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

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

Problem 2: Find the product. 

(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

Problem 3. Simplify.

(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

Summary

Chapter 9 of Class 8 NCERT focuses on algebraic expressions and identities. It covers topics such as multiplication and division of algebraic expressions, identities, and their applications. The chapter introduces students to important algebraic concepts that form the foundation for more advanced mathematics in higher grades. It covers the multiplication and division of algebraic expressions, as well as important algebraic identities such as (a + b)², (a - b)², and (a + b)(a - b). The chapter emphasizes the application of these concepts in simplifying expressions and solving mathematical problems, laying a crucial foundation for more advanced algebraic studies in higher grades.