Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.1 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

## Practice Set 10.1 8th Std Maths Answers Chapter 10 Division of Polynomials

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