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

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NCERT Solutions for Class 8 Maths Chapter 8 (EX 8.4)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.4 and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 8 - Algebraic Expressions and Identities

Exercise:

Exercise - 8.4

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

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  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Every NCERT Solution is provided to make the study simple and interesting. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 8 Science , Maths solutions and solutions of other subjects. You can also download NCERT Solutions for Class 8 Maths to help you to revise complete syllabus and score more marks in your examinations.

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Access NCERT Solutions for Class 8 Chapter 8 - Algebraic Expressions and Identities

1. Multiply the following binomials.

i. $\left( {2x + 5} \right)$ and $\left( {4x - 3} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {2x + 5} \right) \times \left( {4x - 3} \right) = 2x \times \left( {4x - 3} \right) + 5 \times \left( {4x - 3} \right)$

$\left( {2x + 5} \right) \times \left( {4x - 3} \right) = 8{x^2} - 6x + 20x - 15$

On adding the like terms, we get,

$\left( {2x + 5} \right) \times \left( {4x - 3} \right) = 8{x^2} + 14x - 15$

ii. $\left( {y - 8} \right)$ and $\left( {3y - 4} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {y - 8} \right) \times \left( {3y - 4} \right) = y \times \left( {3y - 4} \right) - 8 \times \left( {3y - 4} \right)$

$\left( {y - 8} \right) \times \left( {3y - 4} \right) = 3{y^2} - 4y - 24y + 32$

On adding the like terms, we get,

$\left( {y - 8} \right) \times \left( {3y - 4} \right) = 3{y^2} - 28y + 32$

iii. $\left( {2.5l - 0.5m} \right)$ and $\left( {2.5l + 0.5m} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {2.5l - 0.5m} \right) \times \left( {2.5l + 0.5m} \right) = 2.5l \times \left( {2.5l + 0.5m} \right) - 0.5m \times \left( {2.5l + 0.5m} \right)$

$\left( {2.5l - 0.5m} \right) \times \left( {2.5l + 0.5m} \right) = 6.25{l^2} + 1.251m - 1.25ml - 0.25{m^2}$

On adding the like terms, we get,

$\left( {2.5l - 0.5m} \right) \times \left( {2.5l + 0.5m} \right) = 6.25{l^2} - 0.25{m^2}$

iv. $\left( {a + 3b} \right)$ and $\left( {x + 5} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {a + 3b} \right) \times \left( {x + 5} \right) = a \times \left( {x + 5} \right) + 3b \times \left( {x + 5} \right)$

$\left( {a + 3b} \right) \times \left( {x + 5} \right) = ax + 5a + 3bx + 15b$

v. $\left( {2pq + 3{q^2}} \right)$ and $\left( {3pq - 2{q^2}} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

\[\left( {2pq + 3{q^2}} \right) \times \left( {3pq - 2{q^2}} \right) = 2pq \times \left( {3pq - 2{q^2}} \right) + 3{q^2} \times \left( {3pq - 2{q^2}} \right)\]

\[\left( {2pq + 3{q^2}} \right) \times \left( {3pq - 2{q^2}} \right) = 6{p^2}{q^2} - 4p{q^3} + 9p{q^3} - 6{q^4}\]

On adding the like terms, we get,

\[\left( {2pq + 3{q^2}} \right) \times \left( {3pq - 2{q^2}} \right) = 6{p^2}{q^2} + 5p{q^3} - 6{q^4}\]

vi. $\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right)$ and $4\left( {{a^2} - \dfrac{2}{3}{b^2}} \right)$

Ans: On multiplying the bracket of the second term and simplifying it, we get,

$4\left( {{a^2} - \dfrac{2}{3}{b^2}} \right) = 4{a^2} - \dfrac{8}{3}{b^2}$

We will multiply the given terms by applying the distributive property and then simplify the equation further. 

\[\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) = \dfrac{3}{4}{a^2} \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) + 3{b^2} \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right)\]

\[\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) = 3{a^4} - \dfrac{{24}}{{12}}{a^2}{b^2} + 12{b^2}{a^2} - 8{b^4}\]

\[\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) = 3{a^4} - 2{a^2}{b^2} + 12{b^2}{a^2} - 8{b^4}\]

On adding the like terms, we get,

\[\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) = 3{a^4} + 10{a^2}{b^2} - 8{b^4}\]


2. Find the product of the following binomials.

i. $\left( {5 - 2x} \right)\left( {3 + x} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {5 - 2x} \right)\left( {3 + x} \right) = 5\left( {3 + x} \right) - 2x\left( {3 + x} \right)$

$\left( {5 - 2x} \right)\left( {3 + x} \right) = 15 + 5x - 6x - 2{x^2}$

On adding the like terms, we get,

$\left( {5 - 2x} \right)\left( {3 + x} \right) = 15 - x - 2{x^2}$

ii. $\left( {x + 7y} \right)\left( {7x - y} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {x + 7y} \right)\left( {7x - y} \right) = x\left( {7x - y} \right) + 7y\left( {7x - y} \right)$

$\left( {x + 7y} \right)\left( {7x - y} \right) = 7{x^2} - xy + 49xy - 7{y^2}$

On adding the like terms, we get,

$\left( {x + 7y} \right)\left( {7x - y} \right) = 7{x^2} + 48xy - 7{y^2}$

iii. $\left( {{a^2} + b} \right)\left( {a + {b^2}} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {{a^2} + b} \right)\left( {a + {b^2}} \right) = {a^2}\left( {a + {b^2}} \right) + b\left( {a + {b^2}} \right)$

$\left( {{a^2} + b} \right)\left( {a + {b^2}} \right) = {a^3} + {a^2}{b^2} + ab + {b^3}$

iv. $\left( {{p^2} - {q^2}} \right)\left( {2p + q} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {{p^2} - {q^2}} \right)\left( {2p + q} \right) = {p^2}\left( {2p + q} \right) - {q^2}\left( {2p + q} \right)$

$\left( {{p^2} - {q^2}} \right)\left( {2p + q} \right) = 2{p^3} + {p^2}q - 2{q^2}p - {q^3}$


3. Simplify the following expressions. 

i. $\left( {{x^2} - 5} \right)\left( {x + 5} \right) + 25$

Ans: We will first multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {{x^2} - 5} \right)\left( {x + 5} \right) + 25 = {x^2}\left( {x + 5} \right) - 5\left( {x + 5} \right) + 25$

$\left( {{x^2} - 5} \right)\left( {x + 5} \right) + 25 = {x^3} + {5x^2} - 5x - 25 + 25$

On adding the like terms, we get,

$\left( {{x^2} - 5} \right)\left( {x + 5} \right) + 25 = {x^3} + {5x^2} - 5x$

ii. $\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right) + 5$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right) + 5 = {a^2}\left( {{b^3} + 3} \right) + 5\left( {{b^3} + 3} \right) + 5$

$\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right) + 5 = {a^2}{b^3} + 3{a^2} + 5{b^3} + 15 + 5$

On adding the like terms, we get,

$\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right) + 5 = {a^2}{b^3} + 3{a^2} + 5{b^3} + 20$

iii. $\left( {t + {s^2}} \right)\left( {{t^2} - s} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

\[\left( {t + {s^2}} \right)\left( {{t^2} - s} \right) = t\left( {{t^2} - s} \right) + {s^2}\left( {{t^2} - s} \right)\]

\[\left( {t + {s^2}} \right)\left( {{t^2} - s} \right) = {t^3} - st + {s^2}{t^2} - {s^3}\]

iv. $\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right) = a\left( {c - d} \right) + b\left( {c - d} \right) + a\left( {c + d} \right) - b\left( {c + d} \right) + 2ac + 2bd$

$\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right) = ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd$

On grouping together and adding the like terms, we get,

$\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right) = \left( {ac + ac + 2ac} \right) + \left( {ad - ad} \right) + \left( {bc - bc} \right) + \left( {2bd - bd - bd} \right)$

$\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right) = 4ac$

v. $\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right) = x\left( {2x + y} \right) + y\left( {2x + y} \right) + x\left( {x - y} \right) + 2y\left( {x - y} \right)$

$\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right) = 2{x^2} + xy + 2xy + {y^2} + {x^2} - xy + 2yx - 2{y^2}$

On grouping together and adding the like terms, we get,

$\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right) = \left( {2{x^2} + {x^2}} \right) + \left( {{y^2} - 2{y^2}} \right) + \left( {xy + 2xy - xy + 2xy} \right)$

$\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right) = 3{x^2} - {y^2} + 4xy$

vi. $\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = x\left( {{x^2} - xy + {y^2}} \right) + y\left( {{x^2} - xy + {y^2}} \right)$

$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = {x^3} - {x^2}y + {y^2}x + y{x^2} - x{y^2} + {y^3}$

On grouping together and adding the like terms, we get,

$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = {x^3} + {y^3} + \left( {x{y^2} - x{y^2}} \right) + \left( {{x^2}y - {x^2}y} \right)$

$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = {x^3} + {y^3}$

vii. $\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y = 1.5x\left( {1.5x + 4y + 3} \right) - 4y\left( {1.5x + 4y + 3} \right) - 4.5x + 12y$

$\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y = 2.25{x^2} + 6xy + 4.5x - 6xy - 16{y^2} - 12y - 4.5x + 12y$

On grouping together and adding the like terms, we get,

\[\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y = 2.25{x^2} + \left( {6xy - 6xy} \right) + \left( {4.5x - 4.5x} \right) - 16{y^2} + \left( {12y - 12y} \right)\]

\[\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y = 2.25{x^2} - 16{y^2}\]

viii. $\left( {a + b + c} \right)\left( {a + b - c} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {a + b + c} \right)\left( {a + b - c} \right) = a\left( {a + b - c} \right) + b\left( {a + b - c} \right) + c\left( {a + b - c} \right)$

$\left( {a + b + c} \right)\left( {a + b - c} \right) = {a^2} + ab - ac + ab + {b^2} - bc + ac + bc - {c^2}$

On grouping together and adding the like terms, we get,

$\left( {a + b + c} \right)\left( {a + b - c} \right) = {a^2} + {b^2} - {c^2} + \left( {ab + ab} \right) + \left( {bc - bc} \right) + \left( {ac - ac} \right)$

$\left( {a + b + c} \right)\left( {a + b - c} \right) = {a^2} + {b^2} - {c^2} + 2ab$


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

Opting for the NCERT solutions for Ex 8.4 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 8.4 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.


Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 8 Exercise 8.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.


Besides these NCERT solutions for Class 8 Maths Chapter 8 Exercise 8.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.


Do not delay any more. Download the NCERT solutions for Class 8 Maths Chapter 8 Exercise 8.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.


Class 8 Maths Chapter 8: Exercises Breakdown

Exercise

Number of Questions

Exercise 8.1

2 Questions & solutions

Exercise 8.2

5 Questions & solutions

Exercise 8.3

5 Questions & solutions


Other Study Material for CBSE Class 8 Maths Chapter 8


Chapter-wise NCERT Solutions for Class 8 Maths


Important Related Links for CBSE Class 8 Maths

FAQs on NCERT Solutions for Class 8 Maths Chapter 8: Algebraic Expressions and Identities - Exercise 8.4

1. What should be known before solving NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities (EX 8.4) Exercise 8.4?

Before tackling the NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities (EX 8.4) Exercise 8.4, it is important to understand the four standard algebraic expressions and identities. To complete the task, you must have a basic understanding of these identities. You can obtain their PDF from the Vedantu website.

2. What is the weightage of NCERT Class 8 Maths Chapter 8 Algebraic Expressions and Identities in the final examination?

The weightage of Class 8 Maths Chapter 8 Algebraic Expressions and Identities, in the final examination is around 8-10 marks. This chapter is crucial and easy to comprehend. Important subjects pertaining to algebraic expressions and identities are covered. Higher-level classes will benefit as well. One must thoroughly study the chapter to do well in exams.

3. What Is the Importance of NCERT Solutions for class 8 Maths chapter 8 Algebraic Expressions and Identities (EX 8.4) Exercise 8.4?

NCERT Solutions for class 8 Maths chapter 8 Algebraic Expressions and Identities (EX 8.4) Exercise 8.4 includes well-written examples that explain the concept of algebraic expressions and identities. To enhance comprehension and instil confidence in students, the components of a phrase have undergone extensive analysis. NCERT books' superior calibre is attested to by the CBSE board's requirement that pupils use them. Visit Vendantu.com for the best NCERT solutions.

4. Do I have to go over all of the questions in NCERT Solutions for class 8 Maths chapter 8 Algebraic Expressions and Identities (EX 8.4) Exercise 8.4?

Expressions are independent mathematical objects with a wide range of uses. You have the chance to learn more about this important and practical concept in this chapter. Additionally, expressions can be simple or complex, and although the idea is simple, the applications can be incredibly imaginative. Studying the use of algebraic expressions is essential because of these factors. Students can investigate every basic detail in NCERT Solutions for class 8 Maths chapter 8 Algebraic Expressions and Identities (EX 8.4) Exercise 8.4 because there are numerous examples and practise questions.

5. Is the NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities (EX 8.4) Exercise 8.4 beneficial?

Absolutely, students can benefit much from using the NCERT Solutions for Class 8 Maths, Chapter 8 Algebraic Expressions and Identities (EX 8.4) Exercise 8.4. Using the NCERT solutions for Ex. 8.4 Class 8 Math is recommended for CBSE students preparing for exams. This chapter has many exercises. We provide the Exercise 8.4 Class 8 Maths NCERT solutions on this page in PDF format. You can study this answer straight from the Vedantu website or mobile app, or you can download it as needed.