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NCERT Solutions(Part- 4)- Factorisation - Class 8 PDF Download

EXERCISE 14.4
Question: Find and correct the errors in the following mathematical statements.
1. 4(x – 5) = 4x – 5
2. x(3x + 2) = 3x² + 2
3. 2x + 3y = 5xy
4. x + 2x + 3x = 5x
5. 5y + 2y + y – 7y = 0
6. 3x + 2x = 5x²
7. (2x)²+ 4(2x) + 7 = 2x²+ 8x + 7
8. (2x)² + 5x = 4x + 5x = 9x²
9. (3x + 2)² = 3x² + 6x + 4
10. Substituting x = – 3 in
(a) x² + 5x + 4 gives (–3)² + 5(–3) + 4 = 9 + 2 + 4 = 15
(b) x² – 5x + 4 gives (–3)² – 5(–3) + 4 = 9 – 15 + 4 = – 2
(c) x² + 5x + 4 gives (–3)² + 5(–3) = – 9 – 15 = – 24
11. (y – 3)² = y²– 9
12. (z + 5)² = z² + 25
13. (2a + 3b)(a – b) = 2a² – 3b²
14. (a + 4)(a + 2) = a² + 8
15. (a – 4)(a – 2) = a² – 8
NCERT Solutions(Part- 4)- Factorisation - Class 8

Solution:
1. 4(x – 5) = 4x – 5
The given statement is incorrect.
The correct statement is:
4(x – 5) = 4x – 20 (∵ 4 * 5 = 20)

2. x(3x + 2) = 3x2 + 2
It is an incorrect statement.
The correct statement is:
x(3x + 2) = 3x2 + 2x

3. 2x + 3y = 5xy
It is an incorrect statement.
The correct statement is:
2x + 3y = 2x + 3y

4. x + 2x + 3x = 5x
∵ 1 + 2 + 3 = 5 is an incorrect statement.
∴ The correct statement is:
x + 2x + 3x = 6x

5. 5y + 2y + y – 7y = 0
It is an incorrect statement.
∵ 5y + 2y + y = 8y and 8y – 7y = y
∴ The correct statement is
5y + 2y + y – 7y = y

6. 3x + 2x = 5x²
It is an incorrect statement.
The correct statement is:
3x + 2x = 5x

7. (2x)² + 4(2x) + 7= 2x² + 8x + 7
∵ (2x)² = 4x²
∴ The given statement is incorrect.
The correct statement is:
(2x)² + 4(2x) + 7 = 4x² + 8x + 7

8. (2x)² + 5x = 4x + 5x = 9x, is an incorrect statement.
∵ (2x)² = 4x²
∴ The correct statement is:
(2x)² + 5x = 4x² + 5x

9. (3x + 2)²= 3x² + 6x + 4
The given statement is incorrect.
∵ (3x + 2)² = (3x)² + 2(3x)(2) + (2)²
= 9x² + 6x + 4
∴ The correct statement is:
(3x + 2)² = 9x² + 6x + 4

10. (a) Incorrect statement.
∵ x² + 5x + 4 = (–3)² + 5(–3) + 4
= 9 – 15 + 4
= (9 + 4) – 15
= 13 – 15 = –2
Thus, the correct statement is:
x² + 5x + 4 = (–3)² + 5(–3) + 4
= 9 – 15 + 4 = –2

(b) We have
x² – 5x + 4 = (–3)² – 5(–3) + 4
= 9 + 15 + 4
= 28
∴ The correct statement is
x² – 5x + 4 at x = –3 is
(–3)² – 5(–3) + 4 = 9 + 15 + 4 = 28

(c) ∵ x² + 5x at x = – 3 is
(–3)² + 5(–3) = 9 – 15 = –6
∴ The correct statement is
x2 + 5x at x = –3 is
(–3)2 + 5(–3) = 9 – 15 = –6

11. (y – 3)² = y² – 9
The given statement is incorrect.
∵ (y – 3)² = y² – 2(y)(3) + (3)² = y² – 6y + 9
The correct statement is
(y – 3)² = y² – 6y + 9

12. (z + 5)² = z² + 25
The given statement is incorrect.
∵ (z + 5)²= z² + 2(z)(5) + (5)²
= z² + 10z + 25
∴ The correct statement is
(z + 5)2 = z2 + 10z + 25

13. (2a + 3b)(a – b) = 2a² – 3b²
∵ (2a + 3b) (a – b) = a(2a + 3b) – b (2a – 3b)
= 2a² + 3ab – 2ab – 3b²
= 2a² + ab – 3b²
∴ The correct statement is
(2a + 3b)(a – b) = 2a² + ab + 3b²

14. (a + 4)(a + 2) = a² + 8
Since (a + 4) (a + 2) = a (a + 4) + 2 (a + 4)
= a² + 4a + 2a + 8
= a² + 6a + 8

15. (b – 4)(a – 2) = a²– 8
Since (a – 4)(a – 2) = a(a – 2) – 4(a – 2)
= a² – 2a – 4a + 8
= a² – 6a + 8
∴ The correct statement is
(a – 4)(a – 2) = a² – 6a + 8

NCERT Solutions(Part- 4)- Factorisation - Class 8
It is an incorrect statement.
∵ The correct statement is

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FAQs on NCERT Solutions(Part- 4)- Factorisation - Class 8

1. What is factorisation and why is it important in Class 8 mathematics?

Ans. Factorisation is the process of expressing a number or an algebraic expression as a product of its factors. It is important in Class 8 mathematics as it helps in simplifying algebraic expressions, solving equations, finding the common factors, and understanding the relationship between numbers and their factors.

2. How can factorisation be useful in solving real-life problems?

Ans. Factorisation can be useful in solving real-life problems by helping us simplify complex expressions or equations. For example, in finance, factorisation can be used to simplify interest calculations or loan repayments. In physics, it can be used to simplify equations involving forces or motion. By breaking down a problem into its factors, we can gain a better understanding and find efficient solutions.

3. What are the different methods of factorisation taught in Class 8?

Ans. In Class 8, students are taught various methods of factorisation, including: - Factorisation by taking out the common factor - Factorisation using identities such as (a+b)^2 = a^2 + 2ab + b^2 - Factorisation of quadratic trinomials using the middle term splitting method - Factorisation of perfect square trinomials and difference of squares These methods help students simplify algebraic expressions and solve equations.

4. Can factorisation be applied to both numbers and algebraic expressions?

Ans. Yes, factorisation can be applied to both numbers and algebraic expressions. In the case of numbers, factorisation involves expressing a number as a product of its prime factors. For example, factorising 24 would give us 2 × 2 × 2 × 3. In the case of algebraic expressions, factorisation involves expressing the expression as a product of its factors. For example, factorising x^2 + 5x + 6 would give us (x + 2)(x + 3). Factorisation is a fundamental concept in mathematics that can be applied to various contexts.

5. What is the significance of learning factorisation in Class 8 for higher-level mathematics?

Ans. Learning factorisation in Class 8 lays the foundation for higher-level mathematics. It helps in simplifying complex expressions, solving equations, and understanding the properties of numbers and algebraic expressions. Factorisation is an essential skill in algebra, which is extensively used in higher-level mathematics, including calculus, linear algebra, and number theory. By mastering factorisation at an early stage, students develop a strong mathematical foundation and are better prepared for advanced mathematical concepts and problem-solving.

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