Math Case Study Questions Class 8 on Factorization

Math Case Study Questions Class 8 on Factorization Formulas | Important Questions

Math Case Study Class 8 on Factorization

A Math Case Study Class 8 on Factorization helps students understand how to simplify algebraic expressions using different factorization techniques. It connects algebraic concepts with real-life applications, enhancing problem-solving skills. Moreover, students develop logical reasoning while solving practical problems.

Understanding Factorization Concepts

By solving factorization case studies, students learn how to apply common factors, regrouping, and algebraic identities. This practice strengthens their foundation for Case Study math questions for class 9 and math case study questions class 9. Additionally, it improves mathematical accuracy and analytical ability.

Practice for Mastery

Students should regularly solve math case study questions and worksheets. Furthermore, consistent practice ensures conceptual clarity and confidence during exams.

Case Study 1: Application of Factorisation in a School Garden Design

A group of Class 8 students from Greenfield School decided to design a rectangular garden for their campus. The garden will have two equal-length pathways on opposite sides, and flowerbeds along the remaining edges. The total length of the garden is represented by \((a + 3)\) meters, while the width is \((a – 2)\) meters. The teacher, Mrs. Sharma, asked the students to calculate the area and the difference in areas when certain portions of the garden are redesigned. The students observed that applying standard algebraic identities could simplify their calculations efficiently.

By using suitable factorisation methods, the students were able to find relationships between the total area, the area without pathways, and the remaining space for flowerbeds. This project helped them realize how factorisation and identities like \((a+b)^2\), \((a-b)^2\), and \((a+b)(a-b)\) can be practically applied in planning and design.

1. The area of the garden can be expressed as:
Solution:

Area of a rectangle = length × breadth. Thus,

\(\text{Area} = (a + 3)(a – 2) = a^2 + 3a – 2a – 6 = a^2 + a – 6.\)
2. Simplify \((a + 3)(a – 2)\) using the identity \((a + b)(a – b) = a^2 – b^2\) after suitable rearrangement.
Solution:
\((a + 3)(a – 2) = a^2 – (2 \times 3) – 6 = a^2 – 6.\)

However, correctly expanding gives \(a^2 + a – 6\), matching earlier simplification.

3. If the garden is redesigned such that both length and width increase by 1 meter, the new area becomes:
Solution:

New length = \((a + 3) + 1 = a + 4\); new width = \((a – 2) + 1 = a – 1\). Hence, new area \(= (a + 4)(a – 1)\).

4. The increase in area compared to the original design is:
Solution:

Original area \(= a^2 + a – 6\), new area \(= a^2 + 3a – 4\). Difference \(= (a^2 + 3a – 4) – (a^2 + a – 6) = 2a + 2\).

Thus correct answer is (None of the given options match correctly). Correct answer: \(2a + 2\).

5. The factorised form of the difference of areas between two square flowerbeds of sides \((a+3)\) and \((a-2)\) is:
Solution:

Using identity \((a+b)^2 – (a-b)^2 = (a+b+a-b)(a+b-(a-b)) = (2a)(2b) = 4ab\). Here, \(a=a\) and \(b=3,2\), thus,

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

Therefore, correct simplification matches option (b).

Review your answers and solutions below:

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