NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3

Introduction:

In this exercise/article we will learn about multiplication of algebraic expressions. multiplying a monomials by a polynomial and multiplying a monomials by a binomial and multiplying a monomials by a trinomial and multiply or add and subtract then do this exercise.

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities :

Class 8 Maths Exercise 9.3 (Page-146)

Q1. Carry out the multiplication of the expressions in each of the following pairs:

(i) 4p , q + r

Solution :

we have 4p , q + r

we know very well

So, 4p(q + r )

= 4pq + 4pr

(ii) ab , a - b

Solution :

We have ab , a - b

we know very well

So, ab( a - b )

= a²b - ab²

(iii) a + b , 7a²b²

Solution :

We have a + b , 7a²b²

we know very well

So, ( a + b ) ( 7a²b² )

= 7a³b² + 7a²b³

(iv) a² - 9 , 4a

Solution :

We have a² - 9 , 4a

we know very well

So, 4a( a² - 9 )

= 4a³ - 36a

(v) pq + qr + rp , 0

Solution :

We have pq + qr + rp , 0

we know very well

= 0( pq + qr + rp )

= o + o + o

= o

Q2. Complete the table

Solution :

Multiply the first expression by the second expression

= a ( b + c + d ) = ( a × b ) + ( a × c ) + ( a × d ) = ab + ac + ad

= 5xy ( x + y - 5 ) = ( 5xy × x ) + ( 5xy × y ) - ( 5xy × 5 ) = 5x²y + 5xy² - 25xy

= p ( 6p² - 7p + 5 ) = ( p × 6p² ) - ( p × 7p ) + ( p × 5 ) = 6p³ - 7p² + 5p

= 4p²q² ( p² - q² ) = ( 4p²q² × p² ) - ( 4p²q² × q² ) = 4p⁴q² - 4p²q⁴

= abc ( a + b + c ) = ( abc × a ) + ( abc × b ) + ( abc × c ) = a²bc + ab²c + abc²

Class 8 Maths Chapter 9 Exercise 9.3

Q3. find the product

(i) ( a² ) × ( 2a²² ) × ( 4a²⁶ )

Solution :

We have ( a² ) × ( 2a²² ) × ( 4a²⁶ )

= 2 × 4 × a^{( 2 + 22 + 26 )}

= 8a^{( 50 )}

= 8a⁵⁰

(ii) \(\displaystyle \left[ {\frac{2}{3}xy} \right]\times \left[ {\frac{{-9}}{{10}}{{x}^{2}}{{y}^{2}}} \right]\)

Solution :

We have \(\displaystyle \left[ {\frac{2}{3}xy} \right]\times \left[ {\frac{{-9}}{{10}}{{x}^{2}}{{y}^{2}}} \right]\)

= \(\displaystyle \,\frac{{-3}}{5}\times {{x}^{3}}\times {{y}^{3}}\)

= \(\displaystyle \frac{{-3}}{5}{{x}^{3}}{{y}^{3}}\)

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

Solution :

We have \(\displaystyle \left[ {-\frac{{10}}{3}p{{q}^{3}}} \right]\times \left[ {\frac{6}{5}{{p}^{3}}q} \right]\)

= \(\displaystyle -\,\frac{4}{1}\,\times \,{{p}^{4}}\,\times \,{{q}^{4}}\)

= \(\displaystyle -\,4{{p}^{4}}{{q}^{4}}\)

(iv) \(\displaystyle x\,\times \,{{x}^{2}}\,\times \,{{x}^{3}}\,\times \,{{x}^{4}}\)

Solution :

we have \(\displaystyle x\,\times \,{{x}^{2}}\,\times \,{{x}^{3}}\,\times \,{{x}^{4}}\)

= \(\displaystyle x{{\,}^{{(\,1\,+\,2\,+\,3\,+\,4\,)}}}\)

= \(\displaystyle x{{\,}^{{(\,10\,)}}}\)

= \(\displaystyle x{{\,}^{{10}}}\)

Q4. (a) Simplify 3x ( 4x - 5 ) + 3 and find its values for (i) x = 3 , (ii) x = \(\displaystyle \frac{1}{2}\)

Solution :

(i) We have 3x ( 4x - 5 ) + 3

= 12x² - 15x + 3

= Now , put x = 3

= 12( 3 )² - 15 × 3 + 3

= 12( 9 ) - 45 + 3

= 108 - 45 + 3

= 111 - 45

= 66

(ii) We have 3x ( 4x - 5 ) + 3

= 12x² - 15x + 3

= Now, put x = \(\displaystyle \frac{1}{2}\)

= \(\displaystyle 12{{\left[ {\frac{1}{2}} \right]}^{2}}-\,15\,\times \,\frac{1}{2}\,+\,3\)

= \(\displaystyle 12\,\times \,\frac{1}{4}\,-\,\frac{{15}}{2}\,+\,3\)

= \(\displaystyle 3\,-\,\frac{{15}}{2}\,+\,3\)

= \(\displaystyle 6\,-\,\frac{{15}}{2}\)

= \(\displaystyle \frac{{12\,-\,15}}{2}\)

= \(\displaystyle \frac{{-3}}{2}\)

(b) Simplify a ( a² + a + 1 ) + 5 and find its values for (i) a = 0 , (ii) a = 1

Solution :

We have a ( a² + a + 1 ) + 5

= a³ + a² + 1a + 5

= Now, put x = 0

= 0³ + 0² + 1 × 0 + 5

= 0 + 0 + 0 + 5

= 5

(ii) We have a ( a² + a + 1 ) + 5

= a³ + a² + 1a + 5

= Now, put x = 1

= 1³ + 1² + 1 × 1 + 5

= 1 + 1 + 1 + 5

= 8

Q5. (a) Add : p ( p - q ) , q ( q - r ) and r ( r - p )

Solution :

We have p ( p - q ) , q ( q - r ) and r ( r - p )

So, add all these

p ( p - q ) + q ( q - r ) + r ( r - p )

= p² - pq + q² - qr + r² - rp

= p²+ q² + r²- pq - qr - rp

(b) Add : 2x ( z - x - y ) and 2y ( z - y - x )

Solution :

We have 2x ( z - x - y ) and 2y ( z - y - x )

So, add all these

2x ( z - x - y ) + 2y ( z - y - x )

= 2zx - 2x² - 2xy + 2yz - 2y² - 2xy

= 2zx - 2x² - 4xy + 2yz - 2y²

= - 2x² - 2y² - 4xy + 2yz + 2zx

Solution :

We have 3l ( l - 4m + 5n ) from 4l ( 10n - 3m + 2l )

So, subtract all these

4l ( 10n - 3m + 2l ) - 3l ( l - 4m + 5n )

= 40ln - 12ml + 8l² - 3l² + 12ml - 15nl

= 40ln - 15nl - 12ml + 12ml + 8l² - 3l²

= 25ln + 5l²

= 5l² + 25ln

(d) Subtract : 3a ( a + b + c ) - 2b ( a - b + c ) from 4c ( -a + b + c )

Solution :

We have 3a ( a + b + c ) - 2b ( a - b + c ) from 4c ( -a + b + c )

So, subtract all these

4c ( -a + b + c ) - { 3a ( a + b + c ) - 2b ( a - b + c ) }

= 4c ( -a + b + c ) - 3a ( a + b + c ) + 2b ( a - b + c )

= - 4ac + 4bc + 4c² - 3a² - 3ab - 3ac + 2ab - 2b² + 2bc

= ( -4ac - 3ac ) ( 4bc + 2bc ) ( - 3ab + 2ab ) ( 4c² - 3a² - 2b² )

= ( -7ac ) ( 2bc ) ( -ab ) ( 4c² - 3a² - 2b² )

= - 7ac + 6bc - ab + 4c² - 3a² - 2b²

= - 3a² - 2b² + 4c² - ab + 6bc - 7ac .

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities :

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3

Introduction:

Class 8 Maths Exercise 9.3 (Page-146)

Class 8 Maths Chapter 9 Exercise 9.3

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