What is the main focus of Exercise 8.4 in Class 8 Maths Chapter 8?

Exercise 8.4 deals with algebraic identities and their applications in solving expressions quickly and accurately. Students learn how to apply standard identities to simplify and expand algebraic expressions.

How do NCERT Solutions help in understanding identities?

NCERT Solutions explain each identity with easy-to-follow examples and show step-by-step solutions for each question. This helps students remember the formulas and apply them correctly in various problems.

Are these identities important for competitive exams?

Yes, algebraic identities are frequently used in math competitions and competitive exams. A strong understanding of these identities improves speed and accuracy in solving algebra problems

NCERT Solutions

Class 8

Maths

Chapter 8 Algebraic Expressions and Identities

Exercise 8.4

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NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.4

NCERT Solutions for Class 8 Maths Exercise 8.4 focuses on solving problems using algebraic identities. These identities help simplify and solve expressions easily without expanding them fully. In this part of the chapter, you will learn to apply standard identities quickly solve expressions and equations.

Our NCERT Solutions for Class 8 Maths Exercise 8.4 are designed to help students understand how to apply these identities in a step-by-step and simple manner. These solutions are perfect for clearing doubts, improving speed, and preparing for exams confidently. With clear explanations and examples, students can strengthen their foundation in algebra.

1.0Download NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.4: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.4 PDF with easy, step-by-step solutions to help you learn and revise quickly.

NCERT Solutions Class 8 Maths Chapter 8 Exercise 8.4

2.0Key Concepts in Exercise 8.4 of Class 8 Maths Chapter 8

Exercise 8.4 focuses on the multiplication of algebraic expressions involving binomials, helping students understand how to apply the distributive property and simplify algebraic products.

Multiplying a Monomial by a Binomial
Students learn how to multiply a single-term expression (monomial) with a two-term expression (binomial) using the distributive property.
Using Distributive Law
The formula used is: $a (b + c) = ab + a c$
This helps in breaking down expressions and simplifying them step-by-step.
Multiplying a Binomial by a Binomial
The exercise also introduces how to multiply two binomials by applying the distributive property twice.
Combining Like Terms
After multiplication, students simplify the resulting expressions by combining like terms.
Building Foundation for Identities
This exercise builds the groundwork for understanding standard algebraic identities covered in the next section.

3.0NCERT Class 8 Maths Chapter 8: Other Exercises

NCERT Solutions Class 8 Maths Chapter 8 : Exercise 8.1

NCERT Solutions Class 8 Maths Chapter 8 : Exercise 8.2

NCERT Solutions Class 8 Maths Chapter 8 : Exercise 8.3

NCERT Solutions Class 8 Maths Chapter 8 : Exercise 8.4

4.0NCERT Class 8 Maths Chapter 8 Exercise 8.4: Detailed Solutions

Multiply the binomials: (i) $(2 x + 5)$ and $(4 x - 3)$ (ii) $(y - 8)$ and $(3 y - 4)$ (iii) $(2.5 ℓ - 0.5 m)$ and $(2.5 ℓ + 0.5 m)$ (iv) $(a + 3 b)$ and $(x + 5)$ (v) $(2 pq + 3 q^{2})$ and $(3 pq - 2 q^{2})$ (vi) $(\frac{3}{4} a^{2} + 3 b^{2})$ and $4 (a^{2} - \frac{2}{3} b^{2})$ Sol. (i) $(2 x + 5)$ and $(4 x - 3)$ $= 2 x (4 x - 3) + 5 (4 x - 3)$ $= 8 x^{2} - 6 x + 20 x - 15 = 8 x^{2} + 14 x - 15$ (ii) $(y - 8) \times (3 y - 4) = y (3 y - 4) - 8 (3 y - 4)$ $= 3 y^{2} - 4 y - 24 y + 32$ $= 3 y^{2} - 28 y + 32$ (iii) $(2.5 ℓ - 0.5 m) \times (2.5 ℓ + 0.5 m)$ $= 2.5 ℓ (2.5 ℓ + 0.5 m) - 0.5 m (2.5 ℓ + 0.5 m)$ $= 6.25 ℓ^{2} + 1.25 ℓ m - 1.25 ℓ m - 0.25 m^{2}$ $= 6.25 ℓ^{2} - 0.25 m^{2}$ (iv) $(a + 3 b) \times (x + 5)$ $= a (x + 5) + 3 b (x + 5)$ $= a x + 5 a + 3 b x + 15 b$ $= a x + 3 b x + 5 a + 15 b$ (v) $(2 pq + 3 q^{2}) \times (3 pq - 2 q^{2})$ $= 2 pq (3 pq - 2 q^{2}) + 3 q^{2} (3 pq - 2 q^{2})$ $= 6 p^{2} q^{2} - 4 p q^{3} + 9 p q^{3} - 6 q^{4}$ $= 6 p^{2} q^{2} + 5 p q^{3} - 6 q^{4}$ (vi) $(\frac{3}{4} a^{2} + 3 b^{2})$ and $4 (a^{2} - \frac{2}{3} b^{2})$ $= (\frac{3}{4} a^{2} + 3 b^{2}) \times (4 a^{2} - \frac{8}{3} b^{2})$ $= \frac{3}{4} a^{2} (4 a^{2} - \frac{8}{3} b^{2}) + 3 b^{2} (4 a^{2} - \frac{8}{3} b^{2})$ $= 3 a^{4} - 2 a^{2} b^{2} + 12 a^{2} b^{2} - 8 b^{4}$ $= 3 a^{4} + 10 a^{2} b^{2} - 8 b^{4}$
Find the product: (i) $(5 - 2 x) (3 + x)$ (ii) $(x + 7 y) (7 x - y)$ (iii) $(a^{2} + b) (a + b^{2})$ (iv) $(p^{2} - q^{2}) (2 p + q)$ Sol. (i) $(5 - 2 x) (3 + x) = 5 (3 + x) - 2 x (3 + x)$ $= 15 + 5 x - 6 x - 2 x^{2}$ $= - 2 x^{2} - x + 15$ (ii) $(x + 7 y) (7 x - y)$ $= x (7 x - y) + 7 y (7 x - y) = 7 x^{2} - x y + 49 x y - 7 y^{2} = 7 x^{2} + 48 x y - 7 y^{2}$ (iii) $(a^{2} + b) (a + b^{2})$ $= a^{2} (a + b^{2}) + b (a + b^{2}) = a^{3} + a^{2} b^{2} + ab + b^{3}$ (iv) $(p^{2} - q^{2}) (2 p + q) = p^{2} (2 p + q) - q^{2} (2 p + q)$ $= 2 p^{3} + p^{2} q - 2 p q^{2} - q^{3}$
Simplify : (i) $(x^{2} - 5) (x + 5) + 25$ (ii) $(a^{2} + 5) (b^{3} + 3) + 5$ (iii) $(t + s^{2}) (t^{2} - s)$ (iv) $(a + b) (c - d) + (a - b) (c + d) + 2 (ac + bd)$ (v) $(x + y) (2 x + y) + (x + 2 y) (x - y)$ (vi) $(x + y) (x^{2} - x y + y^{2})$ (vii) $(1.5 x - 4 y) (1.5 x + 4 y + 3) - 4.5 x + 12 y$ (viii) $(a + b + c) (a + b - c)$ Sol. (i) $(x^{2} - 5) (x + 5) + 25$ $= x^{2} (x + 5) - 5 (x + 5) + 25 = x^{3} + 5 x^{2} - 5 x - 25 + 25 = x^{3} + 5 x^{2} - 5 x$ (ii) $(a^{2} + 5) (b^{3} + 3) + 5$ $= a^{2} (b^{3} + 3) + 5 (b^{3} + 3) + 5$ $= a^{2} b^{3} + 3 a^{2} + 5 b^{3} + 15 + 5$ $= a^{2} b^{3} + 3 a^{2} + 5 b^{3} + 20$ (iii) $(t + s^{2}) (t^{2} - s)$ $= t (t^{2} - s) + s^{2} (t^{2} - s)$ $= t^{3} - ts + s^{2} t^{2} - s^{3}$ (iv) $(a + b) (c - d) + (a - b) (c + d) + 2 (a c + b d)$ $= a (c - d) + b (c - d) + a (c + d) - b (c + d)$ $+ 2 ac + 2 bd$ $= a c - a d + b c - b d + a c + a d - b c - b d$ $+ 2 ac + 2 bd$ $= 4 ac$ (v) $(x + y) (2 x + y) + (x + 2 y) (x - y)$ $= x (2 x + y) + y (2 x + y) + x (x - y) + 2 y (x - y)$ $= 2 x^{2} + x y + 2 x y + y^{2} + x^{2} - x y + 2 x y - 2 y^{2}$ $= (2 + 1) x^{2} + (1 + 2 - 1 + 2) x y + (1 - 2) y^{2}$ $= 3 x^{2} + 4 x y - y^{2}$ (vi) $(x + y) (x^{2} - x y + y^{2})$ $= x (x^{2} - x y + y^{2}) + y (x^{2} - x y + y^{2})$ $= x^{3} - x^{2} y + x y^{2} + x^{2} y - x y^{2} + y^{3}$ $= x^{3} + y^{3}$ (vii) $(1.5 x - 4 y) (1.5 x + 4 y + 3) - 4.5 x + 12 y$ $= 1.5 x (1.5 x + 4 y + 3) - 4 y (1.5 x + 4 y + 3)$ $- 4.5 x + 12 y$ $= 1.5 x \times 1.5 x + 1.5 x \times 4 y + 1.5 x \times 3$ $- 4 y \times 1.5 x - 4 y \times 4 y - 4 y \times 3 - 4.5 x$ $+ 12 y$ $= 2.25 x^{2} + 6 x y + 4.5 x - 6 x y - 16 y^{2}$ $- 12 y - 4.5 x + 12 y$ $= 2.25 x^{2} + (6 - 6) x y + (4.5 - 4.5) x$ $- 16 y^{2} + (- 12 + 12) y$ $= 2.25 x^{2} + (0) x y + (0) x - 16 y^{2} + (0) y$ $= 2.25 x^{2} + 0 + 0 - 16 y^{2} + 0$ $= 2.25 x^{2} - 16 y^{2}$ (viii) $(a + b + c) (a + b - c)$ $= a (a + b - c) + b (a + b - c) + c (a + b - c)$ $= a^{2} + ab - a c + ab + b^{2} - b c + a c + b c - c^{2}$ $= a^{2} + 2 ab + b^{2} - c^{2}$

5.0Key Features and Benefits of Class 8 Maths Chapter 8 Exercise 8.4

Simplifies Complex Concepts: Students learn to expand and simplify these expressions with identities, providing opportunities for faster calculations.
Strengthens Algebra Skills: Repeated practice improves algebraic skills and promotes understanding and retention of algebra, which is required for advanced algebra topics.
Step by Step Learning: The questions are put together to promote progressive learning from basic to slightly complex.
Based on NCERT Guidelines: The exercise follows the NCERT textbook exactly, making it perfect for exam preparation and concept clarity.

NCERT Class 8 Maths Ch. 8 Algebraic Expressions and Identities Other Exercises:-

Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

NCERT Solutions for Class 8 Maths Other Chapters:-

Chapter 1: Rational Numbers

Chapter 2: Linear Equations in One variable

Chapter 3: Understanding Quadrilaterals

Chapter 4: Data Handling

Chapter 5: Squares and Square Roots

Chapter 6: Cubes and Cube Roots

Chapter 7: Comparing Quantities

Chapter 8: Algebraic Expressions and Identities

Chapter 9: Mensuration

Chapter 10: Exponents and Powers

Chapter 11: Direct and Inverse Proportions

Chapter 12: Factorisation

Chapter 13: Introduction of Graphs