RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 8 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 8. Students can also read RBSE Class 8 Maths Important Questions for exam preparation. Students can also go through RBSE Class 8 Maths Notes to understand and remember the concepts easily. Practicing the class 8 maths chapter 6 try these solutions will help students analyse their level of preparation.

RBSE Class 8 Maths Solutions Chapter 14 Factorization Ex 14.4

Find and correct the errors in the following mathematical statements:

Question 1.
4(x - 5) = 4x - 5
Answer:
We have,
4(x - 5) = 4x - 5
This is incorrect.
The correct statement is
4(x - 5) = 4x - 4 x 5 4x - 20

Question 2.
x(3x + 2) = 3x2 + 2
Answer:
x(3x + 2) = 3x2 + 2
This is incorrect statement.
Correct is
x (3x + 2) = 3x2 + 2x

RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

Question 3.
2x + 3y = 5xy
Answer:
We have,
2x + 3y = 5xy
The given statement is incorrect.
The correction is
2x + 3y = 2x + 3y

Question 4.
x + 2x + 3x = 5x
Answer:
We have,
x + 2x + 3x = 5x (Incorrect)
Correct is x + 2x + 3x = 6x

Question 5.
5y + 2y + y -7y = 0
Answer:
5y + 2y + y - 7y = 0
The given statement is incorrect,
∵ 5y + 2y + y= 8y
and (8 - 7)y = y
∴ The correct statement is 5y + 2y + y - 7y = y

Question 6.
3x + 2x = 5x2
Answer:
3x + 2x = 5x2
(Incorrect statement)
The correction is
3x + 2x = 5x

Question 7.
(2x)2 + 4(2x) + 7 = 2x2 + 8x + 7
Answer:
(2x)2 + 4(2x) + 1 = 7x2 + 8x + 7
(Incorrect)
∵ (2x)2 = 4x2
The correct statement is
(2x)2 + 4(2x) + 7 = 4x2 + 8x + 7

Question 8.
(2x)2 + 5x = 4x + 5x = 9x
Answer:
(2x)2 + 5x = 4x + 5x = 9x
(Incorrect statement)
∵ (2X)2=4x2
The correct statement is
(2x)2 + 5x= 4x2 + 5x

Question 9.
(3x + 2)2 = 3x2 + 6x +4
Answer:
(3x + 2)2 = 3x2 + 6x + 4
(Incorrect)
Correction is
(3x + 2)2 = (3x2 + 2(3x) (2) + (2)2
= 9x2 + 6x + 4

RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

Question 10.
Substituting x = - 3 in
(a) x2 + 5x +.4 gives (- 3)2 + 5(- 3) + 4 = 9 + 2 + 4 = 15
Answer:
Incorrect statement
∵ x2 + 5x + 4= (- 3)2 + 5(- 3) + 4 = 9 - 15 + 4 = 13 - 15 = - 2
Hence, the correct statement is x2 + 5x + 4 = (- 3)2 + 5(- 3) + 4
= 9 - 15 + 4 = -2

(b) x2 - 5x + 4 gives (- 3)2 - 5(- 3) + 4 = 9 - 15 + 4 = -2
Answer:
x2 - 5x + 4 gives'
= (- 3)2 - 5(- 3) + 4
= 9 - 15 + 4 (Incorrect)
Correction is :
x2 - 5x + 4 gives (- 3)2 - 5(- 3) + 4 = 9 + 15 + 4 = 28

(c) x2 + 5x gives (- 3)2 + 5(- 3) = - 9 - 15 = - 24
Answer:
x2 + 5x gives (- 3)2 + 5(- 3)
= - 9 - 15 = - 24 (Incorrect)
Correction is :
x2 + 5x= (- 3)2 + 5(- 3) = 9 - 15 = - 6

Question 11.
(y - 3)2 = y2 - 9
Answer:
LHS= (y - 3)2 = y2 - 2 × y × 3 + (- 3)2 = y2 - 6y + 9
Thus, (y - 3)2 ≠ y2 - 9,
But (y - 3)2 = y2 - 6y + 9 is correct statement.

Question 12.
(z + 5)2 = z2 + 25
Answer:
LHS= (z + 5)22 = z2 + 2 × z × 5 + 52 = z2 + 10z + 25
Thus, (z + 5)2 ≠ z2 + 25,
But (z + 5)2 = z2 + 10z + 25 is the correct statement.

Question 13.
(2a + 3b)(a - b) = 2a2 - 3b2
Answer:
LHS = (2a + 3b)(a - b)
= 2a(a - b) + 3b(a - b)
= 2a × a - 2a × b + 3b × a - 3b × b
= 2a22 - 2ab + 3ab - 3b2
= 2a2 + ab - 3 b2
Thus, (2a + 3b)(a - b) ≠ 2a2 - 3b2,
But (2a + 3b) (a - b) = 2a2 + ab - 3b2 is the correct statement.

RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

Question 14.
(a + 4)(a + 2) = a2 + 8
Answer:
LHS = (a + 4)(a + 2)
= a2 + (4 + 2) a + 4 × 2 = a2 + 6a + 8
Thus, (a + 4)(a + 2) ≠ a2 + 8,
But (a + 4)(a + 2) = a2 + 6a + 8 is the correct statement.

Question 15.
(a - 4)(a - 2) = a2 - 8
Answer:
LHS= (a - 4)(a - 2)
= a2 - (4 + 2)a + (- 4 × - 2) = a2 - 6a + 8
Thus, (a - 4)(a - 2) ≠ a2 - 8,
But (a - 4)(a - 2) = a2 - 6a + 8 is the correct statement.

Prasanna
Last Updated on May 20, 2022, 4:15 p.m.
Published May 20, 2022