In the previous exercises, we were discussing concepts based on monomial and binomial expressions. In this exercise, we will discuss polynomials and the factorization of quadratic polynomials in one variable. The primary objective is to help students improve their problem-solving abilities, which in turn, helps in boosting their confidence level to achieve high marks in the annual exam. Students can refer to and download RD Sharma Solutions Class 8 Maths Exercise 7.7 Chapter 7 Factorization from the links given below.
RD Sharma Solutions for Class 8 Maths Exercise 7.7 Chapter 7 Factorization
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 7.7 Chapter 7 Factorization
EXERCISE 7.7 PAGE NO: 7.27
Factorize each of the following algebraic expressions:
1. x2 + 12x – 45
Solution:
We have,
x2 + 12x – 45
To factorize the given expression, we have to find two numbers, p and q, such that p+q = 12 and pq = -45
So, we can replace 12x with 15x – 3x
-45 by 15 × 3
x2 + 12x – 45 = x2 + 15x – 3x – 45
= x (x + 15) – 3 (x + 15)
= (x – 3) (x + 15)
2. 40 + 3x – x2
Solution:
We have,
40 + 3x – x2
-(x2 – 3x – 40)
By considering, p+q = -3 and pq = -40
So, we can replace -3x with 5x – 8x
-40 by 5 × -8
-(x2 – 3x – 40) = x2 + 5x – 8x – 40
= -x (x + 5) – 8 (x + 5)
= -(x – 8) (x + 5)
= (-x + 8) (x + 5)
3. a2 + 3a – 88
Solution:
We have,
a2 + 3a – 88
By considering, p+q = 3 and pq = -88
So, we can replace 3a with 11a – 8a
-40 by -11 × 8
a2 + 3a – 88 = a2 + 11a – 8a – 88
= a (a + 11) – 8 (a + 11)
= (a – 8) (a + 11)
4. a2 – 14a – 51
Solution:
We have,
a2 – 14a – 51
By considering, p+q = -14 and pq = -51
So, we can replace -14a with 3a – 17a
-51 by -17 × 3
a2 – 14a – 51 = a2 + 3a – 17a – 51
= a (a + 3) – 17 (a + 3)
= (a – 17) (a + 3)
5. x2 + 14x + 45
Solution:
We have,
x2 + 14x + 45
By considering, p+q = 14 and pq = 45
So, we can replace 14x with 5x + 9x
45 by 5 × 9
x2 + 14x + 45 = x2 + 5x + 9x + 45
= x (x + 5) – 9 (x + 5)
= (x + 9) (x + 5)
6. x2 – 22x + 120
Solution:
We have,
x2 – 22x + 120
By considering, p+q = -22 and pq = 120
So, we can replace -22x with -12x -10x
120 by -12 × -10
x2 – 22x + 120 = x2 – 12x – 10x + 120
= x (x – 12) – 10 (x – 12)
= (x – 10) (x – 12)
7. x2 – 11x – 42
Solution:
We have,
x2 – 11x – 42
By considering, p+q = -11 and pq = -42
So, we can replace -11x with 3x -14x
-42 by 3 × -14
x2 – 11x – 42 = x2 + 3x – 14x – 42
= x (x + 3) – 14 (x + 3)
= (x – 14) (x + 3)
8. a2 + 2a – 3
Solution:
We have,
a2 + 2a – 3
By considering, p+q = 2 and pq = -3
So, we can replace 2a with 3a -a
-3 by 3 × -1
a2 + 2a – 3 = a2 + 3a – a – 3
= a (a + 3) – 1 (a + 3)
= (a – 1) (a + 3)
9. a2 + 14a + 48
Solution:
We have,
a2 + 14a + 48
By considering, p+q = 14 and pq = 48
So, we can replace 14a with 8a + 6a
48 by 8 × 6
a2 + 14a + 48 = a2 + 8a + 6a + 48
= a (a + 8) + 6 (a + 8)
= (a + 6) (a + 8)
10. x2 – 4x – 21
Solution:
We have,
x2 – 4x – 21
By considering, p+q = -4 and pq = -21
So, we can replace -4x with 3x – 7x
-21 by 3 × -7
x2 + 4x – 21 = x2 + 3x – 7x – 21
= x (x + 3) – 7 (x + 3)
= (x – 7) (x + 3)
11. y2 + 5y – 36
Solution:
We have,
y2 + 5y – 36
By considering, p+q = 5 and pq = -36
So, we can replace 5y with 9y – 4y
-36 by 9 × -4
y2 + 5y – 36 = y2 + 9y – 4y – 36
= y (y + 9) – 4 (y + 9)
= (y – 4) (y + 9)
12. (a2 – 5a)2 – 36
Solution:
We have,
(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)
So now we shall factorize the expression (a2 – 5a + 6)
By considering, p+q = -5 and pq = 6
So, we can replace -5a with a -6a
6 by 1 × -6
a2 -5a – 6 = a2 + a – 6a – 6
= a (a + 1) -6(a + 1)
= (a – 6) (a + 1)
So now we shall factorize the expression (a2 – 5a + 6)
By considering, p+q = -5 and pq = -6
So, we can replace -5a with -2a -3a
6 by -2 × -3
a2 -5a + 6 = a2 – 2a – 3a + 6
= a (a – 2) -3 (a – 2)
= (a – 3) (a – 2)
∴ (a2 – 5a)2 – 36 = (a2 – 5a + 6) (a2 – 5a – 6)
= (a + 1) (a – 6) (a – 2) (a – 3)
13. (a + 7) (a – 10) + 16
Solution:
We have,
(a + 7) (a – 10) + 16
a2 – 10a + 7a – 70 + 16
a2 – 3a – 54
By considering, p+q = -3 and pq = -54
So, we can replace -3a with 6a – 9a
-54 by 6 × -9
a2 – 3a – 54 = a2 + 6a – 9a – 54
= a (a + 6) -9 (a + 6)
= (a – 9) (a + 6)
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