Class 8 RD Sharma Solutions Chapter 7 Factorization Exercise 7.7 (original) (raw)
Last Updated : 5 Aug, 2024
**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]
Summary
Exercise 7.7 in Chapter 7 of RD Sharma's Class 8 Mathematics focuses on factorization of algebraic expressions. This exercise covers various techniques of factorization, including finding common factors, grouping terms, using identities like (a+b)² and (a-b)², and factoring quadratic expressions. Students learn to recognize patterns and apply appropriate methods to break down complex expressions into their simplest factored forms.