RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization

The RD Sharma Solutions for Class 8 Maths Chapter 7 – Factorization are a great help for students who find this topic difficult. These RD SHarma solutions include answers to all the questions from the Class 8 Maths textbook, which follows the CBSE syllabus.

Expert teachers at Home-Tution have carefully prepared these solutions to help students score well in their final exams. You can download the RD SHarma solutions class 8 PDF of this chapter for free using the links below. Chapter 7 – Factorization includes nine exercises, and the solutions provided here cover all the questions in those exercises.

Topics Covered in This Chapter:

• What are factors and factorization
• Factors of a monomial, and how to find the common and greatest common factor of monomials
• Factorizing algebraic expressions with a common monomial in every term
• Factorizing when a binomial is the common factor
• Factorization by grouping terms
• Factorizing binomial expressions as the difference of two squares
• Factorizing binomial expressions as a perfect square
• Introduction to polynomials and factorizing quadratic polynomials in one variable
• Using the method of completing the square to factorize quadratic polynomials

These RD Sharma Solutions make it easier for students to understand and practice the key concepts of factorization step by step.

Download Free RD Sharma Solutions PDF for Class 8 Maths Chapter 7 – Factorization

You can download the free PDF of RD Sharma Solutions for Class 8 Maths Chapter 7 – Factorization, created by expert Maths teachers at Home-Tution. These PDFs are available on both Home-Tution website and mobile app. The PDF includes all the exercise questions from Chapter 7 along with detailed solutions. This helps students revise the entire chapter and prepare well for their exams.

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Home-Tution provides free NCERT Solutions PDFs for Class 8 Maths as well. These solutions are prepared by subject experts following the latest CBSE guidelines. Each PDF includes chapter-wise questions and answers to make learning easy and effective.

Chapter 7 – Factorization has nine exercises, and the RD Sharma solutions on this page include answers for all of them. These solutions are especially useful for students who find factorization problems challenging. You can download the free PDFs from the links provided on Home-Tution website and start your preparation today.

Access Answers to RD Sharma Solutions for Class 8 Maths Chapter 7

Q. Find the greatest common factor (GCF/HCF) of the following polynomials: (1-14)

1. 2x² and 12x²

2. 7x, 21x² and 14xy²

3. 6x³y and 18x²y³

4. 12ax², 6a²x³ and 2a³x⁵

Solutions:

1. 2x² and 12x²

Step 1: Find the numerical HCF of coefficients

HCF of 2 and 12 = 2

Step 2: Find the common literal (variable) factor

Both terms have x²

Final Answer:

HCF = 2x²

2. 7x, 21x² and 14xy²

Step 1: Find the numerical HCF of coefficients

HCF of 7, 21, and 14 = 7

Step 2: Find the common variable(s)

All terms have x

7x = 7 × x

21x² = 7 × 3 × x²

14xy² = 7 × 2 × x × y²

The common variable part is x

Final Answer:

HCF = 7x

3. 6x³y and 18x²y³

Step 1: Find HCF of numerical coefficients

HCF of 6 and 18 = 6

Step 2: Compare powers of each common variable

For x: x³ and x² → HCF = x²

For y: y and y³ → HCF = y

Final Answer:

HCF = 6x²y

4. 12ax², 6a²x³ and 2a³x⁵

Step 1: HCF of numerical coefficients

HCF of 12, 6, and 2 = 2

Step 2: Find the HCF of variable a

Powers: a, a², a³ → HCF = a (lowest power)

Step 3: Find the HCF of variable x

Powers: x², x³, x⁵ → HCF = x²

Final Answer:

HCF = 2ax²

Q. Find the greatest common factor of the terms in each of the following expressions:

1. 5a⁴ + 10a³ – 15a²

2. 2xyz + 3x²y + 4y²

3. 3a²b² + 4b²c² + 12a²b²c²

Solutions:

1. 5a⁴ + 10a³ – 15a²

Step 1: Find the GCF of the coefficients:

5, 10, and 15

HCF = 5

Step 2: Find the common variable factors:

5a⁴ = 5 × a⁴

10a³ = 5 × 2 × a³

15a² = 5 × 3 × a²

All terms have a² as the lowest power

Final Answer:

HCF = 5a²

2. 2xyz + 3x²y + 4y²

Step 1: Find the GCF of the coefficients:

2, 3, and 4 → No common factor other than 1

Step 2: Find the common variable factors:

2xyz → contains x, y, z

3x²y → contains x², y

4y² → contains only y²

The only common variable in all three terms is y

The lowest power of y in all terms is y

Final Answer:

HCF = y

3. 3a²b² + 4b²c² + 12a²b²c²

Step 1: Find the GCF of the coefficients:

3, 4, 12 → HCF = 1

Step 2: Find the common variable factors:

3a²b² → a², b²

4b²c² → b², c²

12a²b²c² → a², b², c²

All three terms contain b² as a common factor

Final Answer:

HCF = b²

Q. Resolve each of the following quadratic trinomials into factors:

1. 2x² – 3x – 2

Solution: We are given the expression:

2x² – 3x – 2

Let’s identify the components:

• The coefficient of x² is 2

• The coefficient of x is –3

• The constant term is –2

Now, we'll split the middle term (–3x) into two terms whose sum is –3x and whose product equals the product of the first and last terms (2 × –2 = –4).
We choose –4x and +x for this.

So, we rewrite the expression as:

2x² – 4x + x – 2

Next, group the terms:

= (2x² – 4x) + (x – 2)

Factor each group:

= 2x(x – 2) + 1(x – 2)

Now, factor the common binomial:

= (x – 2)(2x + 1)

2. 7x – 6 – 2x²

Solution: 7x – 6 – 2x²

Let’s rearrange the terms in standard quadratic form (descending powers of x):

–2x² + 7x – 6

Now, for easier factorization, multiply the entire expression by –1 to make the leading coefficient positive:

2x² – 7x + 6

Let’s identify the components:

• Coefficient of x²: 2

• Coefficient of x: –7

• Constant term: 6

We now split the middle term (–7x) into two terms whose product is 2 × 6 = 12 and whose sum is –7.
The suitable pair is –4x and –3x.

So, we rewrite the expression:

2x² – 4x – 3x + 6

Now group the terms:

= (2x² – 4x) – (3x – 6)
= 2x(x – 2) – 3(x – 2)

Factor the common binomial:

= (x – 2)(2x – 3)

3. 28 – 31x – 5x²

Solution:

We have,

28 – 31x – 5x²

– 5x² -31x + 28

5x² + 31x – 28

The coefficient of x² is 5

The coefficient of x is 31

Constant term is -28

So, we express the middle term 31x as -4x + 35x

5x² + 31x – 28 = 5x² – 4x + 35x – 28

= x (5x – 4) + 7 (5x – 4)

= (x + 7) (5x – 4)