Math Case Study Class 8 on Factorization

Math Case Study Class 8 on Factorization | NCERT-Based Questions

Math Case Study Class 8 on Factorization

A Math Case Study Class 8 on Factorization introduces students to methods of simplifying algebraic expressions through real-life mathematical applications. It enhances their problem-solving skills and strengthens their conceptual understanding of algebra. Moreover, such studies help connect theory with practical examples.

Importance of Factorization in Algebra

Understanding factorization helps students identify patterns and simplify expressions effectively. It lays a strong foundation for Case Study math questions for class 9 and math case study questions class 9. Additionally, regular practice boosts confidence and accuracy in exams.

Practice Through Case Studies

Students should solve math case study questions and worksheets regularly. Furthermore, this consistent practice improves logical thinking and prepares them for advanced algebra topics.

Case Study 3: Designing a Park Boundary Using Factorisation

A municipal committee has allocated a rectangular plot near the school to construct a small park. The park designers decide that the length of the park will be \((a+4)\) metres and the width will be \((a-3)\) metres, where the integer parameter \(a\) represents the number of standard modular units available from surplus materials. In addition to the boundary, they plan two identical square play areas inside the rectangle: one of side \((a+4)\) (aligned along the length) and another of side \((a-3)\) (aligned along the width) to illustrate contrasting space usage. During community consultation the team must compute the rectangular area, compare it with the combined area of the two squares, and decide how much land remains for landscaping. They also consider altering the plan by adding a narrow rectangular strip of width 1 metre along the longer side and subtracting a strip of width 1 metre along the shorter side (to accommodate a path), and must evaluate the net change in available landscaping area symbolically. The exercise requires careful application of common-factor extraction, grouping methods to factor quadratic expressions, and standard identities such as \((x+y)^2\), \((x-y)^2\), and \((x+y)(x-y)\) to simplify expressions, check results by numeric substitution for several integer values of \(a\), and present final factorised forms for quick budgeting decisions.

1. The algebraic expression for the area of the rectangular park is:
Solution:

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

\(\text{Area}=(a+4)(a-3)=a^2+4a-3a-12=a^2+a-12.\)
2. The combined area of the two square play areas (one of side \(a+4\) and the other of side \(a-3\)) minus the park area simplifies to:
Solution:

By definition the combined square area minus park area is

\((a+4)^2+(a-3)^2-(a+4)(a-3).\)

This expression can be expanded and simplified. Expand each term:

\((a+4)^2=a^2+8a+16,\quad (a-3)^2=a^2-6a+9,\quad (a+4)(a-3)=a^2+a-12.\)

Summing the squares and subtracting the rectangle:

\((a^2+8a+16)+(a^2-6a+9)-(a^2+a-12)=a^2+(8a-6a-a)+ (16+9+12),\)
\(=a^2+(1a)+(37)=a^2+a+37.\)

So the fully simplified form is \(a^2+a+37\), which is algebraically equivalent to option (a) after expansion. Options (b)-(d) do not correctly represent the simplified difference.

3. Use the identity \((x+y)(x-y)=x^2-y^2\) with \(x=(a+4)\) and \(y=(a-3)\) to compute \((a+4)^2-(a-3)^2\). Which of the following is correct?
Solution:

Let \(x=a+4\) and \(y=a-3\). Then

\(x^2-y^2=(x+y)(x-y)=((a+4)+(a-3))((a+4)-(a-3)).\)

Compute sums and differences:

\((a+4)+(a-3)=2a+1,\qquad (a+4)-(a-3)=7.\)

Hence

\(x^2-y^2=(2a+1)\times 7=7(2a+1),\)

which matches option (a).

4. Factorise the quadratic \(a^2+a-12\) (the park area) into linear factors over integers.
Solution:

We look for integers \(p\) and \(q\) with \(pq=-12\) and \(p+q=1\). The pair \(4\) and \(-3\) satisfies this. Therefore

\(a^2+a-12=(a+4)(a-3),\)

which is option (a).

5. The designers add a strip of width \(1\) metre along the length (so new length \(=a+5\)) and remove a strip of width \(1\) metre along the width (so new width \(=a-4\)). The net change in park area (new minus original) equals:
Solution:

Original area \(A_{\text{orig}}=(a+4)(a-3)=a^2+a-12\). New area

\(A_{\text{new}}=(a+5)(a-4)=a^2+5a-4a-20=a^2+a-20.\)

Change \(=\;A_{\text{new}}-A_{\text{orig}}=(a^2+a-20)-(a^2+a-12)=-8\).

Note: None of the given options equals \(-8\). Therefore the options are incorrect. The correct answer is \(-8\).

Review your answers and solutions below:

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