Class 8 RD Sharma Solutions Chapter 7 Factorization Exercise 7.8 | Set 1 (original) (raw)
Last Updated : 23 Jul, 2025
Exercise 7.8 in Chapter 7 of RD Sharma's Class 8 Mathematics focuses on advanced factorization techniques. This set challenges students to apply various methods of factorization to more complex algebraic expressions. It builds upon the concepts introduced in previous exercises, pushing students to deepen their understanding and problem-solving skills in algebra. The problems in this exercise are carefully crafted to encompass various factorization strategies, including the sum and difference of cubes, perfect square trinomials, difference of squares, and factoring by grouping. By engaging with these more intricate problems, students are encouraged to develop a deeper intuition for algebraic structures and enhance their ability to recognize patterns and relationships within expressions. This exercise serves as a crucial stepping stone in preparing students for more advanced mathematical concepts they will encounter in higher grades.
**Resolve each of the following quadratic trinomials into factors:
**Question 1. 2x 2 + 5x + 3
**Solution:
**Given:
2x2 + 5x + 3
The coefficient of x2 = 2
The coefficient of x = 5
Constant term = 3
Split the centre term, that is '5' into two parts such that their sum p+q is 5 and product pq = 2 × 3 is 6
So, we write the middle term 5x as 2x + 3x
2x2 + 5x + 3 = 2x2 + 2x + 3x + 3
= 2x (x + 1) + 3 (x + 1)
= (2x + 3) (x + 1)
**Question 2. 2x 2 – 3x – 2
**Solution:
**Given:
2x2 – 3x – 2
The coefficient of x2 = 2
The coefficient of x = -3
Constant term = -2
So, we write the middle term -3x as -4x + x
2x2 – 3x – 2 = 2x2 – 4x + x – 2
= 2x (x – 2) + 1 (x – 2)
= (x – 2) (2x + 1)
**Question 3. 3x 2 + 10x + 3
**Solution:
**Given:
3x2 + 10x + 3
The coefficient of x2 = 3
The coefficient of x = 10
Constant term = 3
So, we write the middle term 10x as 9x + x
3x2 + 10x + 3 = 3x2 + 9x + x + 3
= 3x (x + 3) + 1 (x + 3)
= (3x + 1) (x + 3)
**Question 4. 7x – 6 – 2x 2
**Solution:
**Given:
7x – 6 – 2x2
– 2x2 + 7x – 6
2x2 – 7x + 6
The coefficient of x2 = 2
The coefficient of x = -7
Constant term = 6
So, we write the middle term -7x as -4x – 3x
2x2 – 7x + 6 = 2x2 – 4x – 3x + 6
= 2x (x – 2) – 3 (x – 2)
= (x – 2) (2x – 3)
**Question 5. 7x 2 – 19x – 6
**Solution:
**Given:
7x2 – 19x – 6
The coefficient of x2 = 7
The coefficient of x = -19
Constant term = -6
So, we write the middle term -19x as 2x – 21x
7x2 – 19x – 6 = 7x2 + 2x – 21x – 6
= x (7x + 2) – 3 (7x + 2)
= (7x + 2) (x – 3)
**Question 6. 28 – 31x – 5x 2
**Solution:
**Given:
28 – 31x – 5x2
– 5x2 -31x + 28
5x2 + 31x – 28
The coefficient of x2 = 5
The coefficient of x = 31
Constant term = -28
So, we write the middle term 31x as -4x + 35x
5x2 + 31x – 28 = 5x2 – 4x + 35x – 28
= x (5x – 4) + 7 (5x – 4)
= (x + 7) (5x – 4)
**Question 7. 3 + 23y – 8y 2
**Solution:
**Given:
3 + 23y – 8y2
– 8y2 + 23y + 3
8y2 – 23y – 3
The coefficient of y2 = 8
The coefficient of y = -23
Constant term = -3
So, we write the middle term -23y as -24y + y
8y2 – 23y – 3 = 8y2 – 24y + y – 3
= 8y (y – 3) + 1 (y – 3)
= (8y + 1) (y – 3)
**Question 8. 11x 2 – 54x + 63
**Solution:
**Given:
11x2 – 54x + 63
The coefficient of x2 = 11
The coefficient of x = -54
Constant term = 63
So, we write the middle term -54x as -33x – 21x
11x2 – 54x + 63 = 11x2 – 33x – 21x – 63
= 11x (x – 3) – 21 (x – 3)
= (11x – 21) (x – 3)
**Question 9. 7x – 6x 2 + 20
**Solution:
**Given:
7x – 6x2 + 20
– 6x2 + 7x + 20
6x2 – 7x – 20
The coefficient of x2 = 6
The coefficient of x = -7
Constant term = -20
So, we write the middle term -7x as -15x + 8x
6x2 – 7x – 20 = 6x2 – 15x + 8x – 20
= 3x (2x – 5) + 4 (2x – 5)
= (3x + 4) (2x – 5)
**Question 10. 3x 2 + 22x + 35
**Solution:
**Given:
3x2 + 22x + 35
The coefficient of x2 = 3
The coefficient of x = 22
Constant term = 35
So, we write the middle term 22x as 15x + 7x
3x2 + 22x + 35 = 3x2 + 15x + 7x + 35
= 3x (x + 5) + 7 (x + 5)
= (3x + 7) (x+ 5)