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Class 8 RD Sharma Solutions – Chapter 7 Factorization – Exercise 7.8 | Set 1
  • Last Updated : 07 Apr, 2021

Resolve each of the following quadratic trinomials into factors:

Question 1. 2x2 + 5x + 3

Solution:

Given:

2x2 + 5x + 3

The coefficient of x2 = 2

The coefficient of x = 5



Constant term = 3

Split the center 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. 2x2 – 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. 3x2 + 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 – 2x2

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. 7x2 – 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 – 5x2

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 – 8y2

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. 11x2 – 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 – 6x2 + 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. 3x2 + 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)

Chapter 7 Factorization – Exercise 7.8 | Set 2

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