In RD Sharma Solutions for Class 8 Exercise 7.8 Chapter 7 Factorization, we will discuss the polynomials and factorization of quadratic polynomials of a theorem. Subject experts at BYJU’S have designed solutions to help students to learn shortcut methods and simple techniques to solve difficult problems. Practising RD Sharma Solutions is the best way to secure good marks in their final exams. Download the free PDF of RD Sharma Solutions from the links provided below.
RD Sharma Solutions for Class 8 Maths Exercise 7.8 Chapter 7 Factorization
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 7.8 Chapter 7 Factorization
EXERCISE 7.8 PAGE NO: 7.30
Resolve each of the following quadratic trinomials into factors:
1. 2x2 + 5x + 3
Solution:
We have,
2x2 + 5x + 3
The coefficient of x2 is 2
The coefficient of x is 5
The constant term is 3
We shall split up the centre term, i.e., 5, into two parts such that their sum p+q is 5 and product pq = 2 × 3 is 6
So, we express the middle term 5x as 2x + 3x
2x2 + 5x + 3 = 2x2 + 2x + 3x + 3
= 2x (x + 1) + 3 (x + 1)
= (2x + 3) (x + 1)
2. 2x2 – 3x – 2
Solution:
We have,
2x2 – 3x – 2
The coefficient of x2 is 2
The coefficient of x is -3
The constant term is -2
So, we express the middle term -3x as -4x + x
2x2 – 3x – 2 = 2x2 – 4x + x – 2
= 2x (x – 2) + 1 (x – 2)
= (x – 2) (2x + 1)
3. 3x2 + 10x + 3
Solution:
We have,
3x2 + 10x + 3
The coefficient of x2 is 3
The coefficient of x is 10
The constant term is 3
So, we express the middle term 10x as 9x + x
3x2 + 10x + 3 = 3x2 + 9x + x + 3
= 3x (x + 3) + 1 (x + 3)
= (3x + 1) (x + 3)
4. 7x – 6 – 2x2
Solution:
We have,
7x – 6 – 2x2
– 2x2 + 7x – 6
2x2 – 7x + 6
The coefficient of x2 is 2
The coefficient of x is -7
The constant term is 6
So, we express the middle term -7x as -4x – 3x
2x2 – 7x + 6 = 2x2 – 4x – 3x + 6
= 2x (x – 2) – 3 (x – 2)
= (x – 2) (2x – 3)
5. 7x2 – 19x – 6
Solution:
We have,
7x2 – 19x – 6
The coefficient of x2 is 7
The coefficient of x is -19
The constant term is -6
So, we express the middle term -19x as 2x – 21x
7x2 – 19x – 6 = 7x2 + 2x – 21x – 6
= x (7x + 2) – 3 (7x + 2)
= (7x + 2) (x – 3)
6. 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
The 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)
7. 3 + 23y – 8y2
Solution:
We have,
3 + 23y – 8y2
– 8y2 + 23y + 3
8y2 – 23y – 3
The coefficient of y2 is 8
The coefficient of y is -23
The constant term is -3
So, we express the middle term -23y as -24y + y
8y2 – 23y – 3 = 8y2 – 24y + y – 3
= 8y (y – 3) + 1 (y – 3)
= (8y + 1) (y – 3)
8. 11x2 – 54x + 63
Solution:
We have,
11x2 – 54x + 63
The coefficient of x2 is 11
The coefficient of x is -54
The constant term is 63
So, we express the middle term -54x as -33x – 21x
11x2 – 54x + 63 = 11x2 – 33x – 21x + 63
= 11x (x – 3) – 21 (x – 3)
= (11x – 21) (x – 3)
9. 7x – 6x2 + 20
Solution:
We have,
7x – 6x2 + 20
– 6x2 + 7x + 20
6x2 – 7x – 20
The coefficient of x2 is 6
The coefficient of x is -7
The constant term is -20
So, we express the middle term -7x as -15x + 8x
6x2 – 7x – 20 = 6x2 – 15x + 8x – 20
= 3x (2x – 5) + 4 (2x – 5)
= (3x + 4) (2x – 5)
10. 3x2 + 22x + 35
Solution:
We have,
3x2 + 22x + 35
The coefficient of x2 is 3
The coefficient of x is 22
The constant term is 35
So, we express the middle term 22x as 15x + 7x
3x2 + 22x + 35 = 3x2 + 15x + 7x + 35
= 3x (x + 5) + 7 (x + 5)
= (3x + 7) (x+ 5)
11. 12x2 – 17xy + 6y2
Solution:
We have,
12x2 – 17xy + 6y2
The coefficient of x2 is 12
The coefficient of x is -17y
The constant term is 6y2
So, we express the middle term -17xy as -9xy – 8xy
12x2 -17xy+ 6y2 = 12x2 – 9xy – 8xy + 6y2
= 3x (4x – 3y) – 2y (4x – 3y)
= (3x – 2y) (4x – 3y)
12. 6x2 – 5xy – 6y2
Solution:
We have,
6x2 – 5xy – 6y2
The coefficient of x2 is 6
The coefficient of x is -5y
The constant term is -6y2
So, we express the middle term -5xy as 4xy – 9xy
6x2 -5xy- 6y2 = 6x2 + 4xy – 9xy – 6y2
= 2x (3x + 2y) -3y (3x + 2y)
= (2x – 3y) (3x + 2y)
13. 6x2 – 13xy + 2y2
Solution:
We have,
6x2 – 13xy + 2y2
The coefficient of x2 is 6
The coefficient of x is -13y
The constant term is 2y2
So, we express the middle term -13xy as -12xy – xy
6x2 -13xy+ 2y2 = 6x2 – 12xy – xy + 2y2
= 6x (x – 2y) – y (x – 2y)
= (6x – y) (x – 2y)
14. 14x2 + 11xy – 15y2
Solution:
We have,
14x2 + 11xy – 15y2
The coefficient of x2 is 14
The coefficient of x is 11y
The constant term is -15y2
So, we express the middle term 11xy as 21xy – 10xy
14x2 + 11xy- 15y2 = 14x2 + 21xy – 10xy – 15y2
= 2x (7x – 5y) + 3y (7x – 5y)
= (2x + 3y) (7x – 5y)
15. 6a2 + 17ab – 3b2
Solution:
We have,
6a2 + 17ab – 3b2
The coefficient of a2 is 6
The coefficient of a is 17b
The constant term is -3b2
So, we express the middle term 17ab as 18ab – ab
6a2 +17ab– 3b2 = 6a2 + 18ab – ab – 3b2
= 6a (a + 3b) – b (a + 3b)
= (6a – b) (a + 3b)
16. 36a2 + 12abc – 15b2c2
Solution:
We have,
36a2 + 12abc – 15b2c2
The coefficient of a2 is 36
The coefficient of a is 12bc
The constant term is -15b2c2
So, we express the middle term 12abc as 30abc – 18abc
36a2 –12abc– 15b2c2 = 36a2 + 30abc – 18abc – 15b2c2
= 6a (6a + 5bc) – 3bc (6a + 5bc)
= (6a + 5bc) (6a – 3bc)
= (6a + 5bc) 3(2a – bc)
17. 15x2 – 16xyz – 15y2z2
Solution:
We have,
15x2 – 16xyz – 15y2z2
The coefficient of x2 is 15
The coefficient of x is -16yz
The constant term is -15y2z2
So, we express the middle term -16xyz as -25xyz + 9xyz
15x2 -16xyz- 15y2z2 = 15x2 – 25yz + 9yz – 15y2z2
= 5x (3x – 5yz) + 3yz (3x – 5yz)
= (5x + 3yz) (3x – 5yz)
18. (x – 2y)2 – 5 (x – 2y) + 6
Solution:
We have,
(x – 2y)2 – 5 (x – 2y) + 6
The coefficient of (x-2y)2 is 1
The coefficient of (x-2y) is -5
The constant term is 6
So, we express the middle term -5(x – 2y) as -2(x – 2y) -3(x – 2y)
(x – 2y)2 – 5 (x – 2y) + 6 = (x – 2y)2 – 2 (x – 2y) – 3 (x – 2y) + 6
= (x – 2y – 2) (x – 2y – 3)
19. (2a – b)2 + 2 (2a – b) – 8
Solution:
We have,
(2a – b)2 + 2 (2a – b) – 8
The coefficient of (2a-b)2 is 1
The coefficient of (2a-b) is 2
The constant term is -8
So, we express the middle term 2(2a – b) as 4 (2a –b) – 2 (2a – b)
(2a – b)2 + 2 (2a – b) – 8 = (2a – b)2 + 4 (2a – b) – 2 (2a – b) – 8
= (2a – b) (2a – b + 4) – 2 (2a – b + 4)
= (2a – b + 4) (2a – b – 2)
Comments