 # NCERT Exemplar Class 8 Maths Solutions for Chapter 7 - Algebraic Expressions And Identities Factorisation

## NCERT Exemplar Solutions Class 8 Maths Chapter 7 – Free PDF Download

NCERT Exemplar Class 8 Maths Chapter 7 Algebraic Expressions and Identities & Factorisation, presented here for students to prepare for exams. These exemplars solutions are designed by experts in accordance with CBSE Syllabus(2021-2022), which covers all the topics of Class 8 Maths Chapter 7. This chapter is divided into two parts the first part of the chapter is about algebraic expressions and the second part is about factorization. In the first part, the students will learn about different types of algebraic expressions and their identities. In the second part, the students will learn how to factorize given equations.

To learn the concepts in an easy way students are advised to solve the problems provided in the Class 8 NCERT exemplar for chapter 7 algebraic expressions and identities & factorisation. To understand the concepts present in each chapter of Maths as well as Science subject, solve NCERT Exemplars for Class 8, provided here. Let us discuss the topics based on which Exemplars are given for Chapter 7.

• Know about expressions, terms, factors, coefficient, Monomials, Binomials and Polynomials, Like and Unlike Term
• Operations performed on Algebraic Expression
• Identities of algebra and its applications
• Learn to find factors of natural numbers, algebraic expressions, by the method of common factors, by regrouping terms, using identities.
• Division of Algebraic Expressions
• Dividing monomial by monomial
• Dividing polynomial by monomial, etc.

### NCERT Exemplar Class 8 Maths Solutions Chapter 7 Algebraic Expressions And Identities Factorisation:-           ### Access NCERT Exemplar Solutions for Class 8 Maths Chapter 7

Exercise Page: 224

In questions 1 to 33, there are four options out of which one is correct. Write the correct answer.

1. The product of a monomial and a binomial is a

(a) monomial (b) binomial

(c) trinomial (d) none of these

Solution:-

(b) binomial

Let monomial = 2x, binomial = x + y

Then, product of a monomial and a binomial = (2x) × (x + y)

= 2x2 + 2xy

2. In a polynomial, the exponents of the variables are always

(a) integers (b) positive integers

(c) non-negative integers (d) non-positive integers

Solution:-

(b) positive integers

3. Which of the following is correct?

(a) (a – b)2 = a2 + 2ab – b2 (b) (a – b)2 = a2 – 2ab + b2

(c) (a – b)2 = a2 – b2 (d) (a + b)2 = a2 + 2ab – b2

Solution:-

(b) (a – b)2 = a2 – 2ab + b2

We have, = (a – b) × (a – b)

= a × (a – b) – b × (a – b)

= a2 – ab – ba + b2

= a2 – 2ab + b2

4. The sum of –7pq and 2pq is

(a) –9pq (b) 9pq (c) 5pq (d) – 5pq

Solution:-

(d) – 5pq

The given two monomials are like terms.

Then sum of -7pq and 2pg = – 7pq + 2pq

= (-7 + 2) pq

= -5pq

5. If we subtract –3x2y2 from x2y2, then we get

(a) – 4x2y2 (b) – 2x2y2 (c) 2x2y2 (d) 4x2y2

Solution:-

(d) 4x2y2

We have,

The given two monomials are like terms.

Subtract –3x2y2 from x2y2 = x2y2 – (- 3x2y2)

= x2y2 + 3x2y2

= x2y2 (1 + 3)

= 4x2y2

6. Like term as 4m3n2 is

(a) 4m2n2 (b) – 6m3n2 (c) 6pm3n2 (d) 4m3n

Solution:-

(b) – 6m3n2

Like terms are formed from the same variables and the powers of these variables are also the same. But coefficients of like terms need not be the same.

7. Which of the following is a binomial?

(a) 7 × a + a (b) 6a2 + 7b + 2c

(c) 4a × 3b × 2c (d) 6 (a2 + b)

Solution:-

(d) 6 (a2 + b)

Expressions that contain exactly two terms are called binomials.

= 6 (a2 + b)

= 6a2 + b

8. Sum of a – b + ab, b + c – bc and c – a – ac is

(a) 2c + ab – ac – bc (b) 2c – ab – ac – bc

(c) 2c + ab + ac + bc (d) 2c – ab + ac + bc

Solution:-

(a) 2c + ab – ac – bc

We have,

= (a – b + ab) + (b + c – bc) + (c – a – ac)

= a – b + ab + b + c – bc + c – a – ac

Now, grouping like terms

= (a – a) + (-b + b) + (c + c) + ab – bc – ac

= 2c + ab – bc – ac

9. Product of the following monomials 4p, – 7q3, –7pq is

(a) 196 p2q4 (b) 196 pq4 (c) – 196 p2q4 (d) 196 p2q3

Solution:-

(a) 196 p2q4

= 4p × (– 7q3) × (–7pq)

= (4 × (-7) × (-7)) × p × q3 × pq

= 196p2q4

10. Area of a rectangle with length 4ab and breadth 6b2 is

(a) 24a2b2 (b) 24ab3 (c) 24ab2 (d) 24ab

Solution:-

(b) 24ab3

We know that, area of rectangle = length × breadth

Given, length = 4ab, breadth = 6b2

= 4ab × 6b2

= 24ab3

11. Volume of a rectangular box (cuboid) with length = 2ab, breadth = 3ac and height = 2ac is

(a) 12a3bc2 (b) 12a3bc (c) 12a2bc (d) 2ab +3ac + 2ac

Solution:-

(a) 12a3bc2

We know that, volume of cuboid = length × breadth × height

Given, length = 2ab, breadth = 3ac, height = 2ac

= 2ab × 3ac × 2ac

= (2 × 3 × 2) × ab × ac × ac

= 12a3bc2

12. Product of 6a2 – 7b + 5ab and 2ab is

(a) 12a3b – 14ab2 + 10ab (b) 12a3b – 14ab2 + 10a2b2

(c) 6a2 – 7b + 7ab (d) 12a2b – 7ab2 + 10ab

Solution:-

(b) 12a3b – 14ab2 + 10a2b2

Now we have find product of trinomial and monomial,

= (6a2 – 7b + 5ab) × 2ab

= (2ab × 6a2) – (2ab × 7b) + (2ab × 5ab)

= 12a3b – 14ab2 + 10a2b2

13. Square of 3x – 4y is

(a) 9x2 – 16y2 (b) 6x2 – 8y2

(c) 9x2 + 16y2 + 24xy (d) 9x2 + 16y2 – 24xy

Solution:-

(d) 9x2 + 16y2 – 24xy

As per the condition in the question, (3x – 4y)2

The standard identity = (a – b)2 = a2 – 2ab + b2

Where, a = 3x, b = 4y

Then,

(3x – 4y)2 = (3x)2 – (2 × 3x × 4y) + (4y)2

= 9x2 – 24xy + 16y2

14. Which of the following are like terms?

(a) 5xyz2, – 3xy2z (b) – 5xyz2, 7xyz2

(c) 5xyz2, 5x2yz (d) 5xyz2, x2y2z2

Solution:-

(b) – 5xyz2, 7xyz2

Like terms are formed from the same variables and the powers of these variables are also the same. But coefficients of like terms need not be the same.

15. Coefficient of y in the term –y/3 is

(a) – 1 (b) – 3 (c) -1/3 (d) 1/3

Solution:-

(c) -1/3

-y/3 can also be written as y × (-1/3)

So, Coefficient of y is -1/3

16. a2 – b2 is equal to

(a) (a – b)2 (b) (a – b) (a – b)

(c) (a + b) (a – b) (d) (a + b) (a + b)

Solution:-

(c) (a + b) (a – b)

(a2 – b2) = (a + b) (a – b) is one of the standard identity.

17. Common factor of 17abc, 34ab2, 51a2b is

(a) 17abc (b) 17ab (c) 17ac (d) 17a2b2c

Solution:-

(b) 17ab

The given factors can be written in expanded form as,

17abc = 17 × a × b × c

34ab2 = 2 × 17 × a × b × b

51a2b = 3 × 17 × a × a × b

So, common factors in the above is 17 × a × b

= 17ab

18. Square of 9x – 7xy is

(a) 81x2 + 49x2y2 (b) 81x2 – 49x2y2

(c) 81x2 + 49x2y2 –126x2y (d) 81x2 + 49x2y2 – 63x2y

Solution:-

(c) 81x2 + 49x2y2 –126x2y

As per the condition in the question, (9x – 7xy)2

The standard identity = (a – b)2 = a2 – 2ab + b2

Where, a = 9x, b = 7xy

Then,

(9x – 7xy)2 = (9x)2 – (2 × 9x × 7xy) + (7xy)2

= 81x2 – 126x2y + 49x2y2

19. Factorised form of 23xy – 46x + 54y – 108 is

(a) (23x + 54) (y – 2) (b) (23x + 54y) (y – 2)

(c) (23xy + 54y) (– 46x – 108) (d) (23x + 54) (y + 2)

Solution:-

(a) (23x + 54) (y – 2)

Factorised form of 23xy – 46x + 54y – 108 is = 23xy – (2 × 23x) + 54y – (2 × 54)

Take out the common factors,

= 23x (y – 2) + 54 (y – 2)

Again take out the common factor,

= (y – 2) (23x + 54)

20. Factorised form of r2 – 10r + 21 is

(a) (r – 1) (r – 4) (b) (r – 7) (r – 3)

(c) (r – 7) (r + 3) (d) (r + 7) (r + 3)

Solution:-

(b) (r – 7) (r – 3)

Factorised form of r2 – 10r + 21 is = r2 – 7r – 3r + 21

Take out the common factors,

= r (r – 7) – 3 (r – 7)

Again take out the common factor,

= (r – 7) (r – 3)

21. Factorised form of p2 – 17p – 38 is

(a) (p – 19) (p + 2) (b) (p – 19) (p – 2)

(c) (p + 19) (p + 2) (d) (p + 19) (p – 2)

Solution:-

(a) (p – 19) (p + 2)

Factorised form of p2 – 17p – 38 is = p2 – 19p + 2p – 38

Take out the common factors,

= p (p – 19) + 2 (p – 19)

Again take out the common factor,

= (p – 19) (p + 2)

22. On dividing 57p2qr by 114pq, we get

(a) ¼pr (b) ¾pr (c) ½pr (d) 2pr

Solution:-

(c) ½pr

On dividing 57p2qr by 114pq,

It can be expanded as = (57 × p × p × q × r)/(114 × p × q)

= 57pr/114 … [divide both numerator and denominator by 57]

= ½pr

23. On dividing p (4p2 – 16) by 4p (p – 2), we get

(a) 2p + 4 (b) 2p – 4 (c) p + 2 (d) p – 2

Solution:-

(c) p + 2

On dividing p (4p2 – 16) by 4p (p – 2)

= (p((2p)2 – (4)2))/ (4p(p – 2))

= ((2p – 4) × (2p + 4))/(4(p – 2))

Take out the common factors

= ((2(p – 2)) × (2 (p + 4)))/(4(p -2))

= (4(p – 2)(p + 2))/ (4(p – 2))

= p + 2

24. The common factor of 3ab and 2cd is

(a) 1 (b) – 1 (c) a (d) c

Solution:-

(a) 1

Considering the two monomials 3ab and 2cd there is no common factor except 1.

25. An irreducible factor of 24x2y2 is

(a) x2 (b) y2 (c) x (d) 24x

Solution:-

(c) x

An irreducible factor is a factor which cannot be expressed further as a product of factors. Such a factorisation is called an irreducible factorisation.

24x2y2 = 2 × 2 × 2 × 3 × x × x × y × y

Therefore an irreducible factor is x.

26. Number of factors of (a + b)2 is

(a) 4 (b) 3 (c) 2 (d) 1

Solution:-

(c) 2

Number of factors of (a + b)2 is = (a + b) (a + b) no further factorisation is possible.

27. The factorised form of 3x – 24 is

(a) 3x × 24 (b) 3 (x – 8) (c) 24 (x – 3) (d) 3(x – 12)

Solution:-

(b) 3 (x – 8)

The factorised form of 3x – 24 is,

Take out 3 as common,

= 3 (x – 8)

28. The factors of x2 – 4 are

(a) (x – 2), (x – 2) (b) (x + 2), (x – 2)

(c) (x + 2), (x + 2) (d) (x – 4), (x – 4)

Solution:-

(b) (x + 2), (x – 2)

The factors of x2 – 4 are,

X2 – 4 = x2 – 22

= (x + 2) (x – 2)

29. The value of (– 27x2y) ÷ (– 9xy) is

(a) 3xy (b) – 3xy (c) – 3x (d) 3x

Solution:-

(d) 3x

The value of (– 27x2y) ÷ (– 9xy) = (-27 × x × x × y)/(- 9 × x × y)

= (27/9)x … [divide both numerator, denominator by 3]

= 3x

30. The value of (2x2 + 4) ÷ 2 is

(a) 2x2 + 2 (b) x2 + 2 (c) x2 + 4 (d) 2x2 + 4

Solution:-

(b) x2 + 2

The value of (2x2 + 4) ÷ 2 = (2x2 + 4)/2

= (2(x2 + 2))/2

= x2 + 2

31. The value of (3x3 +9x2 + 27x) ÷ 3x is

(a) x2 +9 + 27x (b) 3x3 +3x2 + 27x

(c) 3x3 +9x2 + 9 (d) x2 +3x + 9

Solution:-

(d) x2 +3x + 9

The value of (3x3 +9x2 + 27x) ÷ 3x = (3x3 + 9x2 + 27x)/3x

Takeout 3x as common,

= 3x (x2 + 3x + 9)/3x

= x2 + 3x + 9

32. The value of (a + b)2 + (a – b)2 is

(a) 2a + 2b (b) 2a – 2b (c) 2a2 + 2b2 (d) 2a2 – 2b2

Solution:-

(c) 2a2 + 2b2

(a + b)2 + (a – b)2 = (a2 + b2 + 2ab) + (a2 + b2 – 2ab)

= (a2 + a2) + (b2 + b2) + (2ab – 2ab)

= 2a2 + 2b2

33. The value of (a + b)2 – (a – b)2 is

(a) 4ab (b) – 4ab (c) 2a2 + 2b2 (d) 2a2 – 2b2

Solution:-

(a) 4ab

The value of (a + b)2 – (a – b)2 = (a2 + b2 + 2ab) – (a2 + b2 – 2ab)

= a2 – a2 + b2 – b2 + 2ab + 2ab

= 4ab

In questions 34 to 58, fill in the blanks to make the statements true:

34. The product of two terms with like signs is a term.

Solution:-

The product of two terms with like signs is a positive term.

Let us assume two like terms are, 3p and 2q

= 3p × 2q

= 6pq

35. The product of two terms with unlike signs is a term.

Solution:-

The product of two terms with unlike signs is a negative term.

Let us assume two unlike terms are, – 3p and 2q

= -3p × 2q

= – 6pq

36. a (b + c) = a × ____ + a × _____.

Solution:-

a (b + c) = a × b + a × c. … [by using left distributive law]

= ab + ac

37. (a – b) _________ = a2 – 2ab + b2

Solution:-

(a – b) (a – b) = (a – b)2= a2 – 2ab + b2

(a – b) (a – b)= a × (a – b) – b × (a – b)

= a2 – ab – ba + b2

= a2 – 2ab + b2

38. a2 – b2 = (a + b ) __________.

Solution:-

a2 – b2 = (a + b) (a – b) … [from the standard identities]

39. (a – b)2 + ____________ = a2 – b2

Solution:-

(a – b)2 + (2ab – 2b2) = a2 – b2

= (a – b)2 + (2ab – 2b2)

= a2 + b2 – 2ab + 2ab – 2b2

= a2 – b2

40. (a + b)2 – 2ab = ___________ + ____________

Solution:-

(a + b)2 – 2ab = a2 + b2

= (a + b)2 – 2ab

= a2 + 2ab + b2 – 2ab

= a2 + b2

41. (x + a) (x + b) = x2 + (a + b) x + ________.

Solution:-

(x + a) (x + b) = x2 + (a + b) x + ab

= (x + a) (x + b)

= x × (x + b) + a × (x + b)

= x2 + xb + xa + ab

= x2 + x (b + a) + ab

42. The product of two polynomials is a ________.

Solution:-

The product of two polynomials is a polynomials.

43. Common factor of ax2 + bx is __________.

Solution:-

Common factor of ax2 + bx is x (ax + b)

44. Factorised form of 18mn + 10mnp is ________.

Solution:-

Factorised form of 18mn + 10mnp is 2mn (9 + 5p)

= (2 × 9 × m × n) + (2 × 5 × m × n × p)

= 2mn (9 + 5p)

Apart from these exemplars, students are also made available with exemplar books, NCERT Solutions for 8th standard Maths and question papers to help students practice well for final exams. Students are advised to solve sample papers and previous year question papers which gives an idea of types of questions asked in the board exam from Algebraic expressions and identities and factorisation.

Download BYJU’S-The Learning App and get personalized videos, explaining the concepts of algebra in terms of expression and identities and their factorisation, with the help of pictures and videos experience a new way of learning to understand the theories easily.

## Frequently Asked Questions on NCERT Exemplar Solutions for Class 8 Maths Chapter 7

### List out the important topics covered in NCERT Exemplar Solutions for Class 8 Maths Chapter 7.

The important concepts covered in NCERT Exemplar Solutions for Class 8 Maths Chapter 7 are
1.Know about expressions, terms, factors, coefficient, Monomials, Binomials and Polynomials, Like and Unlike Term
2. Operations performed on Algebraic Expression
3. Identities of algebra and its applications
4. Learn to find factors of natural numbers, algebraic expressions, by the method of common factors, by regrouping terms, using identities.
5. Division of Algebraic Expressions
6. Dividing monomial by monomial
7. Dividing polynomials by monomials, etc.

### What is the meaning of algebraic expressions in NCERT Solutions for Class 8 Maths Chapter 7?

Algebraic expressions are the idea of expressing numbers using letters or alphabets without specifying their actual values. The basics of algebra taught us how to express an unknown value using letters such as x, y, z, etc. These letters are called variables. An algebraic expression can be a combination of both variables and constants. Any value that is placed before and multiplied by a variable is a coefficient.

### How NCERT Exemplar Solutions for Class 8 Maths Chapter 7 helpful for board exam preparation?

Practising NCERT Exemplar Solutions for Class 8 Maths Chapter 7 help the students to top the final exams and ace a subject. These solutions are devised, based on the most updated syllabus, covering all the crucial topics of the respective subjects. Hence, solving these questions will make the students more confident to face the board exams. Topics given in these solutions form the basis for top scores. It also helps the students to get familiar with answering questions of all difficulty levels. These solutions are highly recommended to the students for referencing and to practice for the board exams.
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