NCERT Solutions for Class 8 Maths Chapter 8 - Algebraic Expressions and Identities Exercise 8.3

Exercise 8.3

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

i. $4p$, $q + r$

Ans: The product of given terms is, 

$\left( {4p} \right) \times \left( {q + r} \right) = 4p \times \left( q \right) + 4p \times \left( r \right)$

$= 4pq + 4pr$

ii. $ab$,$a - b$

Ans: The product of given terms is, 

$\left( {ab} \right) \times \left( {a - b} \right) = \left( {ab} \right) \times a - \left( {ab} \right) \times b$

$= {a^2}b - a{b^2}$

iii. $a + b$,$7{a^2}{b^2}$

Ans: The product of given terms is, 

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

$= 7{a^3}{b^2} + 7{a^2}{b^3}$

iv. ${a^2} - 9$,$4a$

Ans:  The product of given terms is, 

$\left( {{a^2} - 9} \right) \times \left( {4a} \right) = \left( {{a^2} \times 4a} \right) + \left( { - 9 \times 4a} \right)$ 

$= 4{a^2} + \left( { - 36a} \right)$

$= 4{a^2} - 36a$

v. $pq + qr + rp$,$0$

Ans:  The product of given terms is, 

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

$= 0 + 0 + 0 = 0$

2. Complete the following table.

Ans: The Complete table is,

3. Find the product.

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

Ans:  The required product is,

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

$= 8{a^{50}}$

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

Ans:  The required product is,

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

$= \left( { - \dfrac{3}{5}} \right) \times x \times {x^2} \times y \times {y^2}$

$= \left( { - \dfrac{3}{5}} \right){x^3}{y^3}$

iii. \[\left( { - \dfrac{{10}}{3}p{q^3}} \right) \times \left( {\dfrac{6}{5}{p^3}q} \right)\]

Ans: The required product is,

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

$= - 2 \times 2 \times {p^4} \times {q^4}$

$= - 4{p^4}{q^4}$

iv. $x \times {x^2} \times {x^3} \times {x^4}$

Ans: The required product is,

$ x \times {x^2} \times {x^3} \times {x^4} = {x^{1 + 2 + 3 + 4}}$

$= {x^{10}}$

4. Do as follows.

a. Simplify $3x\left( {4x - 5} \right) + 3$ and find its values for (i)$x = 3$, (ii) $x = \dfrac{1}{2}$.

Ans: Solve the expression, $3x\left( {4x - 5} \right) + 3$. Multiply the terms in the parenthesis with $3x$.

$3x\left( {4x - 5} \right) + 3 = 3x\left( {4x} \right) - 3x\left( 5 \right) + 3$

$= 12{x^2} - 15x + 3$

i. $x = 3$

Ans:  Substitute $x$ as $3$ in $12{x^2} - 15x + 3$ and simplify.

$12{\left( 3 \right)^2} - 15\left( 3 \right) + 3 = 12\left( 9 \right) - 45 + 3$

$= 108 - 45 + 3$

$= 66$

ii. $x = \dfrac{1}{2}$

Ans: Substitute $x$ as $\dfrac{1}{2}$ in $12{x^2} - 15x + 3$ and simplify by taking LCM.

$12{\left( {\dfrac{1}{2}} \right)^2} - 15\left( {\dfrac{1}{2}} \right) + 3 = 12 \times \left( {\dfrac{1}{4}} \right) - \dfrac{{15}}{2} + 3$

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

$ = 6 - \dfrac{{15}}{2}$

$ = \dfrac{{12 - 15}}{2}$

$ =  - \dfrac{3}{2}$

b. Simplify $a\left( {{a^2} + a + 1} \right) + 5$and find its values for the given values of $a$.

Ans: Solve the expression, $a\left( {{a^2} + a + 1} \right) + 5$. Multiply the terms in the parenthesis with $a$ and simplify.

$ a\left( {{a^2} + a + 1} \right) + 5 = a \times \left( {{a^2}} \right) + a \times a + a \times 1 + 5$

$= {a^3} + {a^2} + a + 5$

i. $a = 0$

Ans: Substitute $a$ as $0$ in ${a^3} + {a^2} + a + 5$ and simplify.

${a^3} + {a^2} + a + 5 = {\left( 0 \right)^3} + {\left( 0 \right)^2} + \left( 0 \right) + 5$ 

$= 5$

ii. $a = 1$

Ans: Substitute $a$ as $1$ in ${a^3} + {a^2} + a + 5$ and simplify.

${a^3} + {a^2} + a + 5 = {\left( 1 \right)^3} + {\left( 1 \right)^2} + \left( 1 \right) + 5$

$= 1 + 1 + 1 + 5$

$= 8$

iii. $a =  - 1$

Ans: Substitute $a$ as $0$ in ${a^3} + {a^2} + a + 5$ and simplify.

${a^3} + {a^2} + a + 5 = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 5$

$=  - 1 + 1 - 1 + 5$

$= 4$

5. Do as follows.

i. Add: $p\left( {p - q} \right)$,$q\left( {q - r} \right)$, and $r\left( {r - p} \right)$.

Ans: Simplify the first expression, $p\left( {p - q} \right) = {p^2} - pq$.

Simplify the second expression, $q\left( {q - r} \right) = {q^2} - qr$.

Simplify the third expression, $r\left( {r - p} \right) = {r^2} - rp$.

Add the three expressions obtained.

$\begin{array}{*{20}{c}}  {}&{{p^2}}& - &{pq}&{}&{}&{}&{}&{}&{}&{}&{} \\   {}&{}&{}&{}& + &{{q^2}}& - &{qr}&{}&{}&{}&{} \\    + &{}&{}&{}&{}&{}&{}&{}& + &{{r^2}}& - &{pq} \\ \hline  {}&{{p^2}}& - &{pq}& + &{{q^2}}& - &{qr}& + &{{r^2}}& - &{pq} \end{array}$

Hence, the the sum of the given expression is ${p^2} - pq + {q^2} - qr + {r^2} - pq$.

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

Ans: Simplify the first expression,$2x\left( {z - x - y} \right)$.

$2x\left( {z - x - y} \right) = 2x \times z - 2x \times x - 2x \times y$

$= 2xz - 2{x^2} - 2xy$

Simplify the second expression, $2y\left( {z - y - x} \right)$.

$2y\left( {z - y - x} \right) = 2y \times z - 2y \times y - 2y \times x$

$= 2yz - 2{y^2} - 2yx$

Now, add the two expressions obtained.

$\,\,2xz - 2{x^2} - 2xy \\  \underline {\left(  +  \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2xy - 2{y^2} - 2yz} \\  2xz - 2{x^2} - 4xy + 2yz - 2{y^2}$

Hence, the sum of the given expression is $2xz - 2{x^2} - 4xy + 2yz - 2{y^2}$.

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

Ans: Simplify the first expression, $3l\left( {l - 4m + 5n} \right)$.

$3l\left( {l - 4m + 5n} \right) = 3l \times l - 3l \times 4m + 3l \times 5n$

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

Simplify the second expression, $4l\left( {10n - 3m + 2l} \right)$.

$ 4l\left( {10n - 3m + 2l} \right) = 4l \times 10n - 4l \times 3m + 4l \times 2l$

$= 40\ln  - 12lm + 8{l^2}$

Subtract the obtained expression.

$\,\,\,\,\,\,\,\,40\ln  - 12lm + 8{l^2} \\  \underline {\left(  -  \right)15\ln  - 12lm + 3{l^2}\,\,\,\,}  \\  \,\,\,\,\,\,\,\,25\ln  + 0 + \,\,\,\,\,\,5{l^2}$

The difference is $5{l^2} + 25\ln $.

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

Ans: Simplify the first expression, $3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right)$.

$3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right) = 3a \times \left( a \right) + 3a \times \left( b \right) + 3a \times \left( c \right) - 2b\left( a \right) + 2b\left( b \right) - 2b\left( c \right)$

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

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

Simplify the second expression, $4c\left( { - a + b + c} \right)$.

$4c\left( { - a + b + c} \right) = 4c \times \left( { - a} \right) + 4c \times \left( b \right) + 4c \times c$

$=  - 4ac + 4cb + 4{c^2}$

Subtract the obtained expression.

$\,\,\,\,\,\,\,\,-4ac + 4cb + 4{c^2} \\  \underline {\left(  -  \right)3ac - 2bc + 3{a^2} + 2{b^2} + ab \,\,\,\,}  \\  \,\,\,- 7ac + 6bc + 4{c^2} - 3{a^2} - 2{b^2} - ab$

The simplified expression is, $ - 7ac + 6bc + 4{c^2} - 3{a^2} - 2{b^2} - ab$.

Conclusion

Exercise 8.3 Class 8 Solutions focuses on multiplying monomials by polynomials and binomials. This exercise has been instrumental in helping students understand and apply the distributive property to simplify algebraic expressions. By tackling problems in class 8 8.3 exercise that involved the multiplication of monomials with binomials and polynomials, students enhanced their ability to manipulate and simplify complex expressions.

Class 8 Maths Chapter 8: Exercises Breakdown

CBSE Class 8 Maths Chapter 8 Other Study Materials

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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