 # Chapter 7: Factorization Exercise – 7.8

### Question: 1

Resolve of the following quadratic equation trinomials into factors:

2x2 + 5x + 3

### Solution:

The given expression is 2x2 + 5x + 3.

(Co-efficient of x2 = 2, co-efficient of x = 5 and the constant term = 3)

We will split the co-efficient of x into two parts such that their sum is 5 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 2 × 3 = 6.

Now,

2 + 3 = 5 And 2 × 3 = 6

Replacing the middle term 5x by 2x + 3x, we have:

2x2 + 5x + 3 = 2x2 + 2x + 3x + 3

= (2x2 + 2x) + (3x + 3)

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

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

### Question: 2

Resolve of the following quadratic equation trinomials into factors:

2x2 - 3x - 2

### Solution:

The given expression is 2x2 – 3x – 2.

(Co-efficient of x2 = 2, co-efficient of x = -3 and the constant term = -2)

We will split the co-efficient of x into two parts such that their sum is -3 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 2 × (-2) = - 4

Now,

(- 4) + 1 = - 3 And (- 4) × 1 = – 4

Replacing the middle term 3x by – 4x + x, we have:

2x2 - 3x - 2 = 2x2 - 4x + x -2

= (2x2 – 4x) + (x - 2)

= 2x(x - 2) + 1(x – 2)

= (x – 2)(2x + 1)

### Question: 3

Resolve of the following quadratic equation trinomials into factors:

3x2 + 10x + 3

### Solution:

The given expression is 3x2 + 10x + 3.

(Co-efficient of x2 = 3, co-efficient of x = 10 and the constant term = 3)

We will split the co-efficient of x into two parts such that their sum is 10 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 3 × 3 = 9

Now,

9 + 1 = 10 And 9 × 1 = 9

Replacing the middle term 10x by 9x + x, we have:

3x2 + 10x + 3 = 3x2 + 9x + x + 3

= (3x2 + 9x) + (x + 3)

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

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

### Question: 4

Resolve of the following quadratic equation trinomials into factors:

7x – 6 – 2x2

### Solution:

The given expression is 7x – 6 – 2x2.

(Co-efficient of x2 = -2, co-efficient of x = 7 and the constant term = – 6)

We will split the co-efficient of x into two parts such that their sum is 7 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., (-2) × (- 6) = 12

Now,

4 + 3 = 7 And 4 × 3 = 12

Replacing the middle term 7x by 4x + 3x, we have:

7x - 6 - 2x2 = -2x2 + 4x + 3x - 6

= (-2x2 + 4x) + (3x – 6)

= 2x (2 - x) – 3(2 - x)

= (2x – 3) (2 – x)

### Question: 5

Resolve of the following quadratic equation trinomials into factors:

7x2 – 19x – 6

### Solution:

The given expression is 7x2 – 19x – 6.

(Co-efficient of x2 = 7, co-efficient of x = -19 and the constant term = – 6)

We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 7 × (– 6) = 9

Now,

(-21) + 2 = -19 And (-21) × 2 = - 42

Replacing the middle term -19x by -21x + 2x, we have:

7x2 – 19x – 6 = 7x2 – 21x + 2x -6

= (7x2 – 21x) + (2x – 6)

= 7x(x – 3) + 2(x – 3)

= (x – 3) (7x + 2)

### Question: 6

Resolve of the following quadratic equation trinomials into factors:

28 – 31x – 5x2

### Solution:

The given expression is 28 – 31x – 5x2.

(Co-efficient of x2 = -5, co-efficient of x = -31 and the constant term = 28)

We will split the co-efficient of x into two parts such that their sum is -31 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., ( -5 ) × ( 28 ) = -140

Now,

(-35) + 4 = -31 And (-35) × 4 = -140

Replacing the middle term -31x by -35x + 4x, we have:

28 – 31x – 5x2 = -5x2 – 35x + 4x + 28

= (-5x2 – 35x) + (4x + 28)

= -5x(x + 7) + 4(x + 7)

= (4 - 5x)(x + 7)

### Question: 7

Resolve of the following quadratic equation trinomials into factors:

3 + 23y – 8y2

### Solution:

The given expression is 3 + 23y – 8y2.

(Co-efficient of y2 = -8, co-efficient of y = 23 and the constant term = 3)

We will split the co-efficient of x into two parts such that their sum is 23 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., ( -8 ) ×  3 = -24

Now,

(-1) + 24 = 23 And (-1) × 24 = -24

Replacing the middle term 23y by -y + 24y, we have:

3 + 23y – 8y2 = -8y2 – y + 24y + 3

= (-8y2 – y) + (24y + 3)

= – y(8y + 1) + 3(8y + 1)

= (8y + 1)( y + z)

### Question: 8

Resolve of the following quadratic equation trinomials into factors:

11x2 – 54x + 63

### Solution:

The given expression is 11x2 – 54x + 63.

(Co-efficient of x2 = 11, co-efficient of x = – 54 and the constant term = 63)

We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 11 × 63 = 693

Now,

(-33) + (-21) =- 54 And (-33) × (-21) = 693

Replacing the middle term -54x by -33x – 21x, we have:

11x2 – 54x + 63 = 11x2 – 33x – 21x + 63

= (11x2 – 33x) + (-21x + 63)

= 11x(x – 3) - 21(x – 3)

= (x - 3)( 11x - 21)

### Question: 9

Resolve of the following quadratic equation trinomials into factors:

7x – 6x2 + 20

### Solution:

The given expression is 7x – 6x2 + 20.

(Co-efficient of x2 = -6, co-efficient of x = 7 and the constant term = 20)

We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., (– 6) × 20 = – 120

Now,

(15) + (- 8) = 7 And (15) × (-8) = -120

Replacing the middle term 7x by 15x - 8x, we have:

7x – 6x2 + 20 = – 6x2 + 15x - 8x + 20

= (-6x2 + 15x) + (- 8x + 20)

= 3x (-2x + 5) + 4(- 2x + 5)

= (-2x + 5)(3x + 4)

### Question: 10

Resolve of the following quadratic equation trinomials into factors:

3x2 + 22x + 35

### Solution:

The given expression is 3x2 + 22x + 35.

(Co-efficient of x2 = 3, co-efficient of x = 22 and the constant term = 35)

We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 3 × 35 = 105

Now,

(15) + (7) = 22 And (15) × (7) = 105

Replacing the middle term 22x by 15x + 7x, we have:

3x2 + 22x + 35 = 3x2 + 15x + 7x + 35

= (3x2 + 15x) + (7x + 35)

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

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

### Question: 11

Resolve of the following quadratic equation trinomials into factors:

12x2 – 17xy + 6y2

### Solution:

The given expression is 12x2 – 17xy + 6y2.

(Co-efficient of x2 = 12, co-efficient of x = -17y and the constant term = 6y2)

We will split the co-efficient of x into two parts such that their sum is -17y and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 12 × 6y2 = 72y2

Now,

(-9y) + (-8y) = -17y And (-9y) × (-8y) = 72y2

Replacing the middle term -17xy by -9xy – 8xy, we have:

12x2 – 17xy + 6y2 = 12x2 – 9xy – 8xy + 6y2

= (12x2 – 9xy) – (8xy + 6y2)

= 3x(4x – 3y) -2y(4x – 3y)

= (4x – 3y)( 3x – 2y)

### Question: 12

Resolve of the following quadratic equation trinomials into factors:

6x2 – 5xy – 6y2

### Solution:

The given expression is 6x2 – 5xy – 6y2.

(Co-efficient of x2 = 6, co-efficient of x = -5y  and the constant term = – 6y2)

We will split the co-efficient of x into two parts such that their sum is -17y and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 6 × (– 6y2) = –36y2

Now,

(-9y) + (4y) = - 5y And (-9y) × (4y) = -36y2

Replacing the middle term -5xy by -9xy + 4xy, we have:

6x2 – 5xy – 6y2 = 6x2 – 9xy + 4xy – 6y2

= (6x2 – 9xy) + (4xy – 6y2)

= 3x(2x – 3y) + 2y(2x – 3y)

= (2x – 3y)(3x + 2y)

### Question: 13

Resolve of the following quadratic equation trinomials into factors:

6x2 – 13xy + 2y2

### Solution:

The given expression is 6x2 – 13xy + 2y2.

(Co-efficient of x2 = 6, co-efficient of x = -13y and the constant term = 2y2)

We will split the co-efficient of x into two parts such that their sum is -13y and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 6 × ( 2y2 ) = 12y2

Now,

(-12y) + (-y) = -13y And (-12y) × (-y) = 12y2

Replacing the middle term -13xy by -12xy – xy, we have:

6x2 – 13xy + 2y2 = 6x2 – 12xy – xy + 2y2

= (6x2 – 12xy) – (xy – 2y2)

= 6x(x – 2y) – y(x – 2y)

= (x – 2y)(6x – y)

### Question: 14

Resolve of the following quadratic equation trinomials into factors:

14x2 + 11xy – 15y2

### Solution:

The given expression is 14x2 + 11xy – 15y2.

(Co-efficient of x2 = 14, co-efficient of x = 11y and the constant term = -15y2)

We will split the co-efficient of x into two parts such that their sum is 11y and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 14 × (-15y2) = -210y2

Now,

(21y) + (-10y) = 11y And (21y) × (-10y) = -210y2

Replacing the middle term -11xy by -10xy + 21 xy, we have:

14x2 + 11xy – 15y2 = 14x2 – 10xy + 21xy – 15y2

= (14x2 – 10xy) + (21xy – 15y2)

= 2x(7x – 5y) + 3y(7x – 5y)

= (7x – 5y)(2x + 3y)

### Question: 15

Resolve of the following quadratic equation trinomials into factors:

6a2 + 17ab – 3b2

### Solution:

The given expression is 6a2 + 17ab – 3b2.

(Co-efficient of a2 = 6, co-efficient of a = 17b and the constant term = – 3b2)

We will split the co-efficient of x into two parts such that their sum is 17b and their product equals to the product of the co-efficient of a2 and the constant term, i.e., 6 × (-3b2) = -18b2

Now,

(18b) + (-b) = 17b And (18b) × (-b) = -18b2

Replacing the middle term 17ab by -ab + 18ab, we have:

6a2 + 17ab – 3b2 = 6a2 -ab + 18ab - 3b2

= (6a2 – ab) + (18ab – 3b2)

= a(6a – b) + 3b(6a – b)

= (a + 3b)(6a – b)

### Question: 16

Resolve of the following quadratic equation trinomials into factors:

36a2 + 12abc – 15b2c2

### Solution:

The given expression is 36a2 + 12abc – 15b2c2.

(Co-efficient of a2 = 36, co-efficient of a = 12bc and the constant term = -15b2 c2)

We will split the co-efficient of x into two parts such that their sum is 17b and their product equals to the product of the co-efficient of a2 and the constant term, i.e., 36 × ( -15b2 c2) = -540b2c2

Now,

(-18bc) + 30bc = 12bc And (-18bc) × (30bc) = - 540b2c2

Replacing the middle term 12abc by -18abc + 30abc, we have:

36a2 + 12abc – 15b2c2 = 36a2 -18abc + 30abc – 15b2c2

= (36a2 -18abc) + (30abc – 15b2c2)

= 18a(2a – bc) + 15bc( 2a – bc)

= 3(6a + 5bc)(2a – bc)

### Question: 17

Resolve of the following quadratic equation trinomials into factors:

15x2 – 16xyz – 15y2z2

### Solution:

The given expression is 15x2 – 16xyz – 15y2z2.

(Co-efficient of x2 = 15, co-efficient of x = -16yz and the constant term = -15y2 z2)

We will split the co-efficient of x into two parts such that their sum is -16yz and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 15 × ( -15y2 z2) = -225y2z2

Now,

(-25yz) + 9yz = -16yx And (-25yz) × (9yz) = -225y2 z2

Replacing the middle term -16xyz by -25xyz + 9xyz, we have:

15x2 - 16xyz – 15y2z2 = 15x2 – 25xyz + 9xyz – 15y2z2

= (15x2 - 25xyz) + (9xyz – 15y2z2)

= 5x(3x – 5yz) + 3yz(3x – 5yz)

= (3x – 5yz)(5x + 3yz)

### Question: 18

Resolve of the following quadratic equation trinomials into factors:

(x – 2y)2 – 5(x – 2y) + 6

### Solution:

The given expression is a2 – 5a + 6.

Assuming a = x – 2y, we have:

(x – 2y)2 – 5( x – 2y) + 6 = a2 – 5a + 6

(Co-efficient of a2 = 1, co-efficient of a = -5 and the constant term = 6)

Now, we will split the co-efficient of  a  into two parts such that their sum is -5 and their product equals to the product of the co-efficient of a2 and the constant term, i.e., 1 × 6 = 6.

Clearly,

(-2) + (-3) = – 5 And, (-2) × (-3) = 6

Replacing the middle term -5a by -2a – 3a, we have:

a2 – 5a + 6 = a2 – 2a – 3a + 6

= (a2 – 2a) – (3a – 6)

= a( a – 2) – 3 (a – 2)

= (a – 2)(a – 3)

Replacing a by (x – 2y), we get:

(a – 3)(a – 2) = (x – 2y – 3)(x -2y – 2)

### Question: 19

Resolve of the following quadratic equation trinomials into factors:

(2a – b )2 + 2(2a – b) – 8

### Solution:

Assuming x = 2a – b, we have:

(2a – b )2 + 2(2a – b) – 8 = x2 + 2x – 8

The given expression becomes x2 + 2x – 8

(Co-efficient of x2 = 1 and that of x = 2; constant term = -8)

Now, we will split the co-efficient of x into two parts such that their sum is 2 and their product equals the product of the co-efficient of x2 and the constant term, i.e., 1 × (- 8) = – 8

Clearly,

(-2) + 4 = 2 And, (-2) × 4 = -8

Replacing the middle term 2x by -2x + 4x, we get:

x2 + 2x – 8 = x2 – 2x + 4x – 8

= (x2 – 2x) + (4x – 8)

= x(x – 2) + 4 (x – 2)

= (x – 2)(x + 4)

Replacing x by 2a – b, we get:

(x + 4)(x – 2) = (2a – b + 4)(2a – b – 2)

### Course Features

• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution