# NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4

## Introduction:

In this exercise/article we will learn about multiplication of algebraic expressions. multiplying a polynomials by a polynomial and multiplying a binomials by a binomials and multiplying a binomials by a trinomials. Only on multiply do in this exercise. Most important thing bracket multiply bracket.

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

- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.1
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.2
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.3
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.4
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.5

**Class 8 Maths Exercise 9.4 (Page-148)**

**Q1. **Multiply the binomials.

**(i)** ( 2x + 5 ) and ( 4x - 3 )

**Solution : **

We have to multiply the first bracket by the second bracket ( 2x + 5 ) × ( 4x - 3 )

= 2x ( 4x - 3 ) + 5 ( 4x - 3 )

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

= 8x^{2} + 14x - 15

So, answer is 8x^{2} + 14x - 15

**(ii)** ( y - 8 ) and ( 3y - 4 )

**Solution :**

We have to multiply the first bracket by the second bracket ( y - 8 ) × ( 3y - 4 )

= y ( 3y - 4 ) - 8 ( 3y - 4 )

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

= 3y^{2} - 28y + 32

So, answer is 3y^{2} - 28y + 32

**(iii)** ( 2.5l - 0.5m ) and ( 2.5l + 0.5m )

**Solution :**

We have to multiply the first bracket by the second bracket ( 2.5l - 0.5 m ) × ( 2.5l + 0.5m )

= 2.5l ( 2.5l + 0.5m ) - 0.5m ( 2.5l + 0.5m )

= 6.25l^{2} + 1.25lm - 1.25lm - 0.25m^{2}

= 6.25l^{2} - 0.25m^{2}

So, answer is 6.25l^{2} + 1.25lm - 1.25lm - 0.25m^{2}

**(iv)** ( a + 3b ) and ( x + 5 )

**Solution :**

We have to multiply the first bracket by the second bracket ( a + 3b ) × ( x + 5 )

= a ( x + 5 ) + 3b ( x + 5 )

= ax + 5a + 3bx + 15b

So, answer is ax + 5a + 3bx + 15b

**(v)** ( 2pq + 3q^{2} ) and ( 3pq - 2q^{2} )

**Solution :**

We have to multiply the first bracket by the second bracket ( 2pq + 3q^{2} ) × ( 3pq - 2q^{2} )

= 2pq ( 3pq - 2q^{2} ) + 3q^{2} ( 3pq - 2q^{2} )

= 6p^{2}q^{2} - 4pq^{3} + 9pq^{3} - 6q^{4}

= 6p^{2}q^{2} + 5pq^{3} - 6q^{4}

So, answer is 6p^{2}q^{2} + 5pq^{3} - 6q^{4}

**(vi)** \(\displaystyle \left( {\frac{3}{4}{{a}^{2}}\,+\,3{{b}^{2}}} \right)\,and\,4\,\left( {{{a}^{2}}\,-\,\frac{2}{3}{{b}^{2}}} \right)\)

**Solution :**

We have to multiply the first bracket by the second bracket\(\displaystyle \left( {\frac{3}{4}{{a}^{2}}\,+\,3{{b}^{2}}} \right)\,\times \,4\,\left( {{{a}^{2}}\,-\,\frac{2}{3}{{b}^{2}}} \right)\)

= \(\displaystyle \left( {\frac{3}{4}{{a}^{2}}\,+\,3{{b}^{2}}} \right)\,\times \,\left( {4{{a}^{2}}\,-\,\frac{8}{3}{{b}^{2}}} \right)\)

= \(\displaystyle \frac{3}{4}{{a}^{2}}\left( {4{{a}^{2}}-\frac{8}{3}{{b}^{2}}} \right)+3{{b}^{2}}\left( {4{{a}^{2}}-\frac{8}{3}{{b}^{2}}} \right)\)

\(\displaystyle =\,\left( {\frac{3}{4}{{a}^{2}}\times 4{{a}^{2}}-\frac{3}{4}{{a}^{2}}\times \frac{8}{3}{{b}^{2}}} \right)\,\,+\,\left( {3{{b}^{2}}\times 4{{a}^{2}}-3{{b}^{2}}\times \frac{8}{3}{{b}^{2}}} \right)\)

=

= \(\displaystyle 3{{a}^{4}}\,-\,2{{a}^{2}}{{b}^{2}}\,+\,12{{a}^{2}}{{b}^{2}}\,-\,8{{b}^{4}}\)

= \(\displaystyle 3{{a}^{4}}\,+\,10{{a}^{2}}{{b}^{2}}\,-\,8{{b}^{4}}\)

So, answer is \(\displaystyle 3{{a}^{4}}\,+\,10{{a}^{2}}{{b}^{2}}\,-\,8{{b}^{4}}\)

**Q2. **Find the product.

**(i)** ( 5 - 2x ) ( 3 + x )

**Solution :**

We have to multiply the first bracket by the second bracket ( 5 - 2x ) × ( 3 + x )

= 5 ( 3 + x ) - 2x ( 3 + x )

= 15 + 5x - 6x - 2x^{2}

= 15 - x - 2x^{2}

So, answer is 15 - x - 2x^{2}

**(ii)** ( x + 7y ) ( 7x - y )

**Solution :**

We have to multiply the first bracket by the second bracket ( x + 7y ) × ( 7x - y )

= x ( 7x - y ) + 7y ( 7x - y )

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

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

So, answer is 7x^{2} + 48xy - 7y^{2}

**(iii)** ( a^{2} + b ) ( a + b^{2} )

**Solution :**

We have to multiply the first bracket by the second bracket ( a^{2} + b ) × ( a + b^{2} )

= a^{2} ( a + b^{2} ) + b ( a + b^{2} )

= a^{3} + a^{2} b^{2} + ab + b^{3}

So, answer is a^{3} + a^{2} b^{2} + ab + b^{3}

**(iv)** ( p^{2} - q^{2} ) ( 2p + q )

**Solution :**

We have to multiply the first bracket by the second bracket ( p^{2} - q^{2} ) × ( 2p + q )

= p^{2} ( 2p + q ) - q^{2} ( 2p + q )

= 2p^{3} + p^{2}q - 2pq^{2} - q^{3}

So, answer is 2p^{3} + p^{2}q - 2pq^{2} - q^{3}

## Class 8 Maths Chapter 9 Exercise 9.4

**Q3. **Simplify.

**(i) **( x^{2} - 5 ) ( x + 5 ) + 25

**Solution :**

We have to multiply the first bracket by the second bracket ( x^{2} - 5 ) ( x + 5 ) + 25

= x^{2} ( x + 5 ) - 5 ( x + 5 ) + 25

= x^{3 } + 5x^{2} - 5x - 25 + 25 [ because, 25 - 25 = 0 ]

= x^{3 } + 5x^{2} - 5x

So, answer is x^{3 } + 5x^{2} - 5x

**(ii) **( a^{2} + 5 ) ( b^{3} + 3 ) + 5

**Solution :**

We have to multiply the first bracket by the second bracket ( a^{2} + 5 ) ( b^{3} + 3 ) + 5

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

= a^{2}b^{3} + 3a^{2} + 5b^{3} + 15 + 5 [ because, 15 + 5 = 20 ]

= a^{2}b^{3} + 3a^{2} + 5b^{3} + 20

= a^{2}b^{3} + 3a^{2} + 5b^{3} + 20

So, answer is a^{2}b^{3} + 3a^{2} + 5b^{3} + 20

**(iii) **( t + s^{2} ) ( t^{2} - s )

**Solution :**

We have to multiply the first bracket by the second bracket ( t + s^{2} ) ( t^{2} - s )

= t ( t^{2} - s ) + s^{2} ( t^{2} - s )

= t^{3} - st + s^{2}t^{2} - s^{3}

So, answer is t^{3} - st + s^{2}t^{2} - s^{3}

**(iv) **( a + b ) ( c - d ) + ( a - b ) ( c + d ) + 2( ac + bd )

**Solution :**

We have to multiply the first bracket by the second bracket ( a + b ) ( c - d ) + ( a - b ) ( c + d ) + 2( ac + bd )

= a ( c - d ) + b ( c - d ) + a ( c + d ) - b ( c + d ) + 2ac + 2bd

= ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd

= ( ac + ac ) ( - ad + ad ) ( bc - bc ) ( - bd - bd ) + 2ac + 2bd [ because - ad + ad = 0 , bc - bc = 0 ]

= ( 2ac + 2ac ) ( - 2bd + 2bd ) [ because - 2bd + 2bd = 0 ]

= 4ac

So, answer is 4ac

**(v) **( x + y ) ( 2x + y ) + ( x + 2y ) ( x - y )

**Solution : **

We have to multiply the first bracket by the second bracket ( x + y ) ( 2x + y ) + ( x + 2y ) ( x - y )

= x ( 2x + y ) + y ( 2x + y ) + x ( x - y ) + 2y ( x - y )

= 2x^{2} + xy + 2xy + y^{2} + x^{2} - xy + 2xy - 2y^{2}

= ( 2x^{2} + x^{2} ) ( 2xy + 2xy ) ( - 2y^{2} + y^{2} ) ( xy - xy ) [ because, xy - xy = 0 ]

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

So, answer is 3x^{2} + 4xy - y^{2}

**(vi)** ( x + y ) ( x^{2} - xy + y^{2} )

**solution :**

We have to multiply the first bracket by the second bracket ( x + y ) ( x^{2} - xy + y^{2} )

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

= x^{3} - x^{2}y + xy^{2} + x^{2}y - xy^{2} + y^{3}

= x^{3} + y^{3} ( - x^{2}y + x^{2}y ) ( xy^{2} - x^{2}y ) [ because, - x^{2}y + x^{2}y = 0 , xy^{2} - x^{2}y = 0 ]

= x^{3} + y^{3}

= So, answer is x^{3} + y^{3}

**(vii) **( 1.5x - 4y ) ( 1.5x + 4y + 3 ) - 4.5x + 12y

**Solution :**

We have to multiply the first bracket by the second bracket ( 1.5x - 4y ) ( 1.5x + 4y + 3 ) - 4.5x + 12y

= 1.5x ( 1.5x + 4y + 3 ) - 4y ( 1.5x + 4y + 3 ) - 4.5x + 12y

= 2.25x^{2} + 6xy + 4.5x - 6xy - 16y^{2} - 12y - 4.5x + 12y

= 2.25x^{2} - 16y^{2} ( 6xy - 6xy ) ( 4.5x - 4.5x ) ( - 12y + 12y ) [ because, 6xy - 6xy = 0 , 4.5x - 4.5x = 0 , - 12y + 12y = 0 ]

= 2.25x^{2} - 16y^{2}

= So, answer is 2.25x^{2} - 16y^{2}

**(viii)** ( a + b + c ) ( a + b - c )

**Solution :**

We have to multiply the first bracket by the second bracket ( a + b + c ) ( a + b - c )

= a ( a + b - c ) + b ( a + b - c ) + c ( a + b - c )

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

= a^{2} + b^{2} - c^{2} + ( ab + ab ) ( - ac + ac ) ( - bc + bc ) [ because, - ac + ac = 0 , - bc + bc = 0 ]

= a^{2} + b^{2} - c^{2} + 2ab

So, answer is a^{2} + b^{2} - c^{2} + 2ab .

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

- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.1
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.2
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.3
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.4
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.5