RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-7) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

Class 8 : RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-7) Class 8 Notes | EduRev

The document RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-7) Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8

Question 1: Factorize each of the following algebraic expression:
x2 + 12x − 45
Answer 1: To factorise x2+12x45, we will find two numbers p and q such that p+q=12 and pq=45. 

Now,

15+(3)=12  
and

15×(3)=45 
Splitting the middle term 12x in the given quadratic as 3x+15x, we get: 
x2+12x45=x23x+15x45 
=(x23x)+(15x45) 
=x(x3)+15(x3) 
=(x+15)(x−3)

Question 2: Factorize each of the following algebraic expression:
40 + 3xx2
Answer 2: We have: 

40+3xx2 
(x23x40)  
To factorise (x23x40), we will find two numbers p and q such that p+q=3 and pq=40. 
Now,   
5+(8)=3  
and 5×(8)=40 Splitting the middle term 3x in the given quadratic as 5x8x, we get: 
40+3x−x2=−(x2−3x−40)
=(x2+5x8x40) 
=[(x2+5x)(8x+40)] 
=[x(x+5)8(x+5)] 
=(x8)(x+5) 
=(x+5)(x+8) 
Question 3: Factorize each of the following algebraic expression:
a2 + 3a − 88
Answer 3: 
To factorise a2+3a88, we will find two numbers p and q such that p+q=3 and pq=88. 
Now,  11+(8)=3 and 11×(8)=88Splitting the middle term 3a in the given quadratic as 11a8a, we get:a2+3a88=a2+11a8a88 =(a2+11a)(8a+88) 
=a(a+11)8(a+11) 
=(a8)(a+11) 
Question 4: Factorize each of the following algebraic expression:
a2 − 14a − 51 

Answer 4: To factorise a214a51, we will find two numbers p and q such that p+q=14 and pq=51. 
Now,  
3+(17)=14  
and

3×(17)=51 
Splitting the middle term 14a in the given quadratic as 3a17a, we get: 
a2