NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.1
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.2
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.3
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.4
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.5
- Algebraic Expressions and Identities Class 8 Extra Questions
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4
Ex 9.4 Class 8 Maths Question 1.
Multiply the binomials:
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q
2
) and (3pq – 2q
2
)
(vi) (\(\frac { 3 }{ 4 }\)a
2
+ 3b
2
) and 4(a
2
– \(\frac { 2 }{ 3 }\) b
2
)
Solution:
(i) (2x + 5) × (4x – 3)
= 2x × (4x – 3) + 5 × (4x – 3)
= (2x × 4x) – (3 × 2x) + (5 × 4x) – (5 × 3)
= 8x
2
– 6x + 20x – 15
= 8x
2
+ 14x – 15
(ii) (y – 8) × (3y – 4)
= y × (3y – 4) – 8 × (3y – 4)
= (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4)
= 3y
2
– 4y – 24y + 32
= 3y
2
– 28y + 32
(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
= 6.25l
2
+ 1.25ml – 1.25ml – 0.25m
2
= 6.25l
2
+ 0 – 0.25m
2
= 6.25l
2
– 0.25m
2
(iv) (a + 3b) × (x + 5)
= a × (x + 5) + 36 × (x + 5)
= (a × x) + (a × 5) + (36 × x) + (36 × 5)
= ax + 5a + 3bx + 15b
(v) (2pq + 3q
2
) × (3pq – 2q
2
)
= 2pq × (3pq – 2q
2
) + 3q
2
(3pq – 2q
2
)
= (2pq × 3pq) – (2pq × 2q
2
) + (3q
2
× 3pq) – (3q
2
× 2q
2
)
= 6p
2
q
2
– 4pq
3
+ 9pq
3
– 6q
4
= 6p
2
q
2
+ 5pq
3
– 6q
4
Ex 9.4 Class 8 Maths Question 2.
Find the product:
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a
2
+ b) (a + b
2
)
(iv) (p
2
– q
2
)(2p + q)
Solution:
(i) (5 – 2x) (3 + x)
= 5(3 + x) – 2x(3 + x)
= (5 × 3) + (5 × x) – (2x × 3) – (2x × x)
= 15 + 5x – 6x – 2x
2
(ii) (x + 7y) (7x – y)
= x(7x – y) + 7y(7x – y)
= (x × 7x) – (x × y) + (7y × 7x) – (7y × y)
= 7x
2
– xy + 49xy – 7y
2
= 7x
2
+ 48xy – 7y
2
(iii) (a
2
+ b) (a + b
2
)
= a
2
(a + b
2
) + b(a + b
2
)
= (a
2
× a) + (a
2
× b
2
) + (b × a) + (b × b
2
)
= a
3
+ a
2
b
2
+ ab + b
3
(iv) (p
2
– q
2
)(2p + q)
= p
2
(2p + q) – q
2
(2p + q)
= (p
2
× 2p) + (p
2
× q) – (q
2
× 2p) – (q
2
× q)
= 2p
3
+ p
2
q – 2pq
2
– q
3
Ex 9.4 Class 8 Maths Question 3.
Simplify:
(i) (x
2
– 5) (x + 5) + 25
(ii) (a
2
+ 5)(b
3
+ 3) + 5
(iii) (t + s
2
) (t
2
– s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y)(x
2
– xy + y
2
)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c) (a + b – c)
Solution:
(i) (x
2
– 5) (x + 5) + 25
= x
2
(x + 5) + 5(x + 5) + 25
= x
3
+ 5x
2
– 5x – 25 + 25
= x
3
+ 5x
2
– 5x + 0
= x
3
+ 5x
2
– 5x
(ii) (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
= a
2
b
3
+ 3a
2
+ 5b
3
+ 20
(iii) (t + s
2
) (t
2
– s)
= t(t
2
– s) + s
2
(t
2
– s)
= t
3
– st + s
2
t
2
– s
3
= t
3
+ s
2
t
2
– st – s
3
(iv) (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 + 2ac + bc – bc – ad + ad – bd – bd + 2bd
= 4ac + 0 + 0 + 0
= 4ac
(v) (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
+ xy + 2xy – xy + 2xy + y
2
– 2y
2
= 3x
2
+ 4xy – y
2
(vi) (x + y)(x
2
– xy + y
2
)
= x(x
2
– xy + y
2
) + y(x
2
– xy + y
2
)
= x
3
– x
2
y + x
2
y + xy
2
– xy
2
+ y
3
= x
3
– 0 + 0 + y
3
= x
3
+ y
3
(vii) (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
+ 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y
2
= 2.25x
2
+ 0 + 0 + 0 – 16y
2
= 2.25x
2
– 16y
2
(viii) (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
+ ab + ab – bc + bc – ac + ac + b
2
– c
2
= a
2
+ 2ab + b
2
– c
2
+ 0 + 0
= a
2
+ 2ab + b
2
– c
2
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