**Resolve each of the following quadratic trinomials into factors:**

**Question 1. 2x**^{2} + 5x + 3

^{2}+ 5x + 3

**Solution:**

Given:2x

^{2}+ 5x + 3The coefficient of x

^{2}= 2The 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

2x

^{2}+ 5x + 3 = 2x^{2}+ 2x + 3x + 3= 2x (x + 1) + 3 (x + 1)

= (2x + 3) (x + 1)

**Question 2. 2x**^{2} – 3x – 2

^{2}– 3x – 2

**Solution:**

Given:2x

^{2}– 3x – 2The coefficient of x

^{2}= 2The coefficient of x = -3

Constant term = -2

So, we write the middle term -3x as -4x + x

2x

^{2}– 3x – 2 = 2x^{2}– 4x + x – 2= 2x (x – 2) + 1 (x – 2)

= (x – 2) (2x + 1)

**Question 3. 3x**^{2} + 10x + 3

^{2}+ 10x + 3

**Solution:**

Given:3x

^{2}+ 10x + 3The coefficient of x

^{2}= 3The coefficient of x = 10

Constant term = 3

So, we write the middle term 10x as 9x + x

3x

^{2}+ 10x + 3 = 3x^{2}+ 9x + x + 3= 3x (x + 3) + 1 (x + 3)

= (3x + 1) (x + 3)

**Question 4. 7x – 6 – 2x**^{2}

^{2}

**Solution:**

Given:7x – 6 – 2x

^{2}– 2x

^{2}+ 7x – 62x

^{2}– 7x + 6The coefficient of x

^{2}= 2The coefficient of x = -7

Constant term = 6

So, we write the middle term -7x as -4x – 3x

2x

^{2}– 7x + 6 = 2x^{2}– 4x – 3x + 6= 2x (x – 2) – 3 (x – 2)

= (x – 2) (2x – 3)

**Question 5. 7x**^{2} – 19x – 6

^{2}– 19x – 6

**Solution:**

Given:7x

^{2}– 19x – 6The coefficient of x

^{2}= 7The coefficient of x = -19

Constant term = -6

So, we write the middle term -19x as 2x – 21x

7x

^{2}– 19x – 6 = 7x^{2}+ 2x – 21x – 6= x (7x + 2) – 3 (7x + 2)

= (7x + 2) (x – 3)

**Question 6. 28 – 31x – 5x**^{2}

^{2}

**Solution:**

Given:28 – 31x – 5x

^{2}– 5x

^{2}-31x + 285x

^{2}+ 31x – 28The coefficient of x

^{2}= 5The coefficient of x = 31

Constant term = -28

So, we write the middle term 31x as -4x + 35x

5x

^{2}+ 31x – 28 = 5x^{2}– 4x + 35x – 28= x (5x – 4) + 7 (5x – 4)

= (x + 7) (5x – 4)

**Question 7. 3 + 23y – 8y**^{2}

^{2}

**Solution:**

Given:3 + 23y – 8y

^{2}– 8y

^{2}+ 23y + 38y

^{2}– 23y – 3The coefficient of y

^{2}= 8The coefficient of y = -23

Constant term = -3

So, we write the middle term -23y as -24y + y

8y

^{2}– 23y – 3 = 8y^{2}– 24y + y – 3= 8y (y – 3) + 1 (y – 3)

= (8y + 1) (y – 3)

**Question 8. 11x**^{2} – 54x + 63

^{2}– 54x + 63

**Solution:**

Given:11x

^{2}– 54x + 63The coefficient of x2 = 11

The coefficient of x = -54

Constant term = 63

So, we write the middle term -54x as -33x – 21x

11x

^{2}– 54x + 63 = 11x^{2}– 33x – 21x – 63= 11x (x – 3) – 21 (x – 3)

= (11x – 21) (x – 3)

**Question 9. 7x – 6x**^{2} + 20

^{2}+ 20

**Solution:**

Given:7x – 6x

^{2}+ 20– 6x

^{2}+ 7x + 206x

^{2}– 7x – 20The coefficient of x

^{2}= 6The coefficient of x = -7

Constant term = -20

So, we write the middle term -7x as -15x + 8x

6x

^{2}– 7x – 20 = 6x^{2}– 15x + 8x – 20= 3x (2x – 5) + 4 (2x – 5)

= (3x + 4) (2x – 5)

**Question 10. 3x**^{2} + 22x + 35

^{2}+ 22x + 35

**Solution:**

Given:3x

^{2}+ 22x + 35The coefficient of x

^{2}= 3The coefficient of x = 22

Constant term = 35

So, we write the middle term 22x as 15x + 7x

3x

^{2}+ 22x + 35 = 3x^{2}+ 15x + 7x + 35= 3x (x + 5) + 7 (x + 5)

= (3x + 7) (x+ 5)

### Chapter 7 Factorization – Exercise 7.8 | Set 2

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.