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

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

The document RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-7) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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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: 
a214a51=a2+3a17a51 
=(a2+3a)(17a+51) 
=a(a+3)17(a+3) 
=(a17)(a+3) 
Question 5: Factorize each of the following algebraic expression:
x2 + 14x + 45
Answer 5: To factorise x2+14x+45, we will find two numbers p and q such that p+q=14 and pq=45. 

Now, 9+5=14 and 9×5=45Splitting the middle term 14x in the given quadratic as 9x+5x, we get:x2+14x+45=x2+9x+5x+45 =(x2+9x)+(5x+45) 
=x(x+9)+5(x+9) 
=(x+5)(x+9) 
Question 6: Factorize each of the following algebraic expression:
x2 − 22x + 120
Answer 6: To factorise x222x+120, we will find two numbers p and q such that p+q=22 and pq=120. 

Now, (12)+(10)=22 and (12)×(10)=120Splitting the middle term 22x in the given quadratic as 12x10x, we get:x222x+12=x212x10x+120 =(x212x)+(10x+120) 
=x(x12)10(x12) 
=(x10)(x12) 
Question 7: Factorize each of the following algebraic expression:
x2 − 11x − 42
Answer 7: To factorise x211x42, we will find two numbers p and q such that p+q=11 and pq=42. 

Now, 
3+(14)=22  
and 3×(14)=42Splitting the middle term 11x in the given quadratic as14x+3x, we get: x211x42=x214x+3x42 =(x214x)+(3x42)=x(x14)+3(x14)=(x+3)(x14)Question 8: Factorize each of the following algebraic expression:
a2 + 2a − 3
Answer 8: To factorise a2+2a3, we will find two numbers p and q such that p+q=2 and pq=3. 

Now, 3+(1)=2 and  3×(1)=3Splitting the middle term 2a in the given quadratic asa+3a, we get:a2+2a3=a2a+3a3 =(a2a)+(3a3)=a(a1)+3(a1)=(a+3)(a1)Question 9: Factorize each of the following algebraic expression:
a2 + 14a + 48
Answer 9: To factorise a2+14a+48, we will find two numbers p and q such that p+q=14 and pq=48. 

Now, 8+6=14 and 8×6=48Splitting the middle term 14a in the given quadratic as 8a+6a, we get:a2+14a+48=a2+8a+6a+48 =(a2+8a)+(6a+48)=a(a+8)+6(a+8)=(a+6)(a+8)Question 10: Factorize each of the following algebraic expression:
x2 − 4x − 21
Answer 10: To factorise x24x21, we will find two numbers p and q such that p+q=4 and pq=21. 

Now,3+(7)=4 and 3×(7)=21Splitting the middle term 4x in the given quadratic as 7x+3x, we get:x24x21=x27x+3x21 =(x27x)+(3x21)=x(x7)+3(x7)=(x+3)(x7)Question 11: Factorize each of the following algebraic expression:
y2 + 5y − 36
Answer 11: To factorise y2+5y36, we will find two numbers p and q such that p+q=5 and pq=36. 

Now,9+(4)=5 and 9×(4)=36Splitting the middle term 5y in the given quadratic as 4y+9y, we get: y2+5y36=y24y+9y36 =(y24y)+(9y36)=y(y4)+9(y4)=(y+9)(y4)Question 12: Factorize each of the following algebraic expression:
(a2 − 5a)2 − 36
Answer 12: 
(a2−5a)2−36

=(a2−5a)2−62=[(a25a)6][(a25a)+6]=(a25a6)(a25a+6) In order to factorise a25a6, we will find two numbers p and q such that p+q=5 and pq=6 
Now, (6)+1=5 and (6)×1=6Splitting the middle term 5 in the given quadratic as 6a+a, we get:a25a6=a26a+a6                   =(a26a)+(a6)                   =a(a6)+(a6)                   =(a+1)(a6) Now, 
In order to factorise a25a+6, we will find two numbers p and q such that p+q=5 and pq=6 
Clearly,(2)+(3)=5 and (2)×(3)=6Splitting the middle term 5 in the given quadrat