Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  RD Sharma MCQs: Factorization of Algebraic Expressions

RD Sharma MCQs: Factorization of Algebraic Expressions | Mathematics (Maths) Class 9 PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


        
      
          
      
    
Question:97
The factors of x
3
 -x
2
y - xy
2 
+ y
3
 are
a
(x + y) (x
2
 - xy + y
2
)
b
(x + y) (x
2
 + xy + y
2
)
c
(x + y)
2
 (x - y)
d
x -y
2
 x +y
Solution:
The given expression to be factorized is
Take common  from the first two terms and  from the last two terms. That is
Finally, take common from the two terms. That is
So, the correct choice is d
.
Question:98
The factors of x
3
 - 1 + y
3
 + 3xy are
Page 2


        
      
          
      
    
Question:97
The factors of x
3
 -x
2
y - xy
2 
+ y
3
 are
a
(x + y) (x
2
 - xy + y
2
)
b
(x + y) (x
2
 + xy + y
2
)
c
(x + y)
2
 (x - y)
d
x -y
2
 x +y
Solution:
The given expression to be factorized is
Take common  from the first two terms and  from the last two terms. That is
Finally, take common from the two terms. That is
So, the correct choice is d
.
Question:98
The factors of x
3
 - 1 + y
3
 + 3xy are
a
(x - 1 + y) (x
2
 + 1 + y
2
 + x + y - xy)
b
(x + y + 1) (x
2
 + y
2
 + 1 -xy - x - y)
c
(x - 1 + y) (x
2
 - 1 - y
2
 + x + y + xy)
d
3(x + y -1) (x
2
 + y
2
 - 1)
Solution:
The given expression to be factorized is 
This can be written in the form
Recall the formula 
Using the above formula, we have 
So, the correct choice is a
.
Question:99
The factors of 8a
3
 + b
3
 - 6ab + 1 are
a
(2a + b - 1) (4a
2
 + b
2
 + 1 - 3ab - 2a)
b
(2a - b + 1) (4a
2
 + b
2
 - 4ab + 1 - 2a + b)
c
(2a + b + 1) (4a
2
 + b
2
 + 1 -2ab - b - 2a)
d
(2a - 1 + b) (4a
2
 + 1 - 4a - b - 2ab)
Solution:
The given expression to be factorized is 
This can be written in the form
Recall the formula 
Using the above formula, we have 
So, the correct choice is c
.
Page 3


        
      
          
      
    
Question:97
The factors of x
3
 -x
2
y - xy
2 
+ y
3
 are
a
(x + y) (x
2
 - xy + y
2
)
b
(x + y) (x
2
 + xy + y
2
)
c
(x + y)
2
 (x - y)
d
x -y
2
 x +y
Solution:
The given expression to be factorized is
Take common  from the first two terms and  from the last two terms. That is
Finally, take common from the two terms. That is
So, the correct choice is d
.
Question:98
The factors of x
3
 - 1 + y
3
 + 3xy are
a
(x - 1 + y) (x
2
 + 1 + y
2
 + x + y - xy)
b
(x + y + 1) (x
2
 + y
2
 + 1 -xy - x - y)
c
(x - 1 + y) (x
2
 - 1 - y
2
 + x + y + xy)
d
3(x + y -1) (x
2
 + y
2
 - 1)
Solution:
The given expression to be factorized is 
This can be written in the form
Recall the formula 
Using the above formula, we have 
So, the correct choice is a
.
Question:99
The factors of 8a
3
 + b
3
 - 6ab + 1 are
a
(2a + b - 1) (4a
2
 + b
2
 + 1 - 3ab - 2a)
b
(2a - b + 1) (4a
2
 + b
2
 - 4ab + 1 - 2a + b)
c
(2a + b + 1) (4a
2
 + b
2
 + 1 -2ab - b - 2a)
d
(2a - 1 + b) (4a
2
 + 1 - 4a - b - 2ab)
Solution:
The given expression to be factorized is 
This can be written in the form
Recall the formula 
Using the above formula, we have 
So, the correct choice is c
.
Question:100
x +y
3
 - x -y
3
 can be factorized as
a
2y (3x
2
 + y
2
)
b
2x (3x
2
 + y
2
)
c
2y (3y
2
 + x
2
)
d
2x (x
2
+ 3y
2
) 
Solution:
The given expression to be factorized is
Recall the formula for difference of two cubes 
Using the above formula, we have,
So, the correct choice is a
.
Question:101
The expression (a - b)
3
 + (b - c)
3
 + (c -a)
3
 can be factorized as
a
(a - b) (b - c) (c -a)
b
3(a - b) (b - c) (c -a)
c
-3(a - b) (b -c) (c - a)
d
(a + b + c) (a
2
 + b
2
 + c
2
 - ab - bc - ca)
Solution:
The given expression is
Let , and . Then the given expression becomes
Note that:
Page 4


        
      
          
      
    
Question:97
The factors of x
3
 -x
2
y - xy
2 
+ y
3
 are
a
(x + y) (x
2
 - xy + y
2
)
b
(x + y) (x
2
 + xy + y
2
)
c
(x + y)
2
 (x - y)
d
x -y
2
 x +y
Solution:
The given expression to be factorized is
Take common  from the first two terms and  from the last two terms. That is
Finally, take common from the two terms. That is
So, the correct choice is d
.
Question:98
The factors of x
3
 - 1 + y
3
 + 3xy are
a
(x - 1 + y) (x
2
 + 1 + y
2
 + x + y - xy)
b
(x + y + 1) (x
2
 + y
2
 + 1 -xy - x - y)
c
(x - 1 + y) (x
2
 - 1 - y
2
 + x + y + xy)
d
3(x + y -1) (x
2
 + y
2
 - 1)
Solution:
The given expression to be factorized is 
This can be written in the form
Recall the formula 
Using the above formula, we have 
So, the correct choice is a
.
Question:99
The factors of 8a
3
 + b
3
 - 6ab + 1 are
a
(2a + b - 1) (4a
2
 + b
2
 + 1 - 3ab - 2a)
b
(2a - b + 1) (4a
2
 + b
2
 - 4ab + 1 - 2a + b)
c
(2a + b + 1) (4a
2
 + b
2
 + 1 -2ab - b - 2a)
d
(2a - 1 + b) (4a
2
 + 1 - 4a - b - 2ab)
Solution:
The given expression to be factorized is 
This can be written in the form
Recall the formula 
Using the above formula, we have 
So, the correct choice is c
.
Question:100
x +y
3
 - x -y
3
 can be factorized as
a
2y (3x
2
 + y
2
)
b
2x (3x
2
 + y
2
)
c
2y (3y
2
 + x
2
)
d
2x (x
2
+ 3y
2
) 
Solution:
The given expression to be factorized is
Recall the formula for difference of two cubes 
Using the above formula, we have,
So, the correct choice is a
.
Question:101
The expression (a - b)
3
 + (b - c)
3
 + (c -a)
3
 can be factorized as
a
(a - b) (b - c) (c -a)
b
3(a - b) (b - c) (c -a)
c
-3(a - b) (b -c) (c - a)
d
(a + b + c) (a
2
 + b
2
 + c
2
 - ab - bc - ca)
Solution:
The given expression is
Let , and . Then the given expression becomes
Note that:
Recall the formula
When , this becomes
So, we have the new formula
, when .
Using the above formula, the value of the given expression is
So, the correct choice is b
.
Question:102
The value of 
2.3)
3
-0.027
2.3)
2
+0.69+0.09
a
2
b
3
c
2.327
d
2.273
Solution:
The given expression is
This can be written in the form
Assume and . Then the given expression can be rewritten as
Recall the formula for difference of two cubes
Using the above formula, the expression becomes
(
(
Page 5


        
      
          
      
    
Question:97
The factors of x
3
 -x
2
y - xy
2 
+ y
3
 are
a
(x + y) (x
2
 - xy + y
2
)
b
(x + y) (x
2
 + xy + y
2
)
c
(x + y)
2
 (x - y)
d
x -y
2
 x +y
Solution:
The given expression to be factorized is
Take common  from the first two terms and  from the last two terms. That is
Finally, take common from the two terms. That is
So, the correct choice is d
.
Question:98
The factors of x
3
 - 1 + y
3
 + 3xy are
a
(x - 1 + y) (x
2
 + 1 + y
2
 + x + y - xy)
b
(x + y + 1) (x
2
 + y
2
 + 1 -xy - x - y)
c
(x - 1 + y) (x
2
 - 1 - y
2
 + x + y + xy)
d
3(x + y -1) (x
2
 + y
2
 - 1)
Solution:
The given expression to be factorized is 
This can be written in the form
Recall the formula 
Using the above formula, we have 
So, the correct choice is a
.
Question:99
The factors of 8a
3
 + b
3
 - 6ab + 1 are
a
(2a + b - 1) (4a
2
 + b
2
 + 1 - 3ab - 2a)
b
(2a - b + 1) (4a
2
 + b
2
 - 4ab + 1 - 2a + b)
c
(2a + b + 1) (4a
2
 + b
2
 + 1 -2ab - b - 2a)
d
(2a - 1 + b) (4a
2
 + 1 - 4a - b - 2ab)
Solution:
The given expression to be factorized is 
This can be written in the form
Recall the formula 
Using the above formula, we have 
So, the correct choice is c
.
Question:100
x +y
3
 - x -y
3
 can be factorized as
a
2y (3x
2
 + y
2
)
b
2x (3x
2
 + y
2
)
c
2y (3y
2
 + x
2
)
d
2x (x
2
+ 3y
2
) 
Solution:
The given expression to be factorized is
Recall the formula for difference of two cubes 
Using the above formula, we have,
So, the correct choice is a
.
Question:101
The expression (a - b)
3
 + (b - c)
3
 + (c -a)
3
 can be factorized as
a
(a - b) (b - c) (c -a)
b
3(a - b) (b - c) (c -a)
c
-3(a - b) (b -c) (c - a)
d
(a + b + c) (a
2
 + b
2
 + c
2
 - ab - bc - ca)
Solution:
The given expression is
Let , and . Then the given expression becomes
Note that:
Recall the formula
When , this becomes
So, we have the new formula
, when .
Using the above formula, the value of the given expression is
So, the correct choice is b
.
Question:102
The value of 
2.3)
3
-0.027
2.3)
2
+0.69+0.09
a
2
b
3
c
2.327
d
2.273
Solution:
The given expression is
This can be written in the form
Assume and . Then the given expression can be rewritten as
Recall the formula for difference of two cubes
Using the above formula, the expression becomes
(
(
Note that both a and b are positive, unequal. So, neither nor any factor of it can be zero.
Therefore we can cancel the term from both numerator and denominator. Then the expression becomes
So, the correct choice is a
.
Question:103
The value of 
0.013)
3
+ 0.007)
3
0.013)
2
-0.013×0.007+ 0.007)
2
is
a
0.006
b
0.02
c
0.0091
d
0.00185
Solution:
The given expression is
Assume and . Then the given expression can be rewritten as
Recall the formula for sum of two cubes
Using the above formula, the expression becomes
Note that both  and b are positive. So, neither nor any factor of it can be zero.
Therefore we can cancel the term from both numerator and denominator. Then the expression becomes
So, the correct choice is b
.
( (
( (
Read More
44 videos|412 docs|54 tests

Top Courses for Class 9

44 videos|412 docs|54 tests
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

ppt

,

Previous Year Questions with Solutions

,

Summary

,

shortcuts and tricks

,

study material

,

video lectures

,

Exam

,

Free

,

Viva Questions

,

mock tests for examination

,

past year papers

,

practice quizzes

,

MCQs

,

Important questions

,

Extra Questions

,

RD Sharma MCQs: Factorization of Algebraic Expressions | Mathematics (Maths) Class 9

,

Objective type Questions

,

RD Sharma MCQs: Factorization of Algebraic Expressions | Mathematics (Maths) Class 9

,

RD Sharma MCQs: Factorization of Algebraic Expressions | Mathematics (Maths) Class 9

,

Semester Notes

,

pdf

,

Sample Paper

;