NCERT Solutions for Class 8 Maths Chapter 8 - Algebraic Expressions and Identities PDF (2024)

• 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

1. Add the following:

(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.

Ans:

The table can be completed as follows.

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

Ans: The table can be completed as follows:

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

Class 8 Maths chapter 8: Exercises Breakdown

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.

Other Study Material for CBSE Class 8 Maths chapter 8

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.

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