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

• Last Updated : 28 Dec, 2020

### Problem 1. Multiply the binomials.

Solution:

When we multiply two binomials, four multiplications must take place. These multiplications can be in any order, although we need to take care of that each of the first two terms is multiplied by each of the second terms.

For example: (2x + 3)(3x – 1), if we have to multiply these two binomial.

Step 1: Multiply the first term of each binomial together.
(2x)(3x) = 6x2

Step 2: Multiply the outer terms together.
(2x)(–1) = –2x

Step 3: Multiply the inner terms together.
(3)(3x) = 9x

Step 4: Multiply the last term of each expression together.
(3)(–1) = –3

### (i) (2x + 5) × (4x – 3)

Solution:

= 2x × (4x – 3) + 5 × (4x – 3) # Here, we used distributive property of multiplication.
= (2x × 4x) – (3 × 2x) + (5 × 4x) – (5 × 3) # Expanding the terms.
= 8x2 – 6x + 20x – 15 # Adding or subtracting the like terms.
= 8x2 + 14x – 15

### (ii) (y – 8) × (3y – 4)

Solution:

= y × (3y – 4) – 8 × (3y – 4) # Here, we used distributive property of multiplication.
= (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4) # Expanding the terms.
= 3y2 – 4y – 24y + 32 # Adding or subtracting the like terms.
= 3y2 – 28y + 32

### (iii) (2.5l – 0.5m) × (2.5l + 0.5m)

Solution:

= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
# Here, we used distributive property of multiplication.
= 6.25l2 + 1.25ml – 1.25ml – 0.25m2
# Expanding the terms.
= 6.25l2 + 0 – 0.25m2 # Subtracting the like terms.
= 6.25l2 – 0.25m2

### (iv) (a + 3b) × (x + 5)

Solution:

= a × (x + 5) + 36 × (x + 5)
# Here, we used distributive property of multiplication.
= (a × x) + (a × 5) + (36 × x) + (36 × 5)
# Expanding the terms.
= ax + 5a + 3bx + 15b # Adding the like terms.

### (v) (2pq + 3q2) × (3pq – 2q2)

Solution:

= 2pq × (3pq – 2q2) + 3q2 (3pq – 2q2
# Here, we used distributive property of multiplication.
= (2pq × 3pq) – (2pq × 2q2) + (3q2 × 3pq) – (3q2 × 2q2
# Expanding the terms.
= 6p2q2 – 4pq3 + 9pq3 – 6q4 # Subtracting the like terms.
= 6p2q2 + 5pq3 – 6q4

### (vi) (3/4 a2  + 3b2) x 4(a2 – 2/3 b2)

Solution:

= 3/4a2 x (4a2 – 8/3b2 ) + 3b2 x (4a2 – 8/3b2 ) # Here , we used distributive property of multiplication.
= 3a4 – 2a2b2 + 12a2b2 – 8b4 # Subtracting the like terms.
= 3a4 + 10a2b2 – 8b4

### (i) (5 – 2x) (3 + x)

Solution:

= 5(3 + x) – 2x(3 + x)
# Here, we used distributive property of multiplication.
= (5 × 3) + (5 × x) – (2x × 3) – (2x × x)
# Expanding the terms.
= 15 + 5x – 6x – 2x2 # Subtracting the like terms.
= 15 – x – 2x2

### (ii) (x + 7y) (7x – y)

Solution:

= x(7x – y) + 7y(7x – y)
# Here, we used distributive property of multiplication.
= (x × 7x) – (x × y) + (7y × 7x) – (7y × y)
# Expanding the terms.
= 7x2 – xy + 49xy – 7y2 # Subtracting the like terms.
= 7x2 + 48xy – 7y2

### (iii) (a2 + b) (a + b2)

Solution:

= a2 (a + b2) + b(a + b2
# Here, we used distributive property of multiplication.
= (a2 × a) + (a2 × b2) + (b × a) + (b × b2
# Expanding the terms.
= a3 + a2b2 + ab + b3

### (iv) (p2 – q2)(2p + q)

Solution:

=p2(2p + q) – q2(2p + q)
# Here, we used distributive property of multiplication.
= (p2 × 2p) + (p2 × q) – (q2 × 2p) – (q2 × q)
# Expanding the terms.
= 2p3 + p2q – 2pq2 – q3

### (i) (x2 – 5) (x + 5) + 25

Solution:

= x2(x + 5) + 5(x + 5) + 25
# Here, we used distributive property of multiplication.
= x3 + 5x2 – 5x – 25 + 25
# Expanding the terms.
= x3 + 5x2 – 5x + 0
# Subtracting the like terms.
= x3 + 5x2 – 5x

### (ii) (a2 + 5)(b3 + 3) + 5

Solution:

= a2(b3 + 3) + 5(b3 + 3) + 5 # Here, we used distributive property of multiplication.
= a2b3 + 3a2 + 5b3 + 15 + 5 # Expanding the terms.
= a2b3 + 3a2 + 5b3 + 20 # Adding the like terms.

### (iii) (t + s2) (t2 – s)

Solution:

= t(t2 – s) + s2(t2 – s) # Here, we used distributive property of multiplication.
= t3 – st + s2t2 – s3 # Expanding the terms.
= t3 + s2t2 – st – s3

### (iv) (a + b)(c – d) + (a – b) (c + d) + 2(ac + bd)

Solution:

= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2ac + 2bd # Here, we used distributive property of multiplication.
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd # Expanding the terms.
= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd # Adding or subtracting the like terms.
= 4ac + 0 + 0 + 0
= 4ac

### (v) (x + y) (2x + y) + (x + 2y) (x – y)

Solution:

= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y) # Here, we used distributive property of multiplication.
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2 # Expanding the terms.
= 2x2 + x2 + xy + 2xy – xy + 2xy + y2 – 2y2 # Adding or subtracting the like terms.
= 3x2 + 4xy – y2

### (vi) (x + y)(x2  – xy + y2)

Solution:

= x(x2 – xy + y2) + y(x2 – xy + y2) # Here, we used distributive property of multiplication.
= x3 – x2y + x2y + xy2 – xy2 + y3 # Expanding the terms.
= x3 – 0 + 0 + y3 # Adding or subtracting the like terms.
= x3 + y3

### (vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x.+ 12y

Solution:

= 1.5x (1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y # Here, we used distributive property of multiplication.
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y # Expanding the terms.
= 2.25x2 + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y2 # Adding or subtracting the like terms.
= 2.25x2 + 0 + 0 + 0 – 16y2
= 2.25x2 – 16y2

### (viii) (a + b + c) (a + b – c)

Solution:

= a(a + b – c) + b(a + b – c) + c(a + b – c) # Here , we used distributive property of multiplication.
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2 # Expanding the terms.
= a2 + ab + ab – bc + bc – ac + ac + b2 – c2 # Adding or subtracting the like terms.
= a2 + 2ab + b2 – c2 + 0 + 0
= a2 + 2ab + b2 – c2

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