NCERT Solutions Class 8 Maths Chapter 9

NCERT Solutions for Class 8 Mathematics Chapter 9 Algebraic Expressions and Identities

Mathematics deals with numbers. It requires a lot of practice to score good marks in Mathematics. Our NCERT Solutions for Class 8 Mathematics Chapter 9 has clarified the explanations required for the understanding of this chapter of Mathematics. 

The chapter covers topics like terms, factors, coefficients, monomials, binomials, polynomials, like and unlike, and standard identities.

Our solution is built based on the CBSE latest 2021-2022 syllabus in which we offer step-by-step answers to the NCERT questions which helps students to comprehend the concepts thoroughly. Students can take full advantage of Extramarks NCERT Solutions for Class 8 Mathematics Chapter 9 for their preparation.

You can regularly visit Extramarks’ website to check the latest updates and access all the study material related to Class 8 Mathematics preparation.

Key Topics Covered In NCERT Solutions for Class 8 Mathematics Chapter 9

In NCERT Solutions for Class 8 Mathematics Chapter 9, students will cover all topics and concepts in a detailed manner through various step-by-step solved answers and examples. 

The Chapter Algebraic Expressions and identities Class 8 covers one of the most important units of Algebra required in higher classes. It includes all the core concepts required to solve the higher-level questions. For a better understanding of this chapter, one must go through NCERT Solutions for Class 8 Mathematics Chapter 9 thoroughly. The Chapter lays the foundation for the upcoming topics based on algebra. Hence, students must try to leverage their conceptual understanding in this chapter.

The best way to become proficient in this chapter is by referring to NCERT Solutions for Class 8 Mathematics Chapter 9. The chapter begins with a basic introduction to expressions. It helps students learn the difference between like and unlike terms, how to apply identities, operations of algebraic expressions, etc. if you study this chapter properly, you will also develop critical and logical thinking skills. 

The key topics covered in this chapter 

Exercise  Topics
9.1 What are Expressions?
9.2 Terms, Factors and Coefficients 
9.3 Monomials, Binomials and Polynomials
9.4 Like and Unlike Terms
9.5 Addition and Subtraction of Algebraic Expressions
9.6 Multiplication of Algebraic Expressions: Introduction
9.7 Multiplying a Monomial by a Monomial
9.8 Multiplying a Monomial by a Polynomial
9.9 Multiplying a Polynomial by a Polynomial
9.10 What is an Identity?
9.11 Standard Identities
9.12 Applying Identities
9.13 Summary

The Chapter helps to bridge the gap between algebra and geometry and builds a strong foundation in various applications of Mathematics.

9.1 What are Expressions?

Expressions include the constants and variables. Based on the variables, the value of expression changes. The examples of expressions are as follows

  • 2y-5
  • 4xy+7. 

For more examples of expression, you can refer to NCERT Solutions for Class 8 Mathematics Chapter 9 from the Extramarks website and make your concepts clear.

9.2 Terms, Factors and Coefficients 

Terms are basic units that make up an expression. Terms are added to form an expression. The different terms are multiplied with a factor in an expression. Coefficients are the numerical factors of a term. The terms, factors and coefficients in a unit make up an expression.

9.3 Monomial, Binomial and Polynomials

An expression consisting of only one term is called Monomial. On the other hand, the binomial consists of two terms, and a polynomial has many terms.

NCERT Solutions for Class 8 Mathematics Chapter 9 has a detailed explanation of differentiating between Monomial, Binomial and Polynomials.

9.4 Like and Unlike terms 

Two expressions which are Monomial or Binomials or Polynomials, are called like terms, whereas two expressions, in which one expression is a Monomial, and the other one is Binomial, are called, unlike terms.

9.5 Addition and Subtraction of Algebra Expressions

Two algebraic expressions can be added or subtracted if they are like terms. For adding or Subtracting, we place each expression in different rows placing each Like Terms one below the other.

9.6 Multiplication of Algebraic expressions -Introduction

In this section, you will learn the fundamentals of the multiplication of algebraic expressions, the steps involved in carrying out multiplication and the ways to execute the results of multiplication. 

You will also get to know the multiplication of Monomial, Binomials and Polynomials separately in detail in NCERT Solutions for Class 8 Mathematics Chapter 9.

9.7 Multiplication of a Monomial by a Monomial

When a Monomial is multiplied by itself, the multiplication is termed as Multiplication of a Monomial by a Monomial. 

It involves only one term i.e. a single constant and a variable.

For example; 

x × 3y

5x × 4y2

To know more about this section, refer to NCERT Solutions for Class 8 Mathematics Chapter 9.

9.8 Multiplication of a Monomial by a Polynomial

When a Monomial is multiplied by a Polynomial, the multiplication is termed as Multiplication of a Monomial by a Polynomial.

It involves two terms i.e. two constants and two variables.

For example;

4xy × 5x2y2

-20x × xyz

You can find more examples on this topic in the NCERT Solutions for Class 8 Mathematics Chapter 9 available on the Extramarks website.

9.9 Multiplication of a Polynomial by a Polynomial 

When a Polynomial is multiplied by a Polynomial, the multiplication is termed as the Multiplication of a Polynomial by a Polynomial.

It involves two or more than two terms i.e a group of constants and variables

For example;

(3a + 4b) × (2a+3b)

(x-4) × (2x+3)

You can practice lots of problems based on it, you can avail the NCERT solutions for Class 8 Mathematics Chapter 9 from the Extramarks website.

9.10 What is an Identity?

When writing an algebraic expression, if the variables of the left-hand side and right-hand side show equality, it is called an identity. Identities are of great help in the algebra section of Mathematics and make complex problems easy.

9.11 Standard Identity

One can get these identities by multiplying one binomial with the other binomial. There are some standard identities in algebra that function as fundamental units. Students must learn these identities on the tip of their tongue in order to excel in this section.

All the identities are listed in a sequential manner in our NCERT Solutions for Class 8 Mathematics Chapter 9, available on the Extramarks website.

9.12 Applying Identities

Students can easily apply identities in various problems applications once they are thorough with standard identities. They just have to note down the rules of application which they can easily find in NCERT Solutions for Class 8 Mathematics Chapter 9.

9.13 Summary

This chapter is considered to be the backbone of algebra and has a long way to go in the Mathematics syllabus of Class 9, Class 10, Class 11 and Class 12.  Students should be thorough with various core concepts of this chapter in order to be strong in Mathematics and face the examinations confidently.

NCERT Solutions for Class 8 Mathematics Chapter 9 Exercise & Answer Solutions

One can find NCERT Solutions for Class 8 Mathematics Chapter 9 on the Extramarks’ website. In-depth explanations to all questions from the NCERT textbook are provided in our NCERT solutions. You should go through it sincerely to approach your CBSE examination to your full potential.

In our solutions guide for the Class 8 Mathematics chapter 9, we have covered all the topics of the chapter mentioned in the textbook. Experts advise students to revise this chapter multiple times as it’s a core chapter. It is used a lot in the advanced level Mathematics syllabus for engineering, architecture and science courses.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions for Class 8 Mathematics. Chapter 9:

Chapter 9 – Algebraic Expressions and Identities Exercises
Exercise 9.1 Algebraic Expressions and Identities Questions & Solutions
Exercise 9.2 Terms, Factors and Coefficients
Monomials, Binomials and Polynomials
Questions & Solutions
Exercise 9.3 Addition and Subtraction of Algebraic Expressions Questions & Solutions
Exercise 9.4 Multiplication of Algebraic Expressions Questions & Solutions
Exercise 9.5 Multiplying two monomials Questions & Solutions

Along with this, students can also refer to other solutions for primary and secondary classes:

NCERT Exemplar Class 8 Mathematics 

NCERT Exemplar Class 8th Mathematics is a complete source of information and covers all the main topics and sub-topics of the chapter. While revising from NCERT Exemplars, students gather complete knowledge of the Mathematics concepts and help connect on a variety of interlinked topics between various chapters. 

Exemplar books have proved beneficial for students in CBSE and other curricula as well. It covers complex theories that confidently prepare students to face the upcoming examinations. Students will get a more profound knowledge of solving complex queries, and it will help them to prepare for boards. 

Key Features of NCERT Solutions for Class 8 Mathematics Chapter 9

To excel in Exams, students must have the right study material. Therefore, NCERT Solutions for Class 8 Mathematics Chapter 9 offers a complete solution for all problems. The key features are provided: 

  • Extramarks NCERT Solutions for Class 8 Mathematics Chapter 9 have been designed as per the latest CBSE syllabus.
  • The solutions help to clear the student’s doubts by building the base and explaining the chapter’s core concepts through detailed answers to all the NCERT questions of this chapter. 
  • With this chapter’s solutions, students will be able to solve many advanced-level problems.

Q.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 + l2,2lm + 2mn + 2nl


(i)abbc bcca + ab +ca¯ 0(ii) ab+ab b c+bc + a +c +ac¯ ab + bc + ac(iii) 2p2q23pq+4+ 3p2q2+7pq+5¯ –1p2q2+4pq+9

iv l2+m2+ m2+n2l2+n2 +2lm+2mn+2nl¯ 2l2+2m2+2n2+2lm+2mn+2nl

Q.2 (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q


(a) 12a 9ab+ 5b 3 4a 7ab+ 3b+ 12(–) (+) (–) (–)¯ 8a 2ab+ 2b 15¯ (b) 5xy 2yz 2zx+ 10xyz 3xy+ 5yz 7zx(–) (–) (+)¯ 2xy – 7yz+5zx+10xyz¯(c) 18 3p 11q+ 5pq 2pq2+ 5p2q 10 8p + 7q 3pq + 5pq2+4p2q (+) (+) () (+) () ()¯ 28 +5p 18q +8pq 7pq2 +1p2q¯

Q.3 Find the product of the following pairs of monomials.

(I) 4, 7p

(ii) – 4p, 7p

(iii) – 4p, 7pq

(iv) 4p3, –3p

(v) 4p, 0


(i) 4 × 7p = 4 × 7 × p = 28p

(ii) –4p × 7p = (–4) × 7 × p × p = –28p2

(iii) –4p × 7pq =(–4) × 7 × p × p × q = –28p2q

(iv) 4p3 × (–3p) = 4 × (–3) × p3 × p = –12p4

(v) 4p × 0 = 4 × p × 0 = 0

Q.4 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)


          Area of rectangle = Length × BreadthArea of first rectangle = p×q =pq Area of second rectangle = 10m× 5n =10×5×m×n =50 mn Area of third rectangle =20x2×5y2 =20×5×x2×y2 =100 x2y2Area of fourth rectangle =4x×3x2 =4×3×x×x2 = 12x3 Area of fifth rectangle =3mn×4np =3×4×m×n×n×p =12 mn2p

Q.5 Complete the table of products.




Q.6 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


Volume of the the rectangular boxes = Length × Breadth × Height

(i) Volume = 5a × 3a2 × 7a4

= 5 ×3 ×7 × a × a2 × a4

= 105a7

(ii) Volume = 2p × 4q × 8r

= 2 × 4 × 8 × p × q × r

= 64pqr

(iii) Volume = xy × 2x2y × 2xy2

= 2 × 2 × x × x2 × x × y × y × y2

= 4x4y4

(iv) Volume = a × 2b × 3c

= 2 × 3 × a × b × c

= 6abc

Q.7 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


(i) xy×yz×zx=x×y×y×z×z×x=x2y2z2 (ii) a×(a2)×a3=a6(iii) 2×4y×8y2×16y3=2×4×8×16 ×y×y2×y3=1024y6(iv) a×2b×3c×6abc=2×3×6×a×b×c×abc=36a2b2c2 (v) m×(mn)×mnp=m×(m)×n×m×n×p=m3n2p

Q.8 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


(i) 4p×(q+r)=(4p×q)+(4p×r)=4pq+4pr(ii) ab×(ab)=ab×aab×b=a2bab2 (iii)(a+b)× 7a2b2= a ×7a2b2+b×7a2b2= 7a3b2 + 7a2b3 (iv)a2 9× 4a=a2× 4a 9×4a=4a336a (v) (pq+qr+rp)× 0=(pq×0)+(qr×0)+(rp ×0)= 0

Q.9 Complete the table.

First ExpressionSecond ExpressionProduct(i)ab+c+d...(ii)x+y55xy...(iii)p6p27p+5...(iv)4p2q2p2q2...(v)a+b+cabc...


First ExpressionSecond ExpressionProduct(i)ab+c+dab+ac+ad(ii)x+y55xy5x2y+5xy225xy(iii)p6p27p+56p37p2+5p(iv)4p2q2p2q24p4q24p2q4(v)a+b+cabca2bc+ab2c+abc2






(a) Simplify 3x(4x5)+3 and find its values fori x=3 ii x=12(b) Simplify a(a2+a+1)+5 and find its value for(i) a=0, (ii) a=1, (iii) a=1


(a)3x (4x – 5) + 3= 3x×4x-3x×5+3= 12x2-15x+3(i) For x = 3, 12x2-15x+3= 1232-15×3+3= 108-45+3= 66

ii For x=12,12x215x+3=12(12)215×12+3=12×14152+3=3152+3=6152=12152=32(b)a(a2+a+1)+5=a×a2+a×a+a×1+5=a3+a2+a+5(i)For a=0,a3+a2+a+5=0+0+0+5=5(ii) For a=1,a3+a2+a+5=13+12+1+5=8iii For a=1,a3+a2+a+5=(1)3+(1)2+(1)+5=(1)+1+(1)+5=4


(a) Add:p(pq),q(qr) and r(rp)(b) Add:2x(zxy) and 2y(zyx)(c) Subtract : 3l(l4m+5n) from 4l(10n3m+2l)(d) Subtract : 3a(a+b+c)2b(ab+c)from 4c(a+b+c)


First we will simplify the given expressions and then add or subtract them.(a)p(pq)=p2pqq(qr)=q2qrr(rp)=r2rpNow,adding all the expressions p2pqq2qr + r2rp ¯ p2pq+q2qr+r2rp           ¯(b) 2x(zxy)=2xz2x22xy2y(zyx)=2yz2y22yxNow,adding all the expressions 2xz2x22xy+ 2yx+2yz2y2¯ 2xz2x24xy+2yz2y2    ¯         (c) 3l(l4m+5n)=3l212lm+15ln4l(10n3m+2l)=40ln12lm+8l2 On subtracting them, we get 40ln12lm+8l2 15ln12lm+3l2()       (+) ()¯ 25ln +5l2

(d) 3a(a+b+c)2b(ab+c)=3a2+ab+3ac+2b22bc4c(a+b+c)=4ac+4cb+4c2 On subtracting them, we get 4ac+4cb+4c2 +3ac2bc +3a2+ab+2b2 () (+) ()​ () ()¯ 7ac + 6bc + 4c23a2ab2b2¯




(i)(2x+5) and (4x3)=2x×(4x3)+5×(4x3)=8x26x+20x15=8x2+14x15(ii)(y8) and (3y4)=y×(3y4)8×(3y4)=3y24y24y+32=3y2+3228y(iii)(2.5l0.5m) and (2.5l + 0.5m)=2.5l×(2.5l+0.5m)0.5m×(2.5l+0.5m)=6.25l2+1.25lm1.25lm0.25m2=6.25l20.25m2(iv)(a+3b) and (x+5)=a×(x+5)+3b×(x+5)=ax+5a+3bx+15b(v)(2pq+3q2) and (3pq2q2)=2pq×(3pq2q2)+3q2×(3pq2q2)=6p2q24pq3+9pq36q4=6p2q2+5pq36q4vi34a2+3b2 and 4(a223b2)=(34a2)×4(a223b2)+(3b2)×4(a223b2)=3a42a2b2+12a2b28b4=3a4+10a2b28b4

Q.14 Find the product.

(i) (5 – 2x) (3 + x)

(ii) (x + 7y) (7x – y)

(iii) (a2 + b) (a + b2)

(iv) (p2 – q2) (2p + q)


i 5 2x 3 +x=5×3 +x2x3 +x                          =15+5x6x2x2                          =15x2x2ii x+ 7y 7xy=x×7xy+7y×7xy                           =7x2xy+49xy7y2                           =7x2+48xy7y2iii a2+b a+b2=a2×a+b2+b×a+b2                          =a3+a2b2+ba+b3iv p2q2 2p+q=p2×2p+qq2×2p+q                           =2p3+p2q2q3q3                           =2p3+p2q3q3

Q.15 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)


(i) (x2 5)(x+ 5) + 25= x2(x+ 5)5 (x+ 5)+25=x3+5x25x25+25=x3+5x25x(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)(t2s)=t(t2s)+s2(t2s)=t3ts+s2t2s3(iv)(a+b)(cd) + (ab)(c+d)+2 (ac+bd)=a(cd)+b(cd)+a(c+d)b(c+d)+2 (ac+bd)=acad+bcbd+ac+adbcbd+2ac+2bd=4ac(v) (x+y)(2x+y) + (x+ 2y)(xy) =x(2x+y)+y(2x+y)+x(xy) +2y(xy) =2x2+xy+2xy+y2+x2xy+2xy2y2=3x2y2+4xy (vi) (x+y)(x2xy+y2)=x(x2xy+y2)+y(x2xy+y2)=x3x2y+xy2+yx2xy2+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=2.25x2+6xy+4.5x6xy16y212y4.5x+12y=2.25x216y2(viii) (a+b+c)(a+bc) =a(a+bc) +b(a+bc)+c (a+bc) =a2+abac+ab+b2bc+ac+cbc2=a2+b2c2+2ab




ix+3x+3=(x+3)2=(x)2+2×3×x+(3)2 {By using the identity(a+b)2=a2+b2+2ab}=x2+6x+9ii2y+52y+5=(2y+5)2=(2y)2+2×5×2y+(5)2 {By using the identity(a+b)2=a2+b2+2ab}=4y2+20y+25iii2a72a7=(2a7)2=(2a)22×2a×7+(7)2 {By using the identity(ab)2=a2+b22ab}=4a228a+49iv3a123a12=(3a12)2=(3a)22×3a×12+(12)2 {By using the identity(ab)2=a2+b22ab}=9a23a+14(v)(1.1m0.4)(1.1m+0.4)=(1.1m)2(0.4)2 {By using the identity(a+b)(ab)=a2b2}=1.21m20.16(vi)(a2+b2)(a2+b2)=(b2)2(a2)2 {By using the identity(a+b)(ab)=a2b2}=b4a4(vii)(6x7)(6x+7)=(6x)2(7)2 {By using the identity(a+b)(ab)=a2b2}=36x249(viii)(a+c)(a+c)=(a+c)2 {By using the identity(a+b)2=a2+b2+2ab} =c2+a22ac

ix (x2+3y4)(x2+3y4)=(x2+3y4)2=(x2)2+(3y4)2+2(x2)(3y4) {By using the identity(a + b)2=a2+b2+2ab}= x24+9y216+3xy4x 7a9b7a9b=(7a9b)2=(7a)2+(9b)22(7a)(9b) {By using the identity(ab)2=a2+b22ab}=49a2+81b2126ab








i(b7)2=(b)22×b×7+(7)2 {By using the identity(a b)2=a2+b22ab}=b214b+49ii(xy+3z)2=(xy)2+2×xy×3z+(3z)2 {By using the identity(a + b)2=a2+b2+2ab}=x2y2+6xyz+9z2iii(6x25y)2=(6x2)22×6x2×5y+(5y)2 {By using the identity(a b)2=a2+b22ab}=36x460x2y+25y2iv(23m+32n)2=(23m)2+2×23m×32n+(32n)2{By using the identity(a + b)2=a2+b2+2ab}=49m2+2mn+94n2v(0.4p0.5q)2=(0.4p)22×0.4p×0.5q+(0.5q)2 {By using the identity(a b)2=a2+b22ab}=0.16p20.4pq+0.25q2vi(2xy+5y)2=(2xy)2+2×2xy×5y+(5y)2 {By using the identity(a + b)2=a2+b2+2ab}=4x2y2+20xy2+25y2




i(a2b2)2=(a2)22×a2×b2+(b2)2{By using the identity(a b)2=a2+b22ab}=(a)42a2b2+(b)4ii(2x+5)2(2x5)2=[(2x)2+2×2x×5+(5)2][(2x)22×2x×5+(5)2] {By using the identity(a + b)2=a2+b2+2ab}{By using the identity(ab)2=a2+b22ab}=4x2+20x+254x2+20x25=40xiii(7m8n)2+(7m+8n)2=[(7m)22×7m×8n+(8n)2]+[(7m)2+2×7m×8n+(8n)2] {By using the identity(a b)2=a2+b22ab}{By using the identity(a + b)2=a2+b2+2ab}=49m2112mn+64n2+49m2+112mn+64n2=98m2+128n2iv(4m+5n)2+(5m+4n)2=[(4m)2+2×4m×5n+(5n)2]+[(5m)2+2×5m×4n+(4n)2] {By using the identity(a + b)2=a2+b2+2ab}=(16m2+40mn+25n2)+(25m2+40mn+16n2)=41m2+80mn+41n2v(2.5p1.5q)2(1.5p2.5q)2=[(2.5p)22×2.5p×1.5q+(1.5q)2][(1.5p)22×1.5p×2.5q+(2.5q)2] {By using the identity(a b)2=a2+b22ab}=6.25p27.5pq+2.25q2[2.25p27.5pq+6.25q2]=6.25p27.5pq+2.25q22.25p2+7.5pq6.25q2=4p24q2

vi(ab+bc)22ab2c=[(ab)2+2×ab×bc+(bc)2]2ab2c {By using the identity(a + b)2=a2+b2+2ab}=a2b2+2ab2c+b2c22ab2c=a2b2+b2c2vii(m2n2m)2+2m3n2=[(m2)22×m2×n2m+(n2m)2]+2m3n2 {By using the identity(a b)2=a2+b22ab}=m42m3n2+n4m2++2m3n2=m4+n4m2


Showthat.i (3x+7)284x=(3x7)2ii (9p5q)2+180pq=(9p+5q)2iii (43m34n)2+2mn=169m2+916n2iv (4pq+3q)2(4pq3q)2=48pq2v (ab)(a+b)+(bc)(b+c)+(ca)(c+a)=0


i L.H.S=(3x+7)284x=(3x)2+72+2×3x×784x=9x2+49+42x84x=9x2+4942x R.H.S=(3x7)2=(3x)2+722×3x×7=9x2+4942xL.H.S=R.H.SHence ProvediiL.H.S=(9p5q)2+180pq=(9p)2+(5q)22×9p×5q+180pq=81p2+25q290pq+180pq=81p2+25q2+90pq R.H.S=(9p+5q)2=(9p)2+(5q)2+2×9p×5q=81p2+25q2+90pqL.H.S=R.H.SHence ProvediiiL.H.S=(43m34n)2+2mn=(43m)2+(34n)22×43m×(34n)+2mn=169m2+916n22mn+2mn=169m2+916n2 R.H.S=169m2+916n2L.H.S=R.H.SHence Proved

iv L.H.S=(4pq+3q)2(4pq3q)2=48pq2=(4pq)2+(3q)2+2×4pq×(3q)[(4pq)2+(3q)22×m4pq×(3q)]=16p2q2+9q2+24pq2[16p2q2+9q224pq2]=16p2q2+9q2+24pq216p2q29q2+24pq2=48pq2 R.H.S=48pq2L.H.S=R.H.SHence Proved(v)L.H.S=(ab)(a+b)+(bc)(b+c)+(ca)(c+a)=a2b2+b2c2+c2a2=0 R.H.S=0L.H.S=R.H.SHence Proved




i712=(70+1)2=(70)2+2×70×1+12 {By using the identity(a + b)2=a2+b2+2ab}=4900+140+1=5041ii992=(1001)2=(100)22×100×1+12 {By using the identity(a b)2=a2+b22ab}=10000200+1=9801iii1022=(100+2)2=(100)2+2×100×2+22 {By using the identity(a + b)2=a2+b2+2ab}=10000+400+4=10404iv9982=(10002)2=(1000)22×1000×2+22 {By using the identity(a b)2=a2+b22ab}=10000004000+4=996004

(v)5.22=(5+0.2)2=(5)2+2×5×0.2+(0.2)2 {By using the identity(a + b)2=a2+b2+2ab}=25+2+0.04=27.04(vi)297×303=(3003)(300+3)=(300)232 {By using the identity(ab)(a+b)=a2b2}=900009=89991(vii)78×82=(802)(80+2)=(80)222 {By using the identity(ab)(a+b)=a2b2}=64004=6396viii8.92=(9.00.1)2=811.8+0.01 {By using the identity(ab)2=a2+b22ab}=79.21(ix)10.5×9.5=10+0.5100.5=[102(0.5)2] {By using the identity(ab)(a+b)=a2b2}=100 0.25 = 99.75


Using a2 b2 = (a + b) (a b), find(i) 512 492 (ii) (1.02)2 (0.98)2 (iii) 1532 1472(iv) 12.12 7.92


(i)512492=(51+49)(5149)=100×2=200(ii) (1.02)2 (0.98)2 =(1.02+0.98)(1.020.98)=2×0.04=0.08(iii)15321472=(153+147)(153147)=300×6=1800(iv)12.127.92=(12.1+7.9)(12.17.9)=20.0×4.2=84




(i)103×104=(100+3)(100+4)=(100)2+3+4×100+3×4=10000+700+12=10712(ii)5.1×5.2=(5+0.1)(5+0.2)=(5)2+0.1+ 0.2×5+0.1×0.2=25+1.5+0.02=26.52(iii)103×98=(100+3)(1002)=(100)2+[3+(2)]×100+3×(2)=10000+1006=10094(iv)9.7×9.8=(100.3)(100.2)=(10)2+[(0.3)+(0.2)]×10+(0.3)×(0.2)=100+(0.5)10+0.06=100.065=95.06

Q.24 Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 – 3zy
(ii) 1 + x + x2
(iii) 4x2y2 – 4x2y2z2 + z2
(iv) 3 – pq + qr – rp


(vi) 0.3a – 0.6ab + 0.5b

Ans Terms Coefficients
(i) 5xyz2 – 3zy 5xyz2 and –3zy 5 and –3
(ii ) 1 + x + x2 1, x and x2 1,1 and 1
(iii) 4x2y2 – 4x2y2z2 + z2 4x2y2 , – 4x2y2z2 and z2 4, –4 and 1
(iv) 3 – pq + qr – rp 3, – pq , qr, – rp 3 ,–1 ,1and –1

x/2 , y/2 , − xy 1/2 , 1/2 , –1
(vi) 0.3a – 0.6ab + 0.5b 0.3a , – 0.6ab , 0.5b 0.3 , – 0.6 , 0.5

Q.25 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 + pq2, 2p + 2q


An expression that contains only one term is called a monomial.
An expression that contains two terms is called a binomial.
An expression containing three terms is a trinomial.

Monomials Binomials Trinomials
1000 , pqr x + y , 2y – 3y2 , 4z – 15z2, p2q + pq2 , 2p + 2q 7 + y + 5x, 2y – 3y2 + 4y3 , 5x – 4y + 3xy

The polynomials x + x2 + x3 + x4 and ab + bc + cd + da do not fit in any of these three categories.

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