Class 8 RD Sharma Solutions – Chapter 7 Factorization – Exercise 7.8 | Set 1

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 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. 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