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NCERT Solutions
Class 8
Maths
Chapter 8 Algebraic Expressions and Identities

NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities

Chapter 8 of Class 8 Maths, "Algebraic Expressions and Identities", is an important topic that forms the foundation for understanding algebraic concepts in higher studies. This chapter introduces students to the principles of algebraic expressions, their operations, and numerous standard identities. Mastering these concepts is essential as they play a significant role in solving complex equations and mathematical problems encountered in advanced studies.

The NCERT Solutions for Class 8 Maths Chapter 8 provides detailed explanations and step-by-step solutions to textbook exercises, guiding students through the process of working with algebraic expressions and applying identities effectively. This blog aims to help students enhance their algebraic skills, boost their problem-solving abilities, and gain confidence in their math knowledge as they progress in their academic journey.

1.0Download Class 8 Maths Chapter 8 NCERT Solutions PDF Online

Downloading the NCERT Solutions of Class 8 Chapter 8 Algebraic Expressions and Identities PDF will help you understand the curriculum's key concepts. Below is the link to download the NCERT solutions for Class 8 Maths Chapter 8 PDF.

NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities

2.0Class 8 Maths Chapter 8: Breakdown of Exercises

Exercise

Total Questions

Exercise 8.1

2 Questions

Exercise 8.2

5 Questions

Exercise 8.3

5 Questions

Exercise 8.4

3 Questions

3.0NCERT Questions with Solutions for Class 8 Maths Chapter 8 - Detailed Solutions

Exercise : 8.1

  • Add the following : (i) ab−bc,bc−ca,ca−ab (ii) a−b+ab,b−c+bc,c−a+ac (iii) 2p2q2−3pq+4,5+7pq−3p2q2 (iv) ℓ2+m2, m2+n2,n2+ℓ2,2ℓ m+2mn+2nℓ Sol. (i) Writing the given expressions in separate rows with like terms one below the other, we have ab−bc+bc−ca−ab+ca0+0+0

(ii) Writing the given expressions in separate rows with like terms one below the other, we have

a−b+ab
+ b-c+bc
-a+c+ac
0+0+ab+ + + bc +ac

or ab+bc+ac (iii) Writing the given expressions in separate rows with like terms one below the other, we have

2p2q2−3pq+4−3p2q2+7pq+5−p2q2+4pq+9​​

(iv) Writing the given expressions in separate rows with like terms one below the other, we have ℓ2+m2 +m2+n2 ℓ2+n2 2ℓ2+2m2+2n2+2ℓm+2mn+2nℓ​+2ℓm+2mn+2nℓ​

  • (i) Subtract 4a−7ab+3b+12 from 12a −9ab+5b−3 (ii) Subtract 3xy+5yz−7zx from 5xy− 2yz−2zx+10xyz (iii) Subtract 4p2q−3pq+5pq2−8p+7q -10 from 18−3p−11q+5pq−2q2 +5p2q. Sol. Rearranging the terms of the given expressions, changing the sign of each term of the expression to be subtracted and adding the two expressions, we get, (i) 12a−9ab+5 b−34a−7ab+3 b+12​
-+--
8 a-2 ab+2 b-15

(ii) Rearranging the terms of the given expressions, changing the sign of each term of the expression to be subtracted and adding the two expressions, we get,

5xy−2yz−2zx+10xyz 3xy+5yz−7zx - - 2xy−7yz+5zx+10xyz

(iii) Rearranging the terms of the given expressions, changing the sign of each term of the expression to be subtracted and adding the two expressions, we get. 5p2q−2pq2+5qp−11q−3p+18 4p2q+5q2−3pq+7q−8p−10 p2q−7q2+8pq−18q+5p+28−−+−++​

  • Identify the terms, their coefficients for each of the following expressions : (i) 5xyz​2−3zy (ii) 1+x+x2 (iii) 4x2y2−4x2y2z2+z2 (iv) 3−pq+qr−rp (v) 2x​+2y​−xy (vi) 0.3a−0.6ab+0.5 b Sol. (i) 5xyz2−3zy In the expression 5x2z2−3zy, the terms are 5x′2z2 and −3zy. Coefficient of xyz2 in the term 5x2z2 is 5 . Coefficient of zy in the term −3zy is -3 . (ii) 1+x+x2 In the expression 1+x+x2, the terms are 1,x and x2. Coefficient of x0 in the term 1 is 1 . Coefficient of x in the term x is 1 . Coefficient of x2 is the term x2 is 1 . (iii) In the expression 4x2y2−4x2y2z2+z2, the terms are 4x2y2,−4x2y2z2 and z2. Coefficient of x2y2 in the term 4x2y2 is 4 . Coefficient of x2y2z2 in the term −4x2y2z2 is -4 . Coefficient of z2 in the term z2 is 1 . (iv) 3 - pq + qr - rp In the expression 3−pq+qr−rp, the terms are 3, -pq, qr and -rp Coefficient of x0 in the term 3 is 3 . Coefficient of pq in the term -pq is -1 . Coefficient of qr in the term qr is 1. Coefficient of rp in the term -rp is -1 . (v) In the expression 2x​+2y​−xy, the terms are 2x​,2y​ and −xy Coefficient of x in the term 2x​ is 21​. Coefficient of y in the term 2y​ is 21​. Coefficient of xy in the term -xy is -1 . (vi) In the expression 0.3a−0.6ab+0.5b, the terms are 0.3a,−0.6ab and 0.5 b Coefficient of a in the term 0.3 a is 0.3 . Coefficient of ab in the term -0.6 ab is - 0.6. Coefficient of b in the term 0.5b is 0.5 .
  • Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories? x+y,1000,x+x2+x3+x4,7+y+5x,2y− 3y2,2y−3y2+4y3,5x−4y+3xy,4z−15z2, ab+bc+cd+da,pqr,p2q+q2,2p+2q. Sol. The given polynomials are classified as under: Monomials : 1000, pqr Binomials : x+y,2y−3y2,4z−15z2,p2q+ pq2,2p+2q. Trinomials: 7+y+5x,2y−3y2+4y3,5x −4y+3xy. Polynomials that do not fit in any category : x+x2+x3+x4,ab+bc+cd+da

Exercise : 8.2

  • Find the product of the following pairs of monomials (i) 4,7p (ii) −4p,7p (iii) −4p,7pq (iv) 4p3,−3p (v) 4p,0 Sol. (i) 4×7p=(4×7)×p=28p (ii) −4p×7p=(−4×7)×(p×p) =−28p1+1 =−28p2 (iii) −4p×7pq=(−4×7)×(p×p×q) =−28p1+1q =−28p2q (iv) 4p3×(−3p)={4×(−3)}×(p3×p) =−12×p3+1 =−12p4 (v) 4p×0=(4×0)×(p) =0×p =0
  • Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively : (p, q); (10m, 5n); (20x2,5y2); (4x, 3x2 ); (3mn, 4np) Sol. We know that the area of a rectangle =ℓ×b, where ℓ= length and b= breadth . Therefore, the areas of rectangles with pair of monomials (p, q); (10m, 5n); (20x2,5y2); (4x,3x2) and ( 3mn,4np ) as their lengths and breadths are given by p×q=pq 10 m×5n=(10×5)×(m×n)=50mn 20x2×5y2=(20×5)×(x2×y2) =100x2y2 4x×3x2=(4×3)×(x×x2) =12x3 and, 3mn×4np=(3×4)×(m×n×n×p) =12mn2p
  • Complete the table of products :
2x−5y3x2−4xy7x2y−9x2y2
2x4x2-----
-5y--−15x2y---
3x2------
−4xy------
7x2y------
−9x2y2------

Sol.

First Monomial → Second Monomial ↓2x−5y3x2−4xy7x2y−9x2y2
2x4x2−10xy6x3−8x2y14x3y−18x3y2
−5y−10xy25y2−15x2y20xy2−35x2y245x2y3
3x26x3−15x2y9x4−12x3y21x4y−27x4y2
−4xy−8x2y20xy2−12x3y16x2y2−28x3y236x3y3
7xy14x3y−35x2y221x4y−28x3y249x4y2−63x4y3
−9x2y2−18x3y245x2y3−27x4y236x3y3−63x4y381x4y4
  • Obtain the volume of rectangular boxes with the following length, breadth and height respectively : (i) 5a,3a2,7a4 (ii) 2p,4q,8r (iii) xy,2x2y,2xy2 (iv) a, 2b, 3c Sol. (i) Required volume =5a×3a2×7a4 =(5×3×7)×(a×a2×a4) =105a1+2+4=105a7 (ii) Required volume =2p×4q×8r =(2×4×8)×(p×q×r) =64pqr (iii) Required volume =xy×2x2y×2xy2 =(1×2×2)×(xy×x2y×x2) =(4)×(x1+2+1×y1+1+2) =4x4y4 (iv) Required volume =a×2b×3c =(1×2×3)×(a×b×c) =6abc
  • Obtain the product of (i) xy,yz,zx (ii) a,−a2,a3 (iii) 2,4y,8y2,16y3 (iv) a, 2b, 3c, 6abc (v) m,−mn,mnp Sol. (i) xy×yz×zx=x×x×y×y×z×z =x1+1×y1+1×z1+1 =x2y2z2 (ii) a×(−a2)×a3 =[1×(−1)×1]×(a×a×a×a×a×a) =(−1)×(a6) =−a6 (iii) 2×(4y)×8y2×16y3 =(2×4×8×16)×(y×y2×y3) =(1024)×(y1+2+3) =1024y6 (iv) a×2 b×3c×6abc =(2×3×6)×(a×b×c×abc) =(36)×(a1+1×b1+1×c1+1) =36a2b2c2 (v) m×−mn×mnp =(1×−1×1)×(m×m×m×n×n×p) =−1×m3×n2×p =−m3n2p

Exercise : 8.3

  • Carry out the multiplication of the expressions in each of the following pairs: (i) 4p,q+r (ii) ab, a - b (iii) a+b,7a2b2 (iv) a2−9,4a (v) pq+qr+rp,0 Sol. (i) 4p×(q+r)=4p×q+4p×r =4pq+4pr (ii) ab×(a−b)=ab×a−ab×(b) =a2b−ab2 (iii) (a+b)×(7a2b2)=7a2b2×a+7a2b2×b =7a3 b2+7a2 b3 (iv) (a2−9)×4a=a2×4a−9×4a =4a3−36a (v) (pq+qr+rp)×0=0
  • Complete the table:
First expressionSecond expressionProduct
(i)ab+c+d-
(ii)x+y−55 xy-
(iii)p6p2−7p+5-
(iv)4p2q2p2−q2-
(v)a+b+cabc-

Sol. (i) a(b+c+d)=a×b+a×c+a×d=ab +ac+ad (ii) (x+y−5)×(5xy) =5xy×x+5xy×y+5xy×(−5) =5x2y+5xy2−25xy (iii) 6p3−7p2+5p (iv) 4p4q2−4p2q4 (v) a2bc+ab2c+abc2

  • Find the product: (i) (a2)×(2a22)×(4a26) (ii) (32​xy)×(10−9​x2y2) (iii) (−310​pq3)×(56​p3q) (iv) x×x2×x3×x4 Sol. (i) (a2)×(2a22)×(4a26)=(1×2×4)× (a2×a22×a26)=8a2+22+26=8a50 (ii) (32​xy)×(10−9​x2y2) =(32​×10−9​)×(x×x2×y×y2) =−53​×x1+2×y1+2 =−53​x3y3 (iii) (−310​pq3)×(56​p3q) =(−310​×56​)×(p×q3×p3×q) =(−4)×(p4×q4)=−4p4q4 (iv) x×x2×x3×x4=x1+2+3+4=x10
  • (a) Simplify: 3x(4x−5)+3 and find its values for (i) x=3, (ii) x=21​. (b) Simplify: a(a2+a+1)+5 and find its value for (i) a=0, (ii) a=1, (iii) a=−1. Sol. (a) We have, 3x(4x−5)+3=3x×4x−3x×5+3 =12x2−15x+3 (i) When x=3, then 12(3)2−15(3)+3=66 (ii) When x=21​, then 12(21​)2−15×21​+3=3−215​+3=−23​ (b) We have, a(a2+a+1)+5 =a×a2+a×a+a×1+5 =a3+a2+a+5 (i) When a=0, then a3+a2+a+5 =03+02+0+5 =5 (ii) When a=1, then a3+a2+a+5 13+12+1+5 =1+1+1+5 =3+5=8 (iii) When a=−1, then a3+a2+a+5 =(−1)3+(−1)2+(−1)+5 =−1+1−1+5=4
  • (a) Add: p(p−q),q(q−r) and r(r−p) (b) Add: 2x(z−x−y) and 2y(z−y−x) (c) Subtract: 3ℓ(ℓ−4m+5n) from 4ℓ(10n−3m+2ℓ) (d) Subtract: 3a(a+b+c)−2 b(a−b+c) from 4c(−a+b+c) Sol. (a) p(p−q)+q(q−r)+r(r−p) =p×p−p×q+q×q−q×r+r×r −r×p =p2−pq+q2−qr+r2−pr =p2+q2+r2−pq−qr−pr (b) 2x(z−x−y)+2y(z−y−x) =(2x×z)−(2x×x)−(2x×y)+(2y×z) −(2y×y)−(2y×x) =2xz−2x2−2xy+2yz−2y2−2xy =−2x2−2y2−4xy+2xz+2yz. (c) 4ℓ(10n−3m+2ℓ)−3ℓ(ℓ−4m+5n) =(4ℓ×10n)−(4ℓ×3 m)+(4ℓ×2ℓ)+ (−3ℓ×ℓ)+(−3ℓ×−4 m)+(−3ℓ×5n) =40ℓn−12ℓ m+8ℓ2−3ℓ2+12ℓ m−15ℓn =5ℓ2+25ℓn (d) 4c(−a+b+c)−{3a(a+b+c)−2 b(a−b+c)} ={4c×(−a)+4c×b+4c×c}−[{(3a ×a+3a×b+3a×c)}+{(−2b×a)+ (−2b×−b)+(−2b×c)}] =−4ac+4bc+4c2−[3a2+3ab+3ac −2ab+2b2−2bc] =−4ac+4bc+4c2−3a2−3ab−3ac+ 2ab−2 b2+2bc =−3a2−2b2+4c2−7ac+6bc−ab

Exercise: 8.4

  • Multiply the binomials: (i) (2x+5) and (4x−3) (ii) (y−8) and (3y−4) (iii) (2.5ℓ−0.5 m) and (2.5ℓ+0.5 m) (iv) (a+3 b) and (x+5) (v) (2pq+3q2) and (3pq−2q2) (vi) (43​a2+3b2) and 4(a2−32​b2) Sol. (i) (2x+5) and (4x−3) =2x(4x−3)+5(4x−3) =8x2−6x+20x−15=8x2+14x−15 (ii) (y−8)×(3y−4)=y(3y−4)−8(3y−4) =3y2−4y−24y+32 =3y2−28y+32 (iii) (2.5ℓ−0.5 m)×(2.5ℓ+0.5 m) =2.5ℓ(2.5ℓ+0.5 m)−0.5 m(2.5ℓ+0.5 m) =6.25ℓ2+1.25ℓ m−1.25ℓ m−0.25 m2 =6.25ℓ2−0.25 m2 (iv) (a+3b)×(x+5) =a(x+5)+3b(x+5) =ax+5a+3bx+15b =ax+3bx+5a+15b (v) (2pq+3q2)×(3pq−2q2) =2pq(3pq−2q2)+3q2(3pq−2q2) =6p2q2−4pq3+9pq3−6q4 =6p2q2+5pq3−6q4 (vi) (43​a2+3b2) and 4(a2−32​b2) =(43​a2+3b2)×(4a2−38​b2) =43​a2(4a2−38​ b2)+3 b2(4a2−38​ b2) =3a4−2a2b2+12a2b2−8b4 =3a4+10a2b2−8b4
  • Find the product: (i) (5−2x)(3+x) (ii) (x+7y)(7x−y) (iii) (a2+b)(a+b2) (iv) (p2−q2)(2p+q) Sol. (i) (5−2x)(3+x)=5(3+x)−2x(3+x) =15+5x−6x−2x2 =−2x2−x+15 (ii) (x+7y)(7x−y) =x(7x−y)+7y(7x−y)=7x2−xy+49xy−7y2=7x2+48xy−7y2 (iii) (a2+b)(a+b2) =a2(a+b2)+b(a+b2)=a3+a2b2+ab+b3 (iv) (p2−q2)(2p+q)=p2(2p+q)−q2(2p+q) =2p3+p2q−2pq2−q3
  • Simplify : (i) (x2−5)(x+5)+25 (ii) (a2+5)(b3+3)+5 (iii) (t+s2)(t2−s) (iv) (a+b)(c−d)+(a−b)(c+d)+2(ac+bd) (v) (x+y)(2x+y)+(x+2y)(x−y) (vi) (x+y)(x2−xy+y2) (vii) (1.5x−4y)(1.5x+4y+3)−4.5x+12y (viii) (a+b+c)(a+b−c) Sol. (i) (x2−5)(x+5)+25 =x2(x+5)−5(x+5)+25=x3+5x2−5x−25+25=x3+5x2−5x (ii) (a2+5)(b3+3)+5 =a2(b3+3)+5(b3+3)+5 =a2b3+3a2+5b3+15+5 =a2b3+3a2+5b3+20 (iii) (t+s2)(t2−s) =t(t2−s)+s2(t2−s) =t3−ts+s2t2−s3 (iv) (a+b)(c−d)+(a−b)(c+d)+2(ac+bd) =a(c−d)+b(c−d)+a(c+d)−b(c+d) +2ac+2bd =ac−ad+bc−bd+ac+ad−bc−bd +2ac+2bd =4ac (v) (x+y)(2x+y)+(x+2y)(x−y) =x(2x+y)+y(2x+y)+x(x−y)+2y(x−y) =2x2+xy+2xy+y2+x2−xy+2xy−2y2 =(2+1)x2+(1+2−1+2)xy+(1−2)y2 =3x2+4xy−y2 (vi) (x+y)(x2−xy+y2) =x(x2−xy+y2)+y(x2−xy+y2) =x3−x2y+xy2+x2y−xy2+y3 =x3+y3 (vii) (1.5x−4y)(1.5x+4y+3)−4.5x+12y =1.5x(1.5x+4y+3)−4y(1.5x+4y+3) −4.5x+12y =1.5x×1.5x+1.5x×4y+1.5x×3 −4y×1.5x−4y×4y−4y×3−4.5x +12y =2.25x2+6xy+4.5x−6xy−16y2 −12y−4.5x+12y =2.25x2+(6−6)xy+(4.5−4.5)x −16y2+(−12+12)y =2.25x2+(0)xy+(0)x−16y2+(0)y =2.25x2+0+0−16y2+0 =2.25x2−16y2 (viii) (a+b+c)(a+b−c) =a(a+b−c)+b(a+b−c)+c(a+b−c) =a2+ab−ac+ab+b2−bc+ac+bc−c2 =a2+2ab+b2−c2

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 to Graphs


CBSE Notes for Class 8 Maths - All Chapters:-

Class 8 Maths Chapter 1 - Rational Numbers Notes

Class 8 Maths Chapter 2 - Linear Equations In One Variable Notes

Class 8 Maths Chapter 3 - Understanding Quadrilaterals Notes

Class 8 Maths Chapter 4 - Data Handling Notes

Class 8 Maths Chapter 5 - Squares and Square Roots Notes

Class 8 Maths Chapter 6 - Cubes and Cube Roots Notes

Class 8 Maths Chapter 7 - Comparing Quantities Notes

Class 8 Maths Chapter 8 - Algebraic Expressions and Identities Notes

Class 8 Maths Chapter 9 - Mensuration Notes

Class 8 Maths Chapter 10 - Exponents and Powers Notes

Class 8 Maths Chapter 11 - Direct and Inverse Proportions Notes

Class 8 Maths Chapter 12 - Factorisation Notes

Class 8 Maths Chapter 13 - Introduction to Graphs Notes

Frequently Asked Questions:

NCERT Solutions for Class 8 Maths Chapter 8 provides detailed and step-by-step solutions to all of the questions. To improve understanding of the subject and give students the confidence they need, the elements that make up an expression have been covered in detail. Because of the NCERT books' outstanding quality, the CBSE board requests that students consult them.

NCERT Class 8 Chapter 8 contains 15 questions in total. In addition to the exercise questions, students can practice the solved examples for better conceptual understanding.

Expressions have many uses and are similar to independent mathematical entities. You have the chance to learn about this crucial and practical idea in this chapter. Furthermore, the concept is simple, but the applications that can be made with it are incredibly creative. Expressions can range from simple to complex. For these reasons, it is important to learn how to apply algebraic expressions. With plenty of questions in NCERT Solutions Class 8 Maths Algebraic Expressions and Identities, the students can explore every basic detail.

The concept of algebraic expressions and identities is covered in detail in NCERT Solutions Class 8 Maths Chapter 8 with the aid of appropriate explanations and, most importantly, by introducing the "terms," or components, which make up an expression. It then focuses on the "coefficient" and "variable," which are the essential components of a term, before taking a macro perspective on the various types of expressions and how they function.

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