A=(3a+2)(a-2)=3a^2-6a+2a-4=3a^2-4a-4.\)
2. Expand and simplify the product \((3a+2)(a-2)\). Which expression is correct?
Solution:

Multiply termwise:

\((3a+2)(a-2)=3a\cdot a+3a\cdot(-2)+2\cdot a+2\cdot(-2)=3a^2-6a+2a-4=3a^2-4a-4.\)
3. The net panel-covered area after removing the storage footprint \((a+1)(a-2)\) from the partitioned rectangle equals:
Solution:

Net area

\(A_{\text{net}}=(3a+2)(a-2)-(a+1)(a-2).\)

Factor common \((a-2)\):

\(A_{\text{net}}=(a-2)\big((3a+2)-(a+1)\big)=(a-2)(2a+1).\)
4. Using the identity \((x+y)(x-y)=x^2-y^2\) compute \((3a+2)^2-(a-2)^2\). Which factorised form is correct?
Solution:

Let \(x=3a+2\), \(y=a-2\). Then

\(x^2-y^2=(x+y)(x-y)=((3a+2)+(a-2))((3a+2)-(a-2)).\)

Compute sums and differences:

\((3a+2)+(a-2)=4a,\qquad (3a+2)-(a-2)=2a+4.\)

Thus

\(x^2-y^2=(4a)(2a+4)=8a^2+16a.\)

Options (a) and (c) are the same as (d) up to ordering; option (d) explicitly matches the product and is presented in correct mathematical form.

5. If the partition length is decreased by \(1\) module and the width increased by \(3\) modules, the new dimensions are \((3a+1)\) and \((a+1)\). The algebraic change in area (new minus original) equals:
Solution:

Original area \(A_{\text{orig}}=3a^2-4a-4\). New area

\(A_{\text{new}}=(3a+1)(a+1)=3a^2+3a+a+1=3a^2+4a+1.\)

Difference

\(A_{\text{new}}-A_{\text{orig}}=(3a^2+4a+1)-(3a^2-4a-4)=8a+5.\)

Hence option (a) is correct.

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