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 Solutions for Class 8 Maths Chapter 7 Algebraic Expressions, Identities and Factorisation, are provided here for students to prepare well for exams. These exemplar solutions are designed by subject experts at BYJU’S in accordance with the latest CBSE Syllabus (2023-2024), which covers all the topics of Class 8 Maths Chapter 7. This chapter is divided into two parts; the first part is about algebraic expressions, and the second part is about factorisation. In the first part, students will learn about different types of algebraic expressions and their identities. In the second part, they will learn how to factorise given equations.

To learn the concepts easily, students are advised to solve the problems provided in the Class 8 NCERT Exemplar for Chapter 7, Algebraic Expressions, Identities and Factorisation. To understand the concepts present in each chapter of Maths as well as Science subjects, solve NCERT Exemplars for Class 8, provided here. Let us discuss the topics based on which exemplar solutions are given for Chapter 7.

• Know about expressions, terms, factors, coefficients, 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 and 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 Solutions for Class 8 Maths Solutions Chapter 7 Algebraic Expressions and Identities Factorisation:-Download PDF Here

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 the area of a 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 to find the 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 the 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, the 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 identities.

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 are 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 and 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 polynomial.

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 the exemplar solutions, students are also provided with exemplar books, NCERT Solutions for Class 8 Maths and question papers to help them practise well for final exams. Students are advised to solve sample papers and previous years’ question papers which gives an idea of the types of questions asked in the annual exam from the topic, Algebraic Expressions, Identities and Factorisation.

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