Class 8 Exam  >  Class 8 Notes  >  NCERT Textbooks & Solutions for Class 8  >  NCERT Solutions: Factorisation- 2

NCERT Solutions for Class 8 Maths - Factorisation- 2

Exercise 14.3 

Question 1: Carry out the following divisions: 

NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2  

Solution:
(i)
NCERT Solutions for Class 8 Maths - Factorisation- 2

(ii)
NCERT Solutions for Class 8 Maths - Factorisation- 2

(iii)
NCERT Solutions for Class 8 Maths - Factorisation- 2

(iv)

NCERT Solutions for Class 8 Maths - Factorisation- 2

(v)
NCERT Solutions for Class 8 Maths - Factorisation- 2


Question 2: Divide the given polynomial by the given monomial: 

NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2

Solution:
(i)
NCERT Solutions for Class 8 Maths - Factorisation- 2

(ii)
NCERT Solutions for Class 8 Maths - Factorisation- 2

(iii)
NCERT Solutions for Class 8 Maths - Factorisation- 2

(iv)
NCERT Solutions for Class 8 Maths - Factorisation- 2

(v)
NCERT Solutions for Class 8 Maths - Factorisation- 2


Question 3: Work out the following divisions:

NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2
Solution: 

NCERT Solutions for Class 8 Maths - Factorisation- 2

NCERT Solutions for Class 8 Maths - Factorisation- 2



Question 4:  Divide as directed: 

NCERT Solutions for Class 8 Maths - Factorisation- 2

Answer 4:

NCERT Solutions for Class 8 Maths - Factorisation- 2

Question 5: Factorize the expressions and divide them as directed: 

NCERT Solutions for Class 8 Maths - Factorisation- 2

Solution: 
NCERT Solutions for Class 8 Maths - Factorisation- 2
                                          = m - 16
NCERT Solutions for Class 8 Maths - Factorisation- 2

NCERT Solutions for Class 8 Maths - Factorisation- 2

NCERT Solutions for Class 8 Maths - Factorisation- 2

NCERT Solutions for Class 8 Maths - Factorisation- 2


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)+ 4(2x) + 7 = 2x+ 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 = y– 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 for Class 8 Maths - Factorisation- 2   
NCERT Solutions for Class 8 Maths - Factorisation- 2

NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2
NCERT Solutions for Class 8 Maths - Factorisation- 2
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 + 12x + 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)= 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 for Class 8 Maths - Factorisation- 2It is an incorrect statement.
∵ The correct statement is

NCERT Solutions for Class 8 Maths - Factorisation- 2 
NCERT Solutions for Class 8 Maths - Factorisation- 2  

NCERT Solutions for Class 8 Maths - Factorisation- 2

The document NCERT Solutions for Class 8 Maths - Factorisation- 2 is a part of the Class 8 Course NCERT Textbooks & Solutions for Class 8.
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FAQs on NCERT Solutions for Class 8 Maths - Factorisation- 2

1. What is factorisation in mathematics?
Ans. Factorisation is a mathematical process of breaking down a given expression or number into its factors, which are smaller expressions or numbers that can be multiplied together to produce the original expression or number.
2. What are the different methods of factorisation?
Ans. The different methods of factorisation include: (i) Common factor method (ii) Factorisation by regrouping terms (iii) Factorisation of quadratic expressions using the formula (iv) Factorisation of perfect square trinomials (v) Factorisation of the difference of two squares.
3. What is the difference between a factor and a multiple?
Ans. A factor is a number that divides another number exactly without leaving a remainder. For example, 3 is a factor of 12 because 12 can be divided by 3 without leaving any remainder. A multiple, on the other hand, is a number that is obtained by multiplying a given number by another whole number. For example, 12 is a multiple of 3 because it is obtained by multiplying 3 by 4.
4. How can factorisation be used to solve quadratic equations?
Ans. Factorisation can be used to solve quadratic equations by breaking down the quadratic expression into two linear factors and equating each factor to 0. This will give us the roots of the quadratic equation, which are the values of x that satisfy the equation.
5. What are the real-life applications of factorisation?
Ans. Factorisation has many real-life applications in fields such as finance, physics, chemistry, and engineering. It is used in cryptography to encrypt and decrypt messages, in signal processing to compress and transmit data, in chemistry to balance chemical equations, and in engineering to design and analyze complex systems. It is also used in finance to calculate compound interest and in statistics to find the prime factors of large numbers.
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