Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Set 5.1 Solutions

Question 1. Expand :
(i) (a + 2)(a – 1)
(ii) (m – 4)(m + 6)
(iii) (p + 8) (p – 3)
(iv) (13 + x)(13 – x)
(v) (3x + 4y) (3x + 5y)
(vi) (9x – 5t) (9x + 3t)
(vii) \( \left(m + \frac{2}{3}\right) \left(m - \frac{7}{3}\right) \)
(viii) \( \left(x + \frac{1}{x}\right) \left(x - \frac{1}{x}\right) \)
(ix) \( \left(\frac{1}{y} + 4\right) \left(\frac{1}{y} - 9\right) \)
Answer:
Solution:
(i) (a + 2)(a – 1)
\( = a^2 + (2 - 1) a + 2 \times (-1) \)
\( \therefore (x + A) (x + B) = x^2 + (A + B)x + AB \)
\( = a^2 + a - 2 \)
In simple words: This expansion uses the algebraic identity \((x+A)(x+B) = x^2 + (A+B)x + AB\), where 'a' is 'x', '2' is 'A', and '-1' is 'B'.

🎯 Exam Tip: Remember to apply the correct identity for binomial expansion, carefully handling the signs of constants A and B.

(ii) (m – 4)(m + 6)
\( = m^2 + (- 4 + 6) m + (-4) \times 6 \)
\( \therefore (x + a) (x + b) = x^2 + (a + b)x + ab \)
\( = m^2 + 2m – 24 \)
In simple words: Here, 'm' is 'x', '-4' is 'a', and '6' is 'b' in the identity \((x+a)(x+b) = x^2 + (a+b)x + ab\). Combine the constant terms for the middle term and multiply them for the last term.

🎯 Exam Tip: Pay close attention to the signs when performing addition and multiplication of the constant terms, as a simple error can lead to an incorrect result.

(iii) (p + 8) (p – 3)
\( = p^2 + (8 – 3) p + 8 \times (-3) \)
\( \therefore (x + a) (x + b) = x^2 + (a + b)x + ab \)
\( = p^2 + 5p - 24 \)
In simple words: Using the identity \((x+a)(x+b) = x^2 + (a+b)x + ab\), where 'p' is 'x', '8' is 'a', and '-3' is 'b'. The product of the constants '8' and '-3' gives '-24'.

🎯 Exam Tip: Double-check the sum and product of the constant terms. A common mistake is miscalculating the product when one of the numbers is negative.

(iv) (13 + x) (13 – x)
\( = (13)^2 + (x - x) 13 + x \times (-x) \)
\( \therefore (x + a) (x + b) = x^2 + (a + b)x + ab \)
\( = 169 + 0 \times 13 – x^2 \)
\( = 169 - x^2 \)
In simple words: This is an application of the difference of squares formula, \((A+B)(A-B) = A^2 - B^2\). Here, \(A = 13\) and \(B = x\), resulting in \(13^2 - x^2\).

🎯 Exam Tip: Recognize special product identities like the difference of squares to simplify calculations and ensure a quick and accurate solution.

(v) (3x + 4y) (3x + 5y)
\( = (3x)^2 + (4y + 5y) 3x + 4y \times 5y \)
\( \therefore (x + a) (x + b) = x^2 + (a + b)x + ab \)
\( = 9x^2 + 9y \times 3x + 20y^2 \)
\( = 9x^2 + 27xy + 20y^2 \)
In simple words: This expands using \((X+A)(X+B)\) where \(X = 3x\), \(A = 4y\), and \(B = 5y\). Square the common term, add the 'A' and 'B' terms multiplied by 'X', then multiply 'A' and 'B'.

🎯 Exam Tip: When the 'x' term in the identity is itself an expression (like 3x), ensure to square it completely and distribute it correctly to the sum of the 'a' and 'b' terms.

(vi) (9x – 5t) (9x + 3t)
\( = (9x)^2 + [(-5t) + 3t] 9x + (-5t) \times 3t \)
\( \therefore (x + a) (x + b) = x^2 + (a + b)x + ab \)
\( = 81x^2 + (-2t) \times 9x – 15t^2 \)
\( = 81x^2 - 18xt – 15t^2 \)
In simple words: Apply the \((X+A)(X+B)\) identity where \(X = 9x\), \(A = -5t\), and \(B = 3t\). Be careful with the negative sign in the 'A' term throughout the calculation.

🎯 Exam Tip: Practice working with negative coefficients to avoid common sign errors, especially when multiplying and adding terms in the expansion.

(vii) \( \left(m + \frac{2}{3}\right) \left(m - \frac{7}{3}\right) \)
\( = m^2 + \left(\frac{2}{3} - \frac{7}{3}\right) m + \frac{2}{3} \times \left(-\frac{7}{3}\right) \)
\( \therefore (x + a) (x + b) = x^2 + (a + b)x + ab \)
\( = m^2 - \frac{5}{3} m - \frac{14}{9} \)
In simple words: This uses the \((x+a)(x+b)\) identity with fractional constants. Add the fractions for the middle term and multiply them for the last term.

🎯 Exam Tip: When dealing with fractions, ensure you find a common denominator for addition/subtraction and multiply numerators and denominators directly for multiplication.

(viii) \( \left(x + \frac{1}{x}\right) \left(x - \frac{1}{x}\right) \)
\( = x^2 + \left(\frac{1}{x} - \frac{1}{x}\right) x + \frac{1}{x} \times \left(-\frac{1}{x}\right) \)
\( \therefore (x + a) (x + b) = x^2 + (a + b)x + ab \)
\( = x^2 + 0 \times x - \frac{1}{x^2} \)
\( = x^2 - \frac{1}{x^2} \)
In simple words: This is another instance of the difference of squares identity, \((A+B)(A-B) = A^2 - B^2\), where \(A = x\) and \(B = \frac{1}{x}\).

🎯 Exam Tip: Recognizing the difference of squares with reciprocal terms allows for a direct solution without complex intermediate steps.

(ix) \( \left(\frac{1}{y} + 4\right) \left(\frac{1}{y} - 9\right) \)
\( = \left(\frac{1}{y}\right)^2 + (4 - 9) \frac{1}{y} + 4 \times (-9) \)
\( \therefore (x + a) (x + b) = x^2 + (a + b)x + ab \)
\( = \frac{1}{y^2} - \frac{5}{y} - 36 \)
In simple words: Apply the \((X+A)(X+B)\) identity, where \(X = \frac{1}{y}\), \(A = 4\), and \(B = -9\). Combine the constants for the middle term and multiply them for the last term.

🎯 Exam Tip: Be careful when the variable part is a fraction; square it correctly, and ensure the sum of constants is multiplied by the fractional variable part.

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.1 Intext Questions And Activities

Question 1. Use the above formulae to fill proper terms in the following boxes. (Textbook pg. no. 23)
1. \( (x + 2y)^2 = x^2 + \boxed{4xy} + 4y^2 \)
2. \( (2x – 5y)^2 = \boxed{4x^2} - 20xy + \boxed{25y^2} \)
3. \( (101)^2 = (100 + 1)^2 = \boxed{100^2} + \boxed{2 \times 100 \times 1} + 1^2 = \boxed{10201} \)
4. \( (98)^2 = (100 – 2)^2 = 10000 - \boxed{400} + \boxed{4} = \boxed{9604} \)
5. \( (5m + 3n) (5m – 3n) = \boxed{(5m)^2} - \boxed{(3n)^2} = \boxed{25m^2} - \boxed{9n^2} \)
Answer:
Solution:
1. \( (x + 2y)^2 = x^2 + 4xy + 4y^2 \)
2. \( (2x – 5y)^2 = 4x^2 – 20xy + 25y^2 \)
3. \( (101)^2 = (100 + 1)^2 = 10000 + 200 + 1^2 = 10201 \)
4. \( (98)^2 = (100 – 2)^2 = 10000 – 400 + 4 = 9604 \)
5. \( (5m + 3n) (5m – 3n) = (5m)^2 – (3n)^2 = 25m^2 – 9n^2 \)
In simple words: These problems apply binomial expansion identities such as \((a+b)^2 = a^2+2ab+b^2\), \((a-b)^2 = a^2-2ab+b^2\), and \((a+b)(a-b) = a^2-b^2\) to fill in missing terms.

🎯 Exam Tip: Familiarity with common algebraic identities is crucial for quickly and accurately solving fill-in-the-blank questions involving expansions.

Question 2. Expand (x + a) (x + b) using formulae for areas of a square and a rectangle. (Textbook pg. no. 23)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ज्यामितीय रूप से \((x+a)(x+b)\) के विस्तार को दर्शाता है। इसमें एक बड़ा वर्ग जिसकी भुजा 'x' है, एक आयत जिसकी भुजाएँ 'x' और 'a' हैं, दूसरा आयत जिसकी भुजाएँ 'x' और 'b' हैं, और एक छोटा आयत जिसकी भुजाएँ 'a' और 'b' हैं, इन सभी के क्षेत्रों का योग करके कुल क्षेत्रफल \((x+a)(x+b)\) के बराबर दिखाया गया है।
(x + a) (x + b) = \(x^2 + ax + bx + ab\)
(x + a) (x + b) = \(x^2 + (a + b) x + ab\)
Answer:
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में, \((x+a)(x+b)\) के ज्यामितीय विस्तार को दिखाया गया है। यह एक बड़े आयत को चार छोटे हिस्सों में विभाजित करके उसका क्षेत्रफल निकालने का तरीका है: एक वर्ग \(x^2\), दो आयत \(ax\) और \(bx\), और एक छोटा आयत \(ab\)। इन सभी हिस्सों के क्षेत्रफलों को जोड़कर कुल क्षेत्रफल प्राप्त होता है।
Area of rectangle is (x + a) (x + b)
\( = \) Area of square is \(x^2\) \( + \) Area of rectangle is \(ax\) \( + \) Area of rectangle is \(bx\) \( + \) Area of rectangle is \(ab\)
\( \therefore (x + a) (x + b) = x^2 + ax + bx + ab \)
\( \therefore (x + a) (x + b) = x^2 + (a + b) x + ab \)
In simple words: The expansion \((x+a)(x+b)\) can be visualized as the total area of a large rectangle, which is divided into a square of area \(x^2\), two rectangles of areas \(ax\) and \(bx\), and a smaller rectangle of area \(ab\). Summing these areas gives the expanded form.

🎯 Exam Tip: Understanding the geometric interpretation of algebraic identities helps build intuition and provides an alternative method to verify expansions, especially for area-based problems.