RD Sharma Class 9 Solutions Chapter 5 - Factorization of Algebraic Expressions Exercise 5.4 - Free PDF
FAQs on RD Sharma Class 9 Solutions Chapter 5 - Factorization of Algebraic Expressions (Ex 5.4) Exercise 5.4
1. What is the main algebraic identity used to solve the problems in RD Sharma Class 9 Solutions for Chapter 5, Exercise 5.4?
The solutions for Exercise 5.4 primarily use the algebraic identity for the sum of cubes, which is: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx). Mastering the application of this formula is the key to solving all the questions in this specific exercise.
2. How do you identify if a polynomial from Exercise 5.4 can be factorised using the x³ + y³ + z³ - 3xyz identity?
To identify if this identity applies, check for the following pattern in the given polynomial:
- The expression should have four terms.
- Three of the terms must be perfect cubes (e.g., 8a³, 27b³, -64c³).
- The fourth term should be the product of -3 and the cube roots of the other three terms. For example, in a³ + b³ + c³ - 3abc, the cube roots are a, b, and c, and the last term is -3(a)(b)(c).
3. Can you outline the step-by-step method to factorise an expression like 8x³ + y³ + 27z³ - 18xyz as shown in the solutions?
Certainly. The step-by-step method to factorise 8x³ + y³ + 27z³ - 18xyz is as follows:
- Step 1: Express each term in the form of a cube. Here, 8x³ = (2x)³, y³ = (y)³, and 27z³ = (3z)³.
- Step 2: Verify the fourth term. Check if -18xyz equals -3(2x)(y)(3z). It does, as -3 × 2x × y × 3z = -18xyz.
- Step 3: Apply the identity (a+b+c)(a²+b²+c²-ab-bc-ca), where a=2x, b=y, and c=3z.
- Step 4: Substitute these values into the identity to get the final factors: (2x + y + 3z)(4x² + y² + 9z² - 2xy - 3yz - 6xz).
4. What is a common mistake students make when applying the x³ + y³ + z³ - 3xyz identity, and how do the solutions help?
A very common mistake is managing the signs, especially when one of the terms is negative (e.g., factorising a³ - b³ + c³ + 3abc). Students often forget to write it as a³ + (-b)³ + c³ - 3(a)(-b)(c). The detailed RD Sharma solutions help by explicitly showing how to rewrite the expression with positive signs between the cubes first, correctly identifying the 'a', 'b', and 'c' terms (where 'b' would be -b) before substituting into the formula. This prevents sign errors in the final factored form.
5. The identity for x³ + y³ + z³ - 3xyz is used in Ex 5.4. What happens to the factors if x + y + z = 0?
This is a critical special case often tested in exams. If x + y + z = 0, the entire identity simplifies significantly. Since the factored form is (x + y + z)(x² + y² + z² - xy - yz - zx), substituting (x + y + z) with 0 makes the whole right-hand side zero. This leads to the important result: If x + y + z = 0, then x³ + y³ + z³ = 3xyz. The solutions for later problems in the chapter often use this corollary directly.
6. How does solving problems from RD Sharma Chapter 5, Exercise 5.4, align with the Class 9 CBSE syllabus?
This exercise directly aligns with the 'Polynomials' unit in the CBSE Class 9 Maths syllabus for the 2025-26 session. The syllabus mandates the study and application of algebraic identities, specifically including the factorization of polynomials of the form x³ + y³ + z³ - 3xyz. Working through these solutions ensures a thorough understanding of this high-order thinking skill (HOTS) topic as prescribed by the NCERT framework.
7. After finding the factors for a problem in Exercise 5.4, how can I verify my answer is correct without checking the provided solutions?
To verify your answer, you can use the reverse process: expansion. Multiply the two factors you obtained. For instance, if you factorised an expression and got (x + y + z)(x² + y² + z² - xy - yz - zx), you can multiply these two brackets term by term. If your factorization is correct, the product after simplification should be exactly the original polynomial, x³ + y³ + z³ - 3xyz. This method confirms your answer and strengthens your understanding of algebraic multiplication.
