RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions

Students preparing for their annual exams can greatly benefit from RD Sharma Solutions for Class 8 Maths Chapter 8 – Division of Algebraic Expressions. These solutions are an excellent resource, designed to help students master key concepts and problem-solving techniques. Expert educators at Home-Tution have developed comprehensive, step-by-step solutions to all questions in this chapter, ensuring a clear and thorough understanding of the topic.

The free downloadable PDFs of Chapter 8 are available below, providing convenient access to well-structured explanations for every exercise. This chapter includes six exercises, each focusing on various methods of dividing algebraic expressions.

Key Concepts Covered in Chapter 8:

• Understanding polynomials and the degree of a polynomial in two variables
• Division of a monomial by another monomial
• Division of polynomials in a single variable
• Dividing a polynomial by a monomial
• Long division method for dividing a polynomial by a binomial
• Using factorization techniques to divide polynomials

These RD Sharma Solutions Class 8 serve as a reliable guide for students aiming to improve their conceptual clarity and score higher in exams.

Download Free RD Sharma Solutions PDF for Class 8 Maths Chapter 8 Division of Algebraic Expressions

Practising questions from the RD Sharma Class 8 Chapter 8 PDF helps students gain hands-on understanding of the concepts. This practical knowledge not only strengthens their grasp of Division of Algebraic Expressions but also enables them to apply similar strategies while solving problems from other chapters and tackling real-life mathematical situations. For added convenience, students can access the Chapter 8 solutions in PDF format anytime from the Home-Tution website, making it easy to study from the comfort of their own space.

Access Answers to RD Sharma Solutions for Class 8 Maths Chapter 8

Q. Write the degree of each of the following polynomials:

(i) 2x³ + 5x² – 7

(ii) 5x² – 3x + 2

(iii) 2x + x² – 8

(iv) 1/2y⁷ – 12y⁶ + 48y⁵ – 10

(v) 3x³ + 1

(vi) 5

(vii) 20x³ + 12x²y² – 10y² + 20

Solutions:

(i) 2x³ + 5x² – 7

Explanation:

The degree of a polynomial is the highest power of the variable.

 Here, the powers of x are: 3, 2, and 0 (constant term -7).

Degree = 3

(ii) 5x² – 3x + 2

Explanation:

The powers of x in each term are: 2, 1, and 0 (constant).

Degree = 2

(iii) 2x + x² – 8

Explanation:

The powers of x are: 1 (2x), 2 (x²), and 0 (-8).

Rearranged, the polynomial is: x² + 2x – 8

Degree = 2

(iv) (1/2)y⁷ – 12y⁶ + 48y⁵ – 10

Explanation:

The powers of y are: 7, 6, 5, and 0. Even though coefficients vary, we focus on the highest exponent.

Degree = 7

(v) 3x³ + 1

Explanation:

Powers of x are 3 (3x³) and 0 (constant 1).

Degree = 3

(vi) 5

Explanation:

This is a constant polynomial with no variable.

By convention, the degree of a constant (≠ 0) is 0. 

Degree = 0

(vii) 20x³ + 12x²y² – 10y² + 20

Explanation:

This is a multivariable polynomial.

Degree of each term is:

20x³ - degree = 3

12x²y² - degree = 2 + 2 = 4

10y² - degree = 2

20 - degree = 0

The highest total degree among the terms is 4.

Degree = 4

Q. Write each of the following polynomials in the standard form. Also, write their degree.

(i) x² + 3 + 6x + 5x⁴

(ii) a² + 4 + 5a⁶

(iii) (x³ – 1) (x³ – 4)

(iv) (y³ – 2) (y³ + 11)

(v) (a³ – 3/8) (a³ + 16/17)

(vi) (a + 3/4) (a + 4/3)

Solutions:

(i) x² + 3 + 6x + 5x⁴

Step 1: Rearranging the terms in descending order of powers:

= 5x⁴ + x² + 6x + 3

Standard Form: 5x⁴ + x² + 6x + 3

Degree: 4 (highest power of x)

(ii) a² + 4 + 5a⁶

Step 1: Rearranging the terms in descending order of powers:

= 5a⁶ + a² + 4

Standard Form: 5a⁶ + a² + 4

Degree: 6 (highest power of a)

(iii) (x³ – 1)(x³ – 4)

Step 1: Use the identity (a – b)(a – c) = a² – (b + c)a + bc

= (x³)(x³) – (x³ × 4) – (1 × x³) + (–1 × –4)

= x⁶ – 4x³ – x³ + 4

= x⁶ – 5x³ + 4

Standard Form: x⁶ – 5x³ + 4

Degree: 6

(iv) (y³ – 2)(y³ + 11)

Step 1: Apply the identity (a – b)(a + b) = a² – b²

= (y³)² – (2)²

= y⁶ – 4

Standard Form: y⁶ – 4

Degree: 6

(v) (a³ – 3/8)(a³ + 16/17)

Step 1: Use identity (a – b)(a + b) = a² – b²

= a⁶ – (3/8 × 16/17)

= a⁶ – (48/136)

= a⁶ – (12/34) (simplified form)

= a⁶ – 6/17

Standard Form: a⁶ – 6/17

Degree: 6

(vi) (a + 3/4)(a + 4/3)

Step 1: Apply the identity (a + b)(a + c) = a² + (b + c)a + bc

= a² + (3/4 + 4/3)a + (3/4 × 4/3)

= a² + (25/12)a + 1

Standard Form: a² + (25/12)a + 1

Degree: 2

Q. Which of the following expressions are not polynomials?

(i) x² + 2x^-2

(ii) √(ax) + x² – x³

(iii) 3y³ – √5y + 9

(iv) ax^1/2 + ax + 9x² + 4

(v) 3x^-3 + 2x^-1 + 4x + 5

Solution:

(i) x² + 2x – 2

All terms have whole number exponents:

x² → power = 2

2x → power = 1

–2 → constant term

This is a polynomial.

(ii) √(ax) + x² – x³

(ax) = (ax)¹ᐟ² → involves a square root → not a whole number power

Not a polynomial because of the radical term.

(iii) 3y³ – √5y + 9

√5 is a constant, not a variable.

√5y = (√5) × y → power of y = 1 (which is valid)

This is a polynomial.

(iv) ax¹ᐟ² + ax + 9x² + 4

ax¹ᐟ² → variable x is raised to a fractional exponent (½)

Not a polynomial due to the fractional exponent.

(v) 3x⁻³ + 2x⁻¹ + 4x + 5

x⁻³ and x⁻¹ are negative exponents, which are not allowed in polynomials

Not a polynomial because of negative powers.

Final Answer:

The following expressions are not polynomials:

(ii) √(ax) + x² – x³

(iv) ax¹ᐟ² + ax + 9x² + 4

(v) 3x⁻³ + 2x⁻¹ + 4x + 5

Q. Divide:

i. 6x³y²z² by 3x²yz

ii. -21abc² by 7abc

iii. 24a³b³ by -8ab

iv. 15m²n³ by 5m²n²

Solution:

i. 6x³y²z² ÷ 3x²yz

Step-by-step calculation:

(6x³y²z²) ÷ (3x²yz) = (6 ÷ 3) × (x³ ÷ x²) × (y² ÷ y) × (z² ÷ z)

= 2 × x × y × z

Final Answer: 2xyz

ii. -21abc² ÷ 7abc

Step-by-step calculation:

(-21abc²) ÷ (7abc) = (-21 ÷ 7) × (a ÷ a) × (b ÷ b) × (c² ÷ c)

= -3 × 1 × 1 × c

Final Answer: -3c

iii. 24a³b³ ÷ -8ab

Step-by-step calculation:

(24a³b³) ÷ (-8ab) = (24 ÷ -8) × (a³ ÷ a) × (b³ ÷ b)
= -3 × a² × b²

Final Answer: -3a²b²

iv. 15m²n³ ÷ 5m²n²

Step-by-step calculation:

(15m²n³) ÷ (5m²n²) = (15 ÷ 5) × (m² ÷ m²) × (n³ ÷ n²)

= 3 × 1 × n

Final Answer: 3n

Q. Divide

1. x + 2x² + 3x⁴ – x⁵ by 2x

Solution:

We have,

(x + 2x² + 3x⁴ – x⁵) / 2x

x/2x + 2x²/2x + 3x⁴/2x – x⁵/2x

By using the formula aⁿ / a^m = a^n-m

1/2 x^1-1 + x^2-1 + 3/2 x^4-1 – 1/2 x^5-1

1/2 + x + 3/2 x³ – 1/2 x⁴

2. y⁴ – 3y³ + 1/2y² by 3y

Solution:

We have,

(y⁴ – 3y³ + 1/2y²)/ 3y

y⁴/3y – 3y³/3y + (½)y²/3y

By using the formula aⁿ / a^m = a^n-m

1/3 y^4-1 – y^3-1 + 1/6 y^2-1

1/3y³ – y² + 1/6y

3. -4a³ + 4a² + a by 2a

Solution:

We have,

(-4a³ + 4a² + a) / 2a

-4a³/2a + 4a²/2a + a/2a

By using the formula aⁿ / a^m = a^n-m

-2a^3-1 + 2a^2-1 + 1/2 a^1-1

-2a² + 2a + ½

4. –x⁶ + 2x⁴ + 4x³ + 2x² by √2x²

Solution:

We have,

(–x⁶ + 2x⁴ + 4x³ + 2x²) / √2x²

-x⁶/√2x² + 2x⁴/√2x² + 4x³/√2x² + 2x²/√2x²

By using the formula aⁿ / a^m = a^n-m

-1/√2 x^6-2 + 2/√2 x^4-2 + 4/√2 x^3-2 + 2/√2 x^2-2

-1/√2 x⁴ + √2x² + 2√2x + √2

5. -4a³ + 4a² + a by 2a

Solution:

We have,

(-4a³ + 4a² + a) / 2a

-4a³/2a + 4a²/2a + a/2a

By using the formula aⁿ / a^m = a^n-m

-2a^3-1 + 2a^2-1 + 1/2a^1-1

-2a² + 2a + ½

6. √3a⁴ + 2√3a³ + 3a² – 6a by 3a

Solution:

We have,

(√3a⁴ + 2√3a³ + 3a² – 6a) / 3a

√3a⁴/3a + 2√3a³/3a + 3a²/3a – 6a/3a

By using the formula aⁿ / a^m = a^n-m

√3/3 a^4-1 + 2√3/3 a^3-1 + a^2-1 – 2a^1-1

1/√3 a³ + 2/√3 a² + a – 2

Also Read:

• RD Sharma Solutions for Class 8 Maths Chapter 1
• RD Sharma Solutions for Class 8 Maths Chapter 2
• RD Sharma Solutions for Class 8 Maths Chapter 3
• RD Sharma Solutions for Class 8 Maths Chapter 4
• RD Sharma Solutions for Class 8 Maths Chapter 5
• RD Sharma Solutions for Class 8 Maths Chapter 6

Advantages of Using RD Sharma Class 8 Solutions - Chapter 8: Division of Algebraic Expressions

RD Sharma Class 8 Chapter 8 Solutions offer students a structured and practical approach to mastering complex algebraic expressions. Along with helping students score well, the solutions provide several academic benefits, including:

• Students can grasp complex topics and learn smart techniques to solve challenging questions accurately. The solutions offer expert-level guidance for better clarity
• While studying Chapter 8, students may face uncertainties or doubts. RD Sharma solutions address these challenges with step-by-step explanations, helping learners overcome confusion confidently.
• Consistent practice using RD Sharma Chapter 8 solutions enables students to revise all key topics of the chapter effectively and reinforces their understanding.
• Practising multiple questions under a structured format trains students to solve problems within a time limit, which is crucial during exams.
• Regular exposure to different types of questions helps students enhance their accuracy, reducing careless mistakes during assessments.
• All key formulas and concepts of Chapter 8 are clearly presented in the solutions PDF, making it easier for students to memorize and recall them during exams.