Class 8 RD Sharma Solutions – Chapter 7 Factorization – Exercise 7.7

Question 1: Solve for factors of x² + 12x – 45

Solution:

Given: x² + 12x – 45

To factorize the following expression we have to find two numbers a and b such that a + b = 12 and ab = – 45

We know that : 

15 + (– 3) = 12 and 15 * (-3) = – 45

Splitting the middle term i.e. 12x in the given quadratic equation we get:           

x² + 12x – 45 = x² + 15x – 3x – 45

= x (x + 15) – 3 (x + 15)        [Taking the similar terms common]

= (x – 3) (x + 15)        [as (x + 15) is common]

Question 2: Solve for factors of 40 + 3x – x²

Solution:

Given: 40 + 3x – x²

40 + 3x – x² = -(x² – 3x – 40)

To factorize the following expression we have to find two numbers a and b such that a+b = -3 and ab = – 40

We know that :

5 + (-8) = -3 and 5 * (-8) = – 40

Splitting the middle term i.e. -3x in the given quadratic equation we get:          

-(x² – 3x – 40) = -(x² + 5x – 8x – 40)

= -(x (x + 5) – 8 (x + 5))       [Taking the similar terms common]

= -(x – 8) (x + 5)        [as (x + 5) is common]

= (-x + 8) (x + 5)

Question 3: Solve for factors of a² + 3a – 88

Solution:

Given: a² + 3a – 88         

To factorize the following expression we have to find two numbers a and b such that a+b = 3  and ab = -88

We know that :

11 + (-3) = 3 and 11 * (-8) = -88

Splitting the middle term i.e. 3a in the given quadratic equation we get:          

a² + 3a – 88 = a² + 11a – 8a – 88

= a (a + 11) – 8 (a + 11)        [Taking the similar terms common]

= (a – 8) (a + 11)        [as (a + 11) is common]

Question 4: Solve for factors of a² – 14a – 51

Solution:

Given: a² – 14a – 51          

To factorize the following expression we have to find two numbers a and b such that a+b = -14  and ab = -51

We know that :

3 + (-17) = -14 and 3 * (-17) = -51

Splitting the middle term i.e. -14a in the given quadratic equation we get:          

a² – 14a – 51 = a² + 3a – 17a – 51

= a (a + 3) – 17 (a + 3)          [Taking the similar terms common]

= (a – 17) (a + 3)          [as (a + 3) is common]

Question 5: Solve for factors of x² + 14x + 45

Solution:

Given: x² + 14x + 45          

To factorize the following expression we have to find two numbers a and b such that a+b = 14  and ab = 45

We know that :

5 + 9 = 14 and 5 * 9 = 45

Splitting the middle term i.e. 14x in the given quadratic equation we get:  

x² + 14x + 45 = x² + 5x + 9x + 45       

= x (x + 5) – 9 (x + 5)         [Taking the similar terms common]

= (x + 9) (x + 5)         [as (x+5) is common]

Question 6: Solve for factors of x² – 22x + 120

Solution:

Given: x² – 22x + 120         

To factorize the following expression we have to find two numbers a and b such that a+b = -22  and ab = 120

We know that :

-12 + (-10) = -22 and -12 * (-10) = 120

Splitting the middle term i.e. -22x in the given quadratic equation we get:          

x² – 22x + 120 = x² – 12x – 10x + 120

= x (x – 12) – 10 (x – 12)          [Taking the similar terms common]

= (x – 10) (x – 12)          [as (x – 12) is common]

Question 7: Solve for factors of x² – 11x – 42

Solution:

Given:  x² – 11x – 42        

To factorize the following expression we have to find two numbers a and b such that a+b = -11   and ab = -42

We know that :

3 + (-14) = -11 and 3 * (-14) = -42

Splitting the middle term i.e. -11x in the given quadratic equation we get:          

x² – 11x – 42 = x² + 3x – 14x – 42

= x (x + 3) – 14 (x + 3)          [Taking the similar terms common]

= (x – 14) (x + 3)          [as (x + 3) is common]

Question 8: Solve for factors of a² + 2a – 3

Solution:

Given: a² + 2a – 3         

To factorize the following expression we have to find two numbers a and b such that a+b = 2  and ab = -3

We know that :

3 + (-1) = 2 and 3 * (-1) = -3

Splitting the middle term i.e. 2a in the given quadratic equation we get: 

a² + 2a – 3 = a² + 3a – a – 3         

= a (a + 3) – 1 (a + 3)          [Taking the similar terms common]

= (a – 1) (a + 3)          [as (a + 3) is common]

Question 9: Solve for factors of a² + 14a + 48

Solution:

Given: a² + 14a + 48    

To factorize the following expression we have to find two numbers a and b such that a+b = 14  and ab = 48

We know that :

8 + 6 = 14 and 8 * 6 = 48

Splitting the middle term i.e. 14a in the given quadratic equation we get:          

a² + 14a + 48 = a² + 8a + 6a + 48

= a (a + 8) + 6 (a + 8)          [Taking the similar terms common]

= (a + 6) (a + 8)          [as (a + 8) is common]

Question 10: Solve for factors of x² – 4x – 21

Solution:

Given: x² – 4x – 21         

To factorize the following expression we have to find two numbers a and b such that a+b = -4  and ab = -21

We know that :

3 + (-7) = -4 and 3 * (-7) = -21

Splitting the middle term i.e. -4x in the given quadratic equation we get:          

x² + 4x – 21 = x² + 3x – 7x – 21

= x (x + 3) – 7 (x + 3)          [Taking the similar terms common]

= (x – 7) (x + 3)          [as (x + 3) is common]

Question 11: Solve for factors of y² + 5y – 36

Solution:

Given: y² + 5y – 36         

To factorize the following expression we have to find two numbers a and b such that a+b = 5 and ab = -36

We know that :

9 + (-4) = 5 and 9 *(-4) = -36

Splitting the middle term i.e. 5y in the given quadratic equation we get:    

y² + 5y – 36 = y² + 9y – 4y – 36      

= y (y + 9) – 4 (y + 9)          [Taking the similar terms common]

= (y – 4) (y + 9)          [as (y + 9) is common]

Question 12: Solve for factors of (a² – 5a)² – 36

Solution:

Given: (a² – 5a)² – 36

(a² – 5a)² – 36 = (a² – 5a)² – 6²

By using the formula (a² – b²) = (a+b) (a-b)

(a² – 5a)² – 6² = (a² – 5a + 6) (a² – 5a – 6)        

Now solving the second part i.e a² – 5a + 6

To factorize the following expression we have to find two numbers a and b such that a+b = -5  and ab = 6

We know that :

-2 + (-3) = -5 and -2 * (-3) = 6

Splitting the middle term i.e. -5x in the given quadratic equation we get:          

a² -5a + 6 = a² – 2a – 3a + 6

= a (a – 2) -3 (a – 2)          [Taking the similar terms common]

= (a – 3) (a – 2)          [as (a – 2) is common]

Now solving the first part i.e a² – 5a – 6 

To factorize the following expression we have to find two numbers a and b such that a+b = -5  and ab = -6

We know that :

1 + (-6) = -5 and 1 * (-6) = -6

Splitting the middle term i.e. -5x in the given quadratic equation we get:   

a² -5a – 6 = a² + a – 6a – 6

= a (a + 1) -6(a + 1)          [Taking the similar terms common]

= (a – 6) (a + 1)          [as (a + 1) is common]

Since, (a2 – 5a)² – 36 = (a² – 5a + 6) (a² – 5a – 6)

Substituting the values of (a² – 5a + 6 ) and (a² – 5a – 6 ) we get:

= (a – 2) (a – 3) (a + 1) (a – 6) 

Question 13: Solve for factors of (a + 7) (a – 10) + 16

Solution:

Given: (a + 7) (a – 10) + 16         

(a + 7) (a – 10) + 16 = a² – 10a + 7a – 70 + 16

(a + 7) (a – 10) + 16 = a² – 3a – 54

To factorize the following expression we have to find two numbers a and b such that a+b = -3  and ab = -54

We know that :

6 + (-9) = -3 and 6 * (-9) = -54

Splitting the middle term i.e. -3x in the given quadratic equation we get:          

a² – 3a – 54 = a² + 6a – 9a – 54

= a (a + 6) -9 (a + 6)          [Taking the similar terms common]

= (a – 9) (a + 6)          [as (a + 6) is common]