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# NCERT Solutions Class 8 Maths chapter 8 Algebraic Expressions and Identities

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## NCERT Solutions for Maths Class 8 Chapter 8 - FREE PDF Download

NCERT Solutions for Class 8 Maths chapter 8 Algebraic Expressions and Identities by Vedantu introduces you to the powerful tool of identities and various types of expressions such as monomials, binomials, and polynomials. Algebraic expressions are combinations of variables, constants, and operators that represent a value.

Table of Content
1. NCERT Solutions for Maths Class 8 Chapter 8 - FREE PDF Download
2. Glance on Maths chapter 8 Class 8 - Algebraic Expressions and Identities
3. Access Exercise wise NCERT Solutions for chapter 8 Maths Class 8
4. Exercises Under NCERT Solutions for Class 8 Maths chapter 8 Algebraic Expressions and Identities
5. Access NCERT Solutions for Class 8 Maths chapter 8 –  Algebraic Expressions and Identities
5.1Exercise - 8.1
5.2Exercise - 8.2
5.3Exercise - 8.3
5.4Exercise - 8.4
6. NCERT Solutions for Class 8 Maths chapter 8 PDF
7. Class 8 Maths chapter 8: Exercises Breakdown
8. Other Study Material for CBSE Class 8 Maths chapter 8
9. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs

In this chapter, you will learn how to simplify, and manipulate algebraic expressions. Additionally, you will delve into important algebraic identities that simplify complex expressions and solve equations efficiently. Vedantu’s Class 8 Maths NCERT Solutions provide step-by-step explanations to help you understand these concepts thoroughly. The clear and concise explanations make it easier to grasp the material and apply it to solve problems.

## Glance on Maths chapter 8 Class 8 - Algebraic Expressions and Identities

• This chapter explains that expressions are combinations of variables, numbers, and mathematical operations (like addition, subtraction, multiplication, division).

• Expressions are of different types based on the number of terms they have:

• Monomials (one term) - $4x,xy^{^{2}}$

• Binomials (two terms) - $8x+9, y^{^{2}}-3$

• Trinomials (three terms) - 3x + 6y - 6

• Identities are equations that are always true, no matter what value you assign to the variable.

• This article algebraic expressions and identities class 8 contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter  - Algebraic Expressions and Identities, which you can download as PDFs.

• There are four exercises (15 fully solved questions) in class 8th Maths chapter 8 Algebraic Expressions and Identities.

## Exercises Under NCERT Solutions for Class 8 Maths chapter 8 Algebraic Expressions and Identities

• Exercise 8.1: This exercise consists of 2 Questions and Solutions. This exercise deals with addition and subtraction of Algebraic Expressions.

• Exercise 8.2: This exercise consists of 5 Questions and Solutions. This exercise deals with multiplication of Algebraic Expressions, multiplying a monomial by a monomial and multiplying three or more monomials.

• Exercise 8.3: This exercise consists of 5 Questions and Solutions. This exercise deals with multiplying a monomial by a polynomial, multiplying a monomial by a binomial, multiplying a monomial by a trinomial.

• Exercise 8.4: This exercise consists of 3 Questions and Solutions. This exercise deals with multiplying a polynomial by a polynomial, multiplying a binomial by a binomial, multiplying a binomial by a trinomial.

## Access NCERT Solutions for Class 8 Maths chapter 8 –  Algebraic Expressions and Identities

### Exercise - 8.1

(i) ${\text{ab - bc,bc - ca,ca - ab}}$

Ans:

${\text{ 12a - 9ab + 5b - 3}}$

Therefore, the sum of the given expressions is o.

(ii) ${\text{a - b + ab,b - c + bc,c - a + ac}}$

Ans:

${\text{ }}a - b + ab$

${\text{ }} + b{\text{ }} - c + bc$

${\text{ }} + \quad - a{\text{ }} + c{\text{ + ac}}$

$\overline {{\text{ ab + bc + ac}}}$

Thus the sum of given expressions is ${\text{ab + bc + ac}}$

(iii) ${\text{2}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ - 3pq + 4,5 + 7pq - 3}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}$

Ans:

${\text{ 2}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ - 3pq + 4}}$

${\text{ + - 3}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ + 7pq + 5}}$

$\overline {{\text{ - }}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ + 4pq + 9}}}$

Therefore, the sum of given expressions is ${\text{ - }}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ + 4pq + 9}}$

(iv) ${{\text{l}}^{\text{2}}}{\text{ + }}{{\text{m}}^{\text{2}}}{\text{,}}{{\text{m}}^{\text{2}}}{\text{ + }}{{\text{n}}^{\text{2}}}{\text{,}}{{\text{n}}^{\text{2}}}{\text{ + }}{{\text{l}}^{\text{2}}}{\text{,2lm + 2mn + 2nl}}$

Ans:

${\text{ }}{{\text{l}}^{\text{2}}}{\text{ + }}{{\text{m}}^{\text{2}}}$

${\text{ + }}{{\text{m}}^{\text{2}}}{\text{ + }}{{\text{n}}^{\text{2}}}$

${\text{ + }}{{\text{l}}^{\text{2}}}{\text{ + }}{{\text{n}}^{\text{2}}}$

${\text{ + 2lm + 2mn + 2nl}}$

$\overline {{\text{ 2}}{{\text{l}}^{^{\text{2}}}}{\text{ + 2}}{{\text{m}}^{\text{2}}}{\text{ + 2}}{{\text{n}}^{\text{2}}}{\text{ + 2lm + 2mn + 2nl}}}$

Therefore, the sum of the given expressions is ${\text{2}}{{\text{l}}^{^{\text{2}}}}{\text{ + 2}}{{\text{m}}^{\text{2}}}{\text{ + 2}}{{\text{n}}^{\text{2}}}{\text{ + 2lm + 2mn + 2nl}}$

2. Solve the following:

(i) Subtract ${\text{4a - 7ab + 3b + 12}}$ from ${\text{12a - 9ab + 5b - 3}}$

Ans:

${12a - 9ab + 5b - 3}$

${4a - 7ab + 3b + 12}$

${( - )\quad ( + )\quad ( - )( - )}$

${\overline {8a - 2ab + 2b - 15} }$

(ii) Subtract ${\text{3xy + 5yz - 7zx}}$ from ${\text{5xy - 2yz - 2zx + 10xyz}}$

Ans:

${\text{5xy - 2yz - 2zx + 10xyz}}$

${\text{3xy + 5yz - 7zx}}$

${\text{( - )( - )}}\quad {\text{( + )}}$

$\overline {{\text{2xy - 7yz + 5zx + 10xyz}}}$

(iii) Subtract ${\text{4p 2q - 3pq + 5pq2 - 8p + 7q - 10}}$from ${\text{18 - 3p - 11q + 5pq - 2pq2 + 5p 2q}}$

Ans:

${\text{18 - 3p - 11q + 5pq - 2p}}{{\text{q}}^{\text{2}}}{\text{ + 5}}{{\text{p}}^{\text{2}}}{\text{q}}$

${\text{ - 10 - 8p + 7q - 3pq + 5p}}{{\text{q}}^{\text{2}}}{\text{ + 4}}{{\text{p}}^{\text{2}}}{\text{q}}$

$\dfrac{{{\text{( + )( + )( - )( + )( - )}}\quad {\text{( - )}}}}{{{\text{28 + 5p - 18q + 8pq - 7p}}{{\text{q}}^{\text{2}}}{\text{ + }}{{\text{p}}^{\text{2}}}{\text{q}}}}$

### Exercise - 8.2

1. Find the product of the following pairs of monomials.

(i) ${\text{4,7p}}$

Ans:  ${{4 \times 7p = 4 \times 7 \times p = 28p}}$

(ii) $\dfrac{{{\text{first monomial}} \to }}{{{\text{second monomial}} \downarrow }}$

Ans:  ${{ - 4p \times 7p = - 4 \times p \times 7 \times p = }}\left( {{{ - 4 \times 7}}} \right){{ \times }}\left( {{{p \times p}}} \right){\text{ = - 28 }}{{\text{p}}^2}$

(iii) ${\text{ - 4p,7pq}}$

Ans:  ${{ - 4p \times 7pq = - 4 \times p \times 7 \times p \times q = }}\left( {{{ - 4 \times 7}}} \right){{ \times }}\left( {{{p \times p \times q}}} \right){\text{ = - 28}}{{\text{p}}^2}{\text{q }}$

(iv) ${\text{4}}{{\text{p}}^{\text{3}}}{\text{ , - 3p }}$

Ans:  ${\text{ 4}}{{\text{p}}^{\text{3}}}{{ \times - 3p = 4 \times }}\left( {{\text{ - 3}}} \right){{ \times p \times p \times p \times p = - 12 }}{{\text{p}}^{\text{4}}}$

(v) ${\text{4p, 0}}$

Ans:  ${{4p \times 0 = 4 \times p \times 0 = 0 }}$

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$\left( {{\text{p, q}}} \right){\text{; }}\left( {{\text{10m, 5n}}} \right){\text{; }}\left( {{\text{20}}{{\text{x}}^{\text{2}}}{\text{ , 5}}{{\text{y}}^{\text{2}}}{\text{ }}} \right){\text{; }}\left( {{\text{4x, 3}}{{\text{x}}^{\text{2}}}{\text{ }}} \right){\text{; }}\left( {{\text{3mn, 4np}}} \right){\text{ }}$

Ans:  We know that,

Area of rectangle = length x breadth

Area of 1st rectangle = p x q = pq

Area of 2nd rectangle = ${{10m \times 5n = 10 \times 5 \times m \times n = 50mn}}$

Area of 3rd rectangle = ${\text{20}}{{\text{x}}^{\text{2}}}{{ \times 5}}{{\text{y}}^{\text{2}}}{{ = 20 \times 5 \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{y}}^{\text{2}}}{\text{ = 100}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}$

Area of 4th rectangle = ${{4x }} \times {\text{ 3}}{{\text{x}}^{\text{2}}}{{ = 4 \times 3}} \times {{x}} \times {{\text{x}}^2}{\text{ = 12}}{{\text{x}}^3}$

Area of 5th rectangle ${{ = 3mn \times 4np = 3 \times 4 \times m \times n \times n \times p = 12m}}{{\text{n}}^{\text{2}}}{\text{p}}$

3. Complete the table of products.

 $\dfrac{{{\text{first monomial}} \to }}{{{\text{second monomial}} \downarrow }}$ 2x -5y 3x2 -4xy 7x2y -9x2y 2x 4x2 … .. … … … -5y … … 15x2 … … … 3x2 … … … … … … -4xy … … … … … … 7x2y … … … … … … -9x2y2 … … … … … …

Ans:

The table can be completed as follows.

 $\dfrac{{{\text{first monomial}} \to }}{{{\text{second monomial}} \downarrow }}$ 2x -5y 3x2 4xy 7x2y -9x2y 2x 4x2 -10xy 6x2 -8x2y 14x3y -18x3y2 -5y -10xy 25y2 -15x2 20xy2 -35x2y2 45x2y3 3x2 6x3 -15x2y 9x4 -12x3 21x4y -27x4y2 -4xy -8x2y 20xy2 -12x3y 16x2y2 -28x3y2 36x3y3 7x2y 14x3y -35x2y2 21x4y -28x3y2 49x4y2 -63x4y3 -9x2y2 -18x3y2 45x2y3 -27x4y2 36x3y3 -63x4y3 81x4y4

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) ${\text{5a,3}}{{\text{a}}^{\text{2}}}{\text{,7}}{{\text{a}}^{\text{4}}}$

Ans:  We know that

Volume= length x breadth x height

Volume =${{5a \times 3}}{{\text{a}}^{\text{2}}}{{ \times 7}}{{\text{a}}^{\text{4}}}{\text{ = 105}}{{\text{a}}^{\text{7}}}$

(ii) ${\text{2p,4q,8r}}$

Ans:  We know that

Volume = length x breadth x height

Volume = ${{2p \times 4q \times 8r = 64pqr}}$

(iii) ${\text{xy,2}}{{\text{x}}^{\text{2}}}{\text{y,2x}}{{\text{y}}^{\text{2}}}$

Ans:  We know that

Volume = length x breadth x height

Volume = ${{xy \times 2}}{{\text{x}}^{\text{2}}}{{y \times 2x}}{{\text{y}}^{\text{2}}}{\text{ = 4}}{{\text{x}}^{\text{4}}}{{\text{y}}^{\text{4}}}$

(iv) ${\text{a,2b,3c}}$

Ans:  We know that

Volume = length x breadth x height

Volume = ${{a}} \times {\text{2b}} \times {\text{3c = 6abc}}$

5. Obtain the product of

(i) ${\text{xy, yz, zx }}$

Ans: ${{xy \times yz \times zx = }}{{\text{x}}^{\text{2}}}{\text{ }}{{\text{y}}^{\text{2}}}{\text{ }}{{\text{z}}^{\text{2}}}$

(ii) ${\text{a, - }}{{\text{a}}^{\text{2}}}{\text{ , }}{{\text{a}}^{\text{3}}}{\text{ }}$

Ans:  ${{a}} \times ({\text{ - }}{{\text{a}}^{{2}}}) \times {\text{ }}{{\text{a}}^{\text{3}}}{\text{ = - }}{{\text{a}}^6}{\text{ }}$

(iii) ${\text{2, 4y, 8}}{{\text{y}}^2}{\text{ , 16}}{{\text{y}}^3}$

Ans:  ${{2}} \times {{ 4y}} \times {\text{8}}{{\text{y}}^2} \times {\text{ 16}}{{\text{y}}^3} = 1024{y^6}$

(iv) ${\text{a, 2b, 3c, 6abc}}$

Ans:  ${{a \times 2b \times 3c \times 6abc = }}$${\text{36}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}$

(v) ${\text{m, - mn, mnp}}$

Ans:  ${{m \times }}\left( {{\text{ - mn}}} \right){{ \times mnp = - }}{{\text{m}}^{\text{3}}}{{\text{n}}^{\text{2}}}$

### Exercise - 8.3

1. Carry out the multiplication of the expressions in each of the following pairs.

(i) ${\text{4p, q + r }}$

Ans:  $\left( {{\text{4p}}} \right){{ \times }}\left( {{\text{q + r}}} \right){\text{ = }}\left( {{{4p \times q}}} \right){\text{ + }}\left( {{{4p \times r}}} \right){\text{ = 4pq + 4pr}}$

(ii) ${\text{ab, a - b }}$

Ans:  $\left( {{\text{ab}}} \right){{ \times }}\left( {{\text{a - b}}} \right){\text{ = }}\left( {{{ab \times a}}} \right){\text{ + }}\left[ {{{ab \times }}\left( {{\text{ - b}}} \right)} \right]{\text{ = }}{{\text{a}}^{\text{2}}}{\text{b - a}}{{\text{b}}^{\text{2}}}$

(iii) ${\text{a + b, 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}$

Ans:  $\left( {{\text{a + b}}} \right){{ \times }}\left( {{\text{7a 2 b 2 }}} \right){\text{ = }}\left( {{{a \times 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ }}} \right){\text{ + }}\left( {{{b \times 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ }}} \right){\text{ = 7}}{{\text{a}}^{\text{3}}}{{\text{b}}^{\text{2}}}{\text{ + 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$

(iv) ${{\text{a}}^{\text{2}}}{\text{ - 9, 4a}}$

Ans:  $\left( {{{\text{a}}^2}{\text{ - 9}}} \right){{ \times }}\left( {{\text{4a}}} \right){\text{ = }}\left( {{{\text{a}}^{\text{2}}}{{ \times 4a}}} \right){\text{ + }}\left( {{\text{ - 9}}} \right){{ \times }}\left( {{\text{4a}}} \right){\text{ = 4}}{{\text{a}}^{\text{3}}}{\text{ - 36a}}$

(v) ${\text{pq + qr + rp, 0}}$

Ans:  $\left( {{\text{pq + qr + rp}}} \right){{ \times 0 = }}\left( {{{pq \times 0}}} \right){\text{ + }}\left( {{{qr \times 0}}} \right){\text{ + }}\left( {{{rp \times 0}}} \right){\text{ = 0 }}$

2. Complete the table

 -- First expression Second expression Product a b+c+d - x+y-5 5xy - p ${\text{6}}{{\text{p}}^{\text{2}}}{\text{ - 7p + 5 }}$ - ${\text{4}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}$ ${{\text{p}}^{\text{2}}}{\text{ - }}{{\text{q}}^{\text{2}}}$ - a+b+c abc -

Ans:  The table can be completed as follows:

 --- First expression Second expression Product a b+c+d ab+ac+ad x+y-5 5xy 5x2y+5xy2-25xy p ${\text{6}}{{\text{p}}^{\text{2}}}{\text{ - 7p + 5 }}$ 6p3-7p2+5p ${\text{4}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}$ ${{\text{p}}^{\text{2}}}{\text{ - }}{{\text{q}}^{\text{2}}}$ 4p4q2-4p2q4 a+b+c abc a2bc+ab2c+abc2

3. Find the product:

(i) $\left( {{{\text{a}}^{\text{2}}}} \right){{ \times }}\left( {{\text{2}}{{\text{a}}^{{\text{22}}}}} \right){{ \times }}\left( {{\text{4}}{{\text{a}}^{{\text{26}}}}} \right)$

Ans:  $\left( {{{\text{a}}^{\text{2}}}} \right){{ \times }}\left( {{\text{2}}{{\text{a}}^{{\text{22}}}}} \right){{ \times }}\left( {{\text{4}}{{\text{a}}^{{\text{26}}}}} \right){{ = 2 \times 4 \times }}{{\text{a}}^{\text{2}}}{{ \times }}{{\text{a}}^{{\text{22}}}}{{ \times }}{{\text{a}}^{{\text{26}}}}{\text{ = 8}}{{\text{a}}^{{\text{50}}}}$

(ii) $\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{xy}}} \right){{ \times }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}} \right)$

Ans:  $\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{xy}}} \right){{ \times }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}} \right){\text{ = }}\left( {\dfrac{{\text{2}}}{{\text{3}}}} \right){{ \times }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}} \right){{ \times x \times y \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{y}}^{\text{2}}}{\text{ = }}\dfrac{{{\text{ - 3}}}}{{\text{5}}}{{\text{x}}^{\text{3}}}{{\text{y}}^{\text{3}}}$

(iii) $\left( {{\text{ - }}\dfrac{{{\text{10}}}}{{\text{3}}}{\text{p}}{{\text{q}}^{\text{3}}}} \right){{ \times }}\left( {\dfrac{{\text{6}}}{{\text{5}}}{{\text{p}}^{\text{3}}}{\text{q}}} \right)$

Ans:  $\left( {{\text{ - }}\dfrac{{{\text{10}}}}{{\text{3}}}{\text{p}}{{\text{q}}^{\text{3}}}} \right){{ \times }}\left( {\dfrac{{\text{6}}}{{\text{5}}}{{\text{p}}^{\text{3}}}{\text{q}}} \right){\text{ = }}\left( {\dfrac{{{\text{ - 10}}}}{{\text{3}}}} \right){{ \times }}\left( {\dfrac{{\text{6}}}{{\text{5}}}} \right){{ \times p}}{{\text{q}}^{\text{3}}}{{ \times }}{{\text{p}}^{\text{3}}}{\text{q = - 4}}{{\text{p}}^{\text{4}}}{{\text{q}}^{\text{4}}}$

(iv) ${{x \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{x}}^{\text{3}}}{{ \times }}{{\text{x}}^{\text{4}}}$

Ans:  ${{x \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{x}}^{\text{3}}}{{ \times }}{{\text{x}}^{\text{4}}}{\text{ = }}{{\text{x}}^{10}}$

4. Solve the following

(a) Simplify ${\text{3x }}\left( {{\text{4x - 5}}} \right){\text{ + 3}}$and find its values for

(i) ${\text{ x = 3}}$

Ans:  ${\text{3x }}\left( {{\text{4x - 5}}} \right){\text{ + 3 = 12}}{{\text{x}}^{\text{2}}}{\text{ - 15x + 3 }}$

${\text{ For x = 3, 12}}{{\text{x}}^{\text{2}}}{\text{ - 15x + 3 = 12 }}{\left( {\text{3}} \right)^{\text{2}}}{\text{ - 15}}\left( {\text{3}} \right){\text{ + 3 }}$

${\text{ = 108 - 45 + 3 }}$

${\text{ = 66 }}$

(ii) ${\text{x = }}\dfrac{{\text{1}}}{{\text{2}}}$

Ans:

${\text{ For x = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{, 12}}{{\text{x}}^{\text{2}}}{\text{ - 15x + 3 = 12 }}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\text{2}}}{\text{ - 15}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + 3 }}$

${\text{ = 3 - }}\dfrac{{{\text{15}}}}{{\text{2}}}{\text{ + 3 }}$

${\text{ = 6 - }}\dfrac{{{\text{15}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{12 - 15}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{ - 3}}}}{{\text{2}}}$

(b) ${\text{a }}\left( {{{\text{a}}^{\text{2}}}{\text{ + a + 1}}} \right){\text{ + 5}}$ and find its value for

(i) ${\text{a = 0}}$

Ans:  ${\text{For a = 0, }}{{\text{a}}^{\text{3}}}{\text{ + }}{{\text{a}}^{\text{2}}}{\text{ + a + 5 = 0 + 0 + 0 + 5 = 5}}$

(ii) ${\text{a = 1}}$

Ans:  ${\text{For a = 1, }}{{\text{a}}^{\text{3}}}{\text{ + }}{{\text{a}}^{\text{2}}}{\text{ + a + 5 = }}{\left( {\text{1}} \right)^{\text{3}}}{\text{ + }}{\left( {\text{1}} \right)^{\text{2}}}{\text{ + 1 + 5}}$

${\text{ = 1 + 1 + 1 + 5 = 8 }}$

(iii) ${\text{a = - 1}}$

Ans:  ${\text{For a = - 1, }}{{\text{a}}^{\text{3}}}{\text{ + }}{{\text{a}}^{\text{2}}}{\text{ + a + 5 = }}{\left( {{\text{ - 1}}} \right)^{\text{3}}}{\text{ + }}{\left( {{\text{ - 1}}} \right)^{\text{2}}}{\text{ + }}\left( {{\text{ - 1}}} \right){\text{ + 5 }}$

${\text{ = - 1 + 1 - 1 + 5 = 4 }}$

5. Solve the following

(i) Add: ${\text{p (p - q), q (q - r)}}$ and ${\text{r (r - p)}}$

Ans:

${\text{First expression = p }}\left( {{\text{p - q}}} \right){\text{ = }}{{\text{p}}^2}{\text{ - pq }}$

${\text{Second expression = q }}\left( {{\text{q - r}}} \right){\text{ = }}{{\text{q}}^2}{\text{ - qr}}$

${\text{Third expression = r }}\left( {{\text{r - p}}} \right){\text{ = }}{{\text{r}}^2}{\text{ - pr}}$

Adding the three expressions, we obtain

${\text{ }}{{\text{p}}^{\text{2}}}{\text{ - pq }}$

${\text{ + }}{{\text{q}}^{\text{2}}}{\text{ - qr}}$

${\text{ + }}{{\text{r}}^{\text{2}}}{\text{ - pr}}$

$\overline {{\text{ }}{{\text{p}}^{\text{2}}}{\text{ - pq + }}{{\text{q}}^{\text{2}}}{\text{ - qr + }}{{\text{r}}^{\text{2}}}{\text{ - pr}}}$

Therefore, the sum is ${{\text{p}}^{\text{2}}}{\text{ - pq + }}{{\text{q}}^{\text{2}}}{\text{ - qr + }}{{\text{r}}^{\text{2}}}{\text{ - pr}}$

(ii) Add: ${\text{2x }}\left( {{\text{z - x - y}}} \right){\text{ and 2y }}\left( {{\text{z - y - x}}} \right){\text{ }}$

Ans:

${\text{First expression = 2x }}\left( {{\text{z - x - y}}} \right){\text{ = 2xz - 2}}{{\text{x}}^{\text{2}}}{\text{ - 2xy }}$

${\text{Second expression = 2y }}\left( {{\text{z - y - x}}} \right){\text{ = 2yz - 2}}{{\text{y}}^{\text{2}}}{\text{ - 2yx }}$

Adding the two expressions, we obtain

${\text{ 2xz - 2}}{{\text{x}}^{\text{2}}}{\text{ - 2xy }}$

${\text{ + - 2yx + 2yz - 2}}{{\text{y}}^{\text{2}}}{\text{ }}$

$\overline {\,\,{\text{ 2xz - 2}}{{\text{x}}^{\text{2}}}{\text{ - 4xy + 2yz - 2}}{{\text{y}}^{\text{2}}}}$

Therefore, the sum is ${\text{2xz - 2}}{{\text{x}}^{\text{2}}}{\text{ - 4xy + 2yz - 2}}{{\text{y}}^{\text{2}}}$

(iii) Subtract ${\text{3l }}\left( {{\text{l - 4m + 5n}}} \right){\text{ from 4l }}\left( {{\text{10n - 3m + 2l}}} \right){\text{ }}$

Ans:

${\text{3l }}\left( {{\text{l - 4m + 5n}}} \right){\text{ = 3}}{{\text{l}}^{\text{2}}}{\text{ - 12lm + 15ln }}$

${\text{4l }}\left( {{\text{10n - 3m + 2l}}} \right){\text{ = 40ln - 12lm + 8}}{{\text{l}}^{\text{2}}}{\text{ }}$

Subtracting these expressions, we obtain

${\text{ 8}}{{\text{l}}^{\text{2}}}{\text{ - 12lm + 40ln}}$

${\text{ 3}}{{\text{l}}^{\text{2}}}{\text{ - 12lm + 15ln}}$

$( - )\,{\text{ }}( + ){\text{ }}( - )$

$\overline {{\text{ 5}}{{\text{l}}^2}{\text{ + 25ln }}} {\text{ }}$

Therefore, the result is ${\text{5}}{{\text{l}}^2}{\text{ + 25ln}}$

(iv) Subtract ${\text{3a }}\left( {{\text{a + b + c}}} \right){\text{ - 2b }}\left( {{\text{a - b + c}}} \right){\text{ from 4c }}\left( {{\text{ - a + b + c}}} \right)$

Ans:

${\text{ + 4}}{{\text{c}}^{\text{2}}}{\text{ - 4ac + 4bc }}$

${\text{ 3}}{{\text{a}}^{\text{2}}}{\text{ + 2}}{{\text{b}}^{{\text{2 }}}}{\text{ + ab + 3ac - 2bc}}$

${\text{( - ) ( - ) ( - ) ( - ) ( + )}}$

$\overline {{\text{ - 3}}{{\text{a}}^{\text{2}}}{\text{ - 2}}{{\text{b}}^{\text{2}}}\,{\text{ + 4}}{{\text{c}}^{\text{2}}}{\text{ + ab - 7ac + 6bc}}}$

Therefore, the result is ${\text{ - 3}}{{\text{a}}^{\text{2}}}{\text{ - 2}}{{\text{b}}^{\text{2}}}\,{\text{ + 4}}{{\text{c}}^{\text{2}}}{\text{ + ab - 7ac + 6bc}}$

### Exercise - 8.4

1. Multiply the binomials.

(i) ${\text{(2x + 5)}}$and ${\text{(4x - 3)}}$

Ans: ${\text{(2x + 5) }} \times {\text{ (4x - 3) = 2x }} \times {\text{(4x - 3) + 5}} \times {\text{(4x - 3)}}$

${\text{ = 8}}{{\text{x}}^2}{\text{ - 6x + 20x - 15}}$

${\text{ = 8x2 + 14x - 15 (By adding like terms)}}$

(ii) ${\text{(y - 8)}}$and ${\text{(3y - 4)}}$

Ans: ${{ (y - 8) \times (3y - 4) = y \times (3y - 8) - 8 \times (3y - 4)}}$

${\text{ = 3}}{{\text{y}}^2}{\text{ - 4y - 24y + 32}}$

${\text{ = 3}}{{\text{y}}^{\text{2}}}{\text{ - 28y + 32 (By adding like terms)}}$

(iii) ${\text{(2}}{\text{.5l - 0}}{\text{.5m)}}$and ${\text{(2}}{\text{.5l + 0}}{\text{.5m)}}$

Ans: ${\text{(2}}{\text{.5l - 0}}{\text{.5m)(2}}{\text{.5l + 0}}{\text{.5m) = 2}}{{.5l \times (2}}{\text{.5l + 0}}{\text{.5m) - 0}}{\text{.5m(2}}{\text{.5l + 0}}{\text{.5m)}}$

${\text{ = 6}}{\text{.25}}{{\text{l}}^2}{\text{ + 1}}{\text{.25lm - 1}}{\text{.25lm - 0}}{\text{.25}}{{\text{m}}^2}$

${\text{ = 6}}{\text{.25}}{{\text{l}}^2}{\text{ - 0}}{\text{.25}}{{\text{m}}^2}$

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

Ans:  ${\text{(a + 3b) }} \times {\text{ (x + 5) = a}} \times {{(x + 5) + 3b }} \times {\text{(x + 5)}}$

${\text{ = ax + 5a + 3bx + 15b}}$

(v) ${\text{(2pq + 3}}{{\text{q}}^2}{\text{)}}$and ${\text{(3pq - 2}}{{\text{q}}^2}{\text{)}}$

Ans:  ${\text{(2pq + 3}}{{\text{q}}^2}{\text{)}} \times {\text{(3pq - 2}}{{\text{q}}^2}{\text{) = 2pq }} \times {\text{(3pq - 2}}{{\text{q}}^2}{\text{) + 3}}{{\text{q}}^2} \times {\text{(3pq - 2}}{{\text{q}}^2}{\text{)}}$

${\text{ = 6p2}}{{\text{q}}^2}{\text{ - 4p}}{{\text{q}}^3}{\text{ + 9p}}{{\text{q}}^3}{\text{ - 6}}{{\text{q}}^4}$

${\text{ = 6p2}}{{\text{q}}^2}{\text{ + 5p}}{{\text{q}}^3}{\text{ - 6}}{{\text{q}}^4}$

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

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

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

${\text{ = 3}}{{\text{a}}^{\text{4}}}{\text{ - 2}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ + 12}}{{\text{b}}^{\text{2}}}{{\text{a}}^{\text{2}}}{\text{ - 8}}{{\text{b}}^{\text{4}}}$

${\text{ = 3}}{{\text{a}}^{\text{4}}}{\text{ + 10}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ - 8}}{{\text{b}}^{\text{4}}}$

2. Find the product.

(i) ${\text{(5 - 2x) (3 + x)}}$

Ans:  ${\text{(5 - 2x) (3 + x) = 5 (3 + x) - 2x (3 + x)}}$

${\text{ = 15 + 5x - 6x - 2}}{{\text{x}}^2}$

${\text{ = 15 - x - 2}}{{\text{x}}^2}$

(ii) ${\text{(x + 7y) (7x - y)}}$

Ans:  ${\text{(x + 7y) (7x - y) = x (7x - y) + 7y (7x - y)}}$

${\text{ = 7}}{{\text{x}}^2}{\text{ - xy + 49xy - 7}}{{\text{y}}^2}$

${\text{ = 7}}{{\text{x}}^2}{\text{ + 48xy - 7}}{{\text{y}}^2}$

(iii) ${\text{(}}{{\text{a}}^2}{\text{ + b) (a + }}{{\text{b}}^2}{\text{)}}$

Ans:  ${\text{(}}{{\text{a}}^2}{\text{ + b) (a + }}{{\text{b}}^2}{\text{) = }}{{\text{a}}^2}{\text{ (a + }}{{\text{b}}^2}{\text{) + b (a + }}{{\text{b}}^2}{\text{)}}$

${\text{ = }}{{\text{a}}^3}{\text{ + }}{{\text{a}}^2}{{\text{b}}^2}{\text{ + ab + }}{{\text{b}}^3}$

(iv) ${\text{(}}{{\text{p}}^2}{\text{ - }}{{\text{q}}^2}{\text{) (2p + q)}}$

Ans:  ${\text{(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0}}$

${\text{ = 2}}{{\text{p}}^3}{\text{ + }}{{\text{p}}^2}{\text{q - 2p}}{{\text{q}}^2}{\text{ - }}{{\text{q}}^3}$

3. Simplify.

(i) ${\text{(}}{{\text{x}}^2}{\text{ - 5) (x + 5) + 25}}$

Ans:  ${\text{(}}{{\text{x}}^2}{\text{ - 5) (x + 5) + 25}}$

${{\text{x}}^2}{\text{ (x + 5) - 5 (x + 5) + 25}}$

${\text{ = }}{{\text{x}}^3}{\text{ + 5}}{{\text{x}}^2}{\text{ - 5x - 25 + 25}}$

${\text{ = }}{{\text{x}}^3}{\text{ + 5}}{{\text{x}}^2}{\text{ - 5x}}$

(ii) ${\text{(}}{{\text{a}}^2}{\text{ + 5) (}}{{\text{b}}^3}{\text{ + 3) + 5}}$

Ans:   ${\text{(}}{{\text{a}}^2}{\text{ + 5) (}}{{\text{b}}^3}{\text{ + 3) + 5}}$

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

${\text{ = }}{{\text{a}}^2}{{\text{b}}^3}{\text{ + 3}}{{\text{a}}^2}{\text{ + 5}}{{\text{b}}^3}{\text{ + 15 + 5}}$

${\text{ = }}{{\text{a}}^2}{{\text{b}}^3}{\text{ + 3}}{{\text{a}}^2}{\text{ + 5}}{{\text{b}}^3}{\text{ + 20}}$

(iii) ${\text{(t + }}{{\text{s}}^2}{\text{) (}}{{\text{t}}^2}{\text{ - s)}}$

Ans:  ${\text{(t + }}{{\text{s}}^2}{\text{) (}}{{\text{t}}^2}{\text{ - s)}}$

${\text{ = t (}}{{\text{t}}^2}{\text{ - s) + }}{{\text{s}}^2}{\text{ (}}{{\text{t}}^2}{\text{ - s)}}$

${\text{ = }}{{\text{t}}^3}{\text{ - st + }}{{\text{s}}^2}{{\text{t}}^2}{\text{ - }}{{\text{s}}^3}$

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

Ans:  ${\text{(a + b) (c - d) + (a - b) (c + d) + 2 (ac + bd)}}$

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

${\text{ = ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd}}$

${\text{ = (ac + ac + 2ac) + (ad - ad) + (bc - bc) + (2bd - bd - bd)}}$

${\text{ = 4ac}}$

(v) ${\text{(x + y) (2x + y) + (x + 2y) (x - y)}}$

Ans:  ${\text{(x + y) (2x + y) + (x + 2y) (x - y)}}$

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

${\text{ = 2}}{{\text{x}}^2}{\text{ + xy + 2xy + }}{{\text{y}}^2}{\text{ + }}{{\text{x}}^2}{\text{ - xy + 2xy - 2}}{{\text{y}}^2}$

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

${\text{ = 3}}{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}{\text{ + 4xy}}$

(vi) ${\text{(x + y) (}}{{\text{x}}^2}{\text{ - xy + }}{{\text{y}}^2}{\text{)}}$

Ans:  ${\text{(x + y) (}}{{\text{x}}^2}{\text{ - xy + }}{{\text{y}}^2}{\text{)}}$

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

${\text{ = }}{{\text{x}}^3}{\text{ - }}{{\text{x}}^2}{\text{y + x}}{{\text{y}}^2}{\text{ + }}{{\text{x}}^2}{\text{y - x}}{{\text{y}}^2}{\text{ + }}{{\text{y}}^3}$

${\text{ = }}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + (x}}{{\text{y}}^2}{\text{ - x}}{{\text{y}}^2}{\text{) + (}}{{\text{x}}^2}{\text{y - }}{{\text{x}}^2}{\text{y)}}$

${\text{ = }}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}$

(vii) ${\text{(1}}{\text{.5x - 4y) (1}}{\text{.5x + 4y + 3) - 4}}{\text{.5x + 12y}}$

Ans:  ${\text{(1}}{\text{.5x - 4y) (1}}{\text{.5x + 4y + 3) - 4}}{\text{.5x + 12y}}$

${\text{ = 1}}{\text{.5x (1}}{\text{.5x + 4y + 3) - 4y (1}}{\text{.5x + 4y + 3) - 4}}{\text{.5x + 12y}}$

${\text{ = 2}}{\text{.25 }}{{\text{x}}^2}{\text{ + 6xy + 4}}{\text{.5x - 6xy - 16}}{{\text{y}}^2}{\text{ - 12y - 4}}{\text{.5x + 12y}}$

${\text{ = 2}}{\text{.25 }}{{\text{x}}^2}{\text{ + (6xy - 6xy) + (4}}{\text{.5x - 4}}{\text{.5x) - 16}}{{\text{y}}^2}{\text{ + (12y - 12y)}}$

${\text{ = 2}}{\text{.25}}{{\text{x}}^2}{\text{ - 16}}{{\text{y}}^2}$

(viii) ${\text{(a + b + c) (a + b - c)}}$

Ans:  ${\text{(a + b + c) (a + b - c)}}$

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

${\text{ = }}{{\text{a}}^2}{\text{ + ab - ac + ab + }}{{\text{b}}^2}{\text{ - bc + ca + bc - }}{{\text{c}}^2}$

${\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - }}{{\text{c}}^2}{\text{ + (ab + ab) + (bc - bc) + (ca - ca)}}$

${\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - }}{{\text{c}}^2}{\text{ + 2ab}}$

## NCERT Solutions for Class 8 Maths chapter 8 PDF

### Overview of Deleted Syllabus for CBSE Class 8 Maths Algebraic Expressions and Identities

 Chapter Dropped Topics Algebraic Expressions and Identities 8.1 Introduction 8.2 Terms, Factors and Coefficients 8.3 Monomials, Binomials and Polynomials 8.4 Like and Unlike Terms 8.10 What is an Identity? 8.11 Standard Identities 8.12 Applying Identities

## Class 8 Maths chapter 8: Exercises Breakdown

 Exercise Number of Questions Exercise 8.1 2 Questions and Solutions Exercise 8.2 5 Questions and Solutions Exercise 8.3 5 Questions and Solutions Exercise 8.4 3 Questions with Solutions

## Conclusion

NCERT Solutions for Maths Algebraic Expressions Class 8 Chapter 8  by Vedantu are essential for building a strong foundation in algebra. This chapter introduces you to the basics of forming and simplifying algebraic expressions, and understanding and applying various algebraic identities.

In previous year exams, around 3–4 questions have been asked from this chapter, highlighting its significance in the overall curriculum. By thoroughly practising the problems and understanding the step-by-step solutions provided by Vedantu, you can confidently tackle algebraic expressions and identities.

## Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 8 Maths chapter 8 Algebraic Expressions and Identities

Q1.What are the Topics Covered in the Class 8 Chapter 9 Maths?

Ans: Chapter 9 of Grade 8 Maths is Algebraic Expressions and Identities.

The topics covered in this chapter in the NCERT book are as follows:

• Meaning of algebraic expressions

• What are the terms, factors, and coefficients

• What are monomials, binomials, and polynomials

• Like and Unlike terms

• Addition and subtraction of algebraic terms

• Multiplication of algebraic terms

• Identities, standard identities, and application of identities.

All these topics are the core topics covered. Subtopics like multiplication of monomial to monomial, polynomial to polynomial, monomial to polynomial are also covered in the chapter.

Q2. Are these NCERT Solutions Helpful in Scoring Good Marks in School Exams?

Ans: Our 8th standard Maths algebra NCERT solutions are prepared by experts who are highly qualified and experienced. These solutions are prepared in such a manner that the methods and mathematical steps can be easily understood by the students. When students can gain the concept they will be able to solve the questions related to algebra that may come in their school exams and this will increase their score in their exams. All the Exercises given in the NCERT book in Chapter 9 are included in our algebraic expressions NCERT Class 8 solutions.

Q3. What are algebraic expressions in Class 8?

Ans: In Class 8, algebraic expressions are equations or expressions formed by the combination of variables, numbers and algebraic operations (addition, subtraction, multiplication etc.). An Algebraic expression is a term consisting of variables, coefficients and a constant. For example: 4x + 7y - 2. Here, 4 and 7 are coefficients, x and y are variables, and ‘2’ is a constant. Algebraic expressions are of three types - monomial, binomial, and trinomial.

Q4. What is the formula for algebraic expression?

Ans: Class 8 Maths introduces the concept of algebraic expressions and algebraic identities to the students. Algebraic identities are algebraic equations that are valid for all variable values in them.  Following are the formulas for algebraic expressions:

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

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

(a+b)(a-b)= a2-b2

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

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

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

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

(a+b)3= a3+3ab(a+b) +b3

(a-b)3= a3-3ab(a-b) -b3