Official MSBSHSE Solutions for Class 7 Maths: Chapter 14 Set 53 Algebraic Formulae Expansion of Squares
Access comprehensive textbook solutions for Chapter 14 Set 53 Algebraic Formulae Expansion of Squares using the official curriculum guides for Class 7 Maths. Designed to align with the 2026-27 MSBSHSE standards, these detailed answers help students reinforce core academic concepts.
Chapter-wise Solutions for Maths: Chapter 14 Set 53 Algebraic Formulae Expansion of Squares
View or download the dedicated Chapter 14 Set 53 Algebraic Formulae Expansion of Squares solution resource below. Engaging with these textbook answers under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum for Maths.
Question 1. Factorize the following expressions:
(i) p² - q²
(ii) 4x² - 25y²
(iii) y² - 4
(iv) p² - \( \frac{1}{25} \)
(v) 9x² - \( \frac{1}{16} \)y²
(vi) x² - \( \frac{1}{x^2} \)
(vii) a²b - ab
(viii) 4x²y - 6x²
(ix) \( \frac{1}{2} \)y² - 8z²
(x) 2x² - 8y²
Answer:
(i) p² - q²
Here, a = p, b = q
\( \implies \) p² - q² = (p + q)(p - q)
....[(a² - b²) = (a + b)(a - b)]
(ii) 4x² - 25y²
= (2x)² - (5y)²
Here, a = 2x, b = 5y
\( \implies \) (2x)² - (5y)² = (2x + 5y)(2x - 5y)
....[(a² - b²) = (a + b)(a - b)]
(iii) y² - 4
= y² - 2²
Here, a = y, b = 2
\( \implies \) y² - 2² = (y + 2)(y - 2)
....[(a² - b²) = (a + b)(a - b)]
(iv) p² - \( \frac{1}{25} \)
Here a = p, b = \( \frac{1}{5} \)
\( \implies \) p² - \( (\frac{1}{5})^2 \) = \( (p + \frac{1}{5}) (p - \frac{1}{5}) \)
....[(a² - b²) = (a + b)(a - b)]
(v) 9x² - \( \frac{1}{16} \)y²
Here a = 3x, b = \( \frac{1}{4} \)y
\( \implies \) (3x)² - \( (\frac{1}{4}y)^2 \) = \( (3x + \frac{1}{4}y) (3x - \frac{1}{4}y) \)
....[(a² - b²) = (a + b)(a - b)]
(vi) x² - \( \frac{1}{x^2} \)
Here a = x, b = \( \frac{1}{x} \)
\( \implies \) x² - \( (\frac{1}{x})^2 \) = \( (x + \frac{1}{x})(x - \frac{1}{x}) \)
....[(a² - b²) = (a + b)(a - b)]
(vii) a²b - ab
= a (ab - b)
= ab (a - 1)
(viii) 4x²y - 6x²
= 2 (2x²y - 3x²)
= 2x² (2y - 3)
(ix) \( \frac{1}{2} \)y² - 8z²
= \( \frac{1}{2} \)y² - \( \frac{2 \times 8}{2} \)z²
= \( \frac{1}{2} \)y² - \( \frac{16}{2} \)z²
= \( \frac{1}{2} \)(y² - 16z²)
= \( \frac{1}{2} \)[y² - (4z)²]
= \( \frac{1}{2} \)[(y + 4z)(y - 4z)]
(x) 2x² - 8y²
= 2 (x² - 4y²)
= 2 [x² - (2y)²]
= 2(x + 2y)(x - 2y)
....[(a² - b²) = (a + b)(a - b)]
In simple words: This question asks to factorize algebraic expressions by identifying common factors or applying the difference of squares formula, \( (a^2 - b^2) = (a - b)(a + b) \), to simplify each given expression into its factors.
🎯 Exam Tip: Mastering the difference of squares formula and identifying common factors is crucial for scoring well in factorization problems. Show each step clearly, especially the identification of 'a' and 'b' terms.
MSBSHSE Solutions for Class 7 Maths Chapter 14 Set 53 Algebraic Formulae Expansion of Squares
Official MSBSHSE Solutions for Chapter 14 Set 53 Algebraic Formulae Expansion of Squares
Access structured MSBSHSE textbook solutions for Chapter 14 Set 53 Algebraic Formulae Expansion of Squares. Designed in alignment with the latest academic curriculum for Class 7 Maths, these answers cover all end-of-chapter exercises to support daily learning and homework completion.
Step-by-Step Explanations for Chapter 14 Set 53 Algebraic Formulae Expansion of Squares
Each solution includes detailed reasoning to foster genuine comprehension of Chapter 14 Set 53 Algebraic Formulae Expansion of Squares concepts. Reviewing these step-by-step breakdowns allows learners to master both analytical and descriptive questions expected in school evaluations.
Next Steps in Your Maths Revision
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