Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.1 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.
Division of Polynomials Class 8 Maths Chapter 10 Practice Set 10.1 Solutions Maharashtra Board
Std 8 Maths Practice Set 10.1 Chapter 10 Solutions Answers
Question 1.
Divide and write the quotient and the remainder.
i. 21m² ÷ 7m
ii. 40a³ ÷ (-10a)
iii. (- 48p
4
) ÷ (- 9p
2
)
iv. 40m
5
÷ 30m
3
v. (5x
3
– 3x
2
) ÷ x²
vi. (8p
3
– 4p
2
) ÷ 2p
2
vii. (2y
3
+ 4y
2
+ 3 ) ÷ 2y
2
viii. (21x
4
– 14x
2
+ 7x) ÷ 7x
3
ix. (6x
5
– 4x
4
+ 8x
3
+ 2x
2
) ÷ 2x
2
x. (25m
4
– 15m
3
+ 10m + 8) ÷ 5m
3
Solution:
i. 21m² ÷ 7m
∴ Quotient = 3m
Remainder = 0
ii. 40a³ ÷ (-10a)
∴ Quotient = -4a²
Remainder = 0
iii. (- 48p
4
) ÷ (- 9p
2
)
∴ Quotient = \(\frac { 16 }{ 3 }\) p²
Remainder = 0
iv. 40m
5
÷ 30m
3
∴ Quotient = \(\frac { 4 }{ 3 }\) m²
Remainder = 0
v. (5x
3
– 3x
2
) ÷ x²
∴ Quotient = 5x – 3
Remainder = 0
vi. (8p
3
– 4p
2
) ÷ 2p
2
∴ Quotient = 4p – 2
Remainder = 0
vii. (2y
3
+ 4y
2
+ 3 ) ÷ 2y
2
∴ Quotient = y + 2
Remainder = 3
viii. (21x
4
– 14x
2
+ 7x) ÷ 7x
3
∴ Quotient = 3x
Remainder = -14x² + 7x
ix. (6x
5
– 4x
4
+ 8x
3
+ 2x
2
) ÷ 2x
2
∴ Quotient = 3x³ – 2x² + 4x + 1
Remainder = 0
x. (25m
4
– 15m
3
+ 10m + 8) ÷ 5m
3
∴ Quotient = 5m – 3
Remainder = 10m + 8
Maharashtra Board Class 8 Maths Chapter 10 Division of Polynomials Practice Set 10.1 Intext Questions and Activities
Question 1.
Fill in the blanks in the following examples. (Textbook pg. no. 61)
- 2a + 3a = __
- 7b – 4b = __
- 3p × p² = __
- 5m² × 3m² = __
- (2x + 5y) × \(\frac { 3 }{ x }\) = __
- (3x² + 4y) × (2x + 3y) = __
Solution:
- 2a + 3a = 5a
- 7b – 4b = 3b
- 3p × p² = 3p³
- 5m² × 3m² = 15m 4
- (2x + 5y) × \(\frac { 3 }{ x }\) = \(6+\frac { 15y }{ x }\)
- (3x² + 4y) × (2x + 3y) = 6x³ + 9x²y + 8xy + 12y²