NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths Class 9 Notes | EduRev

Created by: Indu Gupta

Class 9 : NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths Class 9 Notes | EduRev

The document NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths Class 9 Notes | EduRev is a part of Class 9 category.
All you need of Class 9 at this link: Class 9

Exercise 2.5

 1. Use suitable identities to find the following products:

(i) (x + 4) (x + 10)                     

(ii) (x + 8) (x - 10)                      

(iii) (3x + 4) (3x - 5)

(iv) (y+ 3/2) (y- 3/2)             

(v) (3 - 2x) (3 + 2x)

Answer

(i) Using identity, (+ a) (x + b) = x2 + (a + b) x + ab 
In (x + 4) (x + 10), a = 4 and b = 10
Now,
(x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10)
x2 + 14x + 40

(ii) (x + 8) (x - 10)
Using identity, (+ a) (x + b) = x2 + (a + b) x + ab
Here, a = 8 and b = -10
(x + 8) (x - 10) = x2 + {8 +(- 10)}x + {8×(- 10)}
x2 + (8 - 10)x - 80
x2 - 2x - 80

(iii) (3x + 4) (3x - 5)
Using identity, (+ a) (x + b) = x2 + (a + b) x + ab
Here, x = 3x , a = 4 and b = -5
(3x + 4) (3x - 5) = (3x2 + {4 + (-5)}3x + {4×(-5)}
= 9x2 + 3x(4 - 5) - 20
= 9x2 - 3x - 20

(iv) (y+ 3/2) (y- 3/2)
Using identity, (+ y) (x -y) = x2 - y2
Here, x = y2 and y = 3/2
(y+ 3/2) (y- 3/2) = (y2)- (3/2)2
 y4 - 9/4

(v) (3 - 2x) (3 + 2x)
Using identity, (+ y) (x -y) = x2 - y2
Here, x = 3 and y = 2x
(3 - 2x) (3 + 2x) = 32 - (2x)2
=  9 - 4x2
 

2. Evaluate the following products without multiplying directly:
 (i) 103 × 107

(ii) 95 × 96               

(iii) 104 × 96

Answer

(i) 103 × 107 = (100 + 3) (100 + 7)
Using identity, (+ a) (x + b) = x2 + (a + b) x + ab

Here, x = 100, a = 3 and b = 7
103 × 107 = (100 + 3) (100 + 7) = (100)2 + (3 + 7)10 + (3 × 7)
= 10000 + 100 + 21 = 10121

(ii) 95 × 96 = (90 + 5) (90 + 4)
Using identity, (+ a) (x + b) = x2 + (a + b) x + ab 
Here, x = 90, a = 5 and b = 4
95 × 96 = (90 + 5) (90 + 4) = 902 + 90(5 + 6) + (5 × 6) 
= 8100 + (11 × 90) + 30
= 8100 + 990 + 30 = 9120


(iii) 104 × 96 = (100 + 4) (100 - 4)
Using identity, (+ y) (x -y) = x2 - y2
Here, x = 100 and y = 4
104 × 96 = (100 + 4) (100 - 4) = (100)2 - (4)= 10000 - 16 = 9984 

3. Factorise the following using appropriate identities:

(i) 9x2 + 6xyy2                 

(ii) 4y2 - 4y + 1              

(iii) xy2/100

Answer
(i) 9x2 + 6xy + y2  = (3x) 2 + (2×3x×y) + y2
Using identity, (a + b)2 = a2 + 2ab + b2
Here, a = 3x and b = y
9x2 + 6xy + y2  = (3x) 2 + (2×3x×y) + y= (3x + y)= (3x + y) (3x + y)

(ii) 4y2 - 4y + 1 = (2y)2 - (2×2y×1) + 12
Using identity, (a - b)2 = a2 - 2ab + b2
Here, a = 2y and b = 1
4y2 - 4y + 1 = (2y)2 - (2×2y×1) + 1= (2y - 1)= (2y - 1) (2y - 1)

(iii) x- y2/100 = x- (y/10)2
Using identity, a2 - b2 = (a + b) (a - b)
Here, a = x and b = (y/10)
x- y2/100 = x- (y/10)= (x - y/10) (x + y/10)

4. Expand each of the following, using suitable identities:

(i) (x + 2y + 4z)2                     

(ii) (2xyz)2                    

(iii) (-2x + 3y + 2z)2

(iv) (3a - 7bc)2                         

(v) (-2x + 5y - 3z)2                   

(vi) [1/4 a - 1/2 b + 1]2   

Answer

(i) (x + 2y + 4z)2
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = x, b = 2and c = 4z
(x + 2y + 4z)x2 + (2y)2 + (4z)2 + (2×x×2y) + (2×2y×4z) + (2×4z×x)
x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(ii)  (2xyz)2
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = 2x, b = -and c = z
(2xyz)= (2x)2 + (-y)2 + z2 + (2×2x×-y) + (2×-y×z) + (2×z×2x) 
4x2 + y2 + z2 - 4xy - 2yz + 4xz

(iii) (-2x + 3y + 2z)2 
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = -2x, b = 3and c = 2z
(-2x + 3y + 2z)(-2x)2 + (3y)2 + (2z)2 + (2×-2x×3y) + (2×3y×2z) + (2×2z×-2x) 
4x2 + 9y2 + 4z2 - 12xy + 12yz - 8xz

(iv) (3a - 7bc)2
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = 3a, b = -7b and c = -c
(3a - 7b - c)2 (3a)2 + (-7b)2 + (-c)2 + (2×3a×-7b) + (2×-7b×-c) + (2×-c×3a) 
9a2 + 49b2 + c2 - 42ab + 14bc - 6ac

(v) (-2x + 5y - 3z)2  
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = -2x, b = 5y and c = -3z
(-2x + 5y - 3z)2 (-2x)2 + (5y)2 + (-3z)2 + (2×-2x×5y) + (2×5y×-3z) + (2×-3z×-2x) 
4x2 + 25y2 + 9z2 - 20xy - 30yz + 12xz

(vi) [1/4 a - 1/2 b + 1]2 
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = 1/4 a, b = -1/2 b and c = 1
[1/4 a - 1/2 b + 1]2 (1/4 a)2 + (-1/2 b)2 + 12 + (2×1/4 a×-1/2 b) + (2×-1/2 b×1) + (2×1×1/4 a) 
1/16 a2 + 1/4 b2 + 1 - 1/4 ab - b + 1/2 a

5. Factorise:
 (i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
 (ii) 2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz


Answer

(i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca 
4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
= (2x)2 + (3y)2 + (-4z)2 + (2×2x×3y) + (2×3y×-4z) + (2×-4z×2x)
= (2x + 3y - 4z)2
 (2x + 3y - 4z) (2x + 3y - 4z)

(ii) 2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca
2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz 
= (-√2x)2 + (y)2 + (2√2z)2 + (2×-√2x×y) + (2×y×2√2z) + (2×2√2z×-√2x)
(-√2x + y + 2√2z)2
 (-√2x + y + 2√2z) (-√2x + y + 2√2z)


6. Write the following cubes in expanded form:

(i) (2x + 1)3                 

(ii) (2a - 3b)3                

(iii) [3/2 x + 1]3           

(iv) [x - 2/3 y]3

Answer

(i) (2x + 1)3
Using identity, (a + b)= a3 + b3 + 3ab(a + b)
(2x + 1)= (2x)3 + 13 + (3×2x×1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
8x3 + 12x2 + 6x + 1

(ii) (2a - 3b)3
Using identity, (a - b)= a3 - b3 - 3ab(a - b) 
(2a - 3b)= (2a)3 - (3b)3 - (3×2a×3b)(2a - 3b)
= 8a3 - 27b3 - 18ab(2a - 3b)
8a3 - 27b3 - 36a2b + 54ab2

(iii) [3/2 x + 1]3 
Using identity, (a + b)= a3 + b3 + 3ab(a + b)
[3/2 x + 1]= (3/2 x)3 + 13 + (3×3/2 x×1)(3/2 x + 1)
= 27/8 x+ 1 + 9/2 x(3/2 x + 1)
27/8 x+ 1 + 27/4 x2 + 9/2 x
27/8 x+ 27/4 x2 + 9/2 x + 1

(iv) [x - 2/3 y]3
Using identity, (a - b)= a3 - b3 - 3ab(a - b)
[x - 2/3 y]3 = (x)3 - (2/3 y)3 - (3×x×2/3 y)(x - 2/3 y)
= x3 - 8/27y3 - 2xy(x - 2/3 y)
x3 - 8/27y3 - 2x2y + 4/3xy2

7. Evaluate the following using suitable identities:
 (i) (99)3            

(ii) (102)3             

(iii) (998)3

Answer

(i) (99)3 = (100 - 1)3
Using identity, (a - b)= a3 - b3 - 3ab(a - b) 
(100 - 1)= (100)3 - 13 - (3×100×1)(100 - 1)
= 1000000 - 1 - 300(100 - 1)
= 1000000 - 1 - 30000 + 300
= 970299

(ii) (102)3 = (100 + 2)3
Using identity, (a + b)= a3 + b3 + 3ab(a + b)
(100 + 2)= (100)3 + 23 + (3×100×2)(100 + 2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208

(iii) (998)3 
Using identity, (a - b)= a3 - b3 - 3ab(a - b) 
(1000 - 2)= (1000)3 - 23 - (3×1000×2)(1000 - 2)
= 100000000 - 8 - 6000(1000 - 2)
= 100000000 - 8- 600000 + 12000
= 994011992

8. Factorise each of the following: 
 (i) 8a3 + b3 + 12a2b + 6ab2                           

(ii) 8a3 - b3 - 12a2b + 6ab2
 (iii) 27 - 125a3 - 135a + 225a2                      

(iv) 64a3 - 27b3 - 144a2b + 108ab2
 (v) 27p3 - 1/216 - 9/2 p2 + 1/4 p

 Answer

(i) 8a3 + b3 + 12a2b + 6ab2
Using identity, (a + b)= a3 + b3 + 3a2b + 3ab2
8a3 + b3 + 12a2b + 6ab2 
= (2a)3 + b3 + 3(2a)2b + 3(2a)(b)2
= (2a + b)3
= (2a + b)(2a + b)(2a + b)

(ii) 8a3 - b3 - 12a2b + 6ab2
Using identity, (a - b)= a3 - b3 - 3a2b + 3ab2
8a3 - b3 - 12a2b + 6ab2= (2a)3 - b3 - 3(2a)2b + 3(2a)(b)2
= (2a - b)3
= (2a - b)(2a - b)(2a - b)

(iii) 27 - 125a3 - 135a + 225a2
Using identity, (a - b)= a3 - b3 - 3a2b + 3ab2
27 - 125a3 - 135a + 225a2= 33 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2
= (3 - 5a)3
(3 - 5a)(3 - 5a)(3 - 5a)

(iv) 64a3 - 27b3 - 144a2b + 108ab2
Using identity, (a - b)= a3 - b3 - 3a2b + 3ab2
64a3 - 27b3 - 144a2b + 108ab2= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2
= (4a - 3b)3
= (4a - 3b)(4a - 3b)(4a - 3b)

(v) 27p3 - 1/216 - 9/2 p2 + 1/4 p 
Using identity, (a - b)= a3 - b3 - 3a2b + 3ab2
 27p3 - 1/216 - 9/2 p2 + 1/4 p
(3p)3 - (1/6)3 - 3(3p)2(1/6) + 3(3p)(1/6)2
= (3p - 1/6)3
= (3p - 1/6)(3p - 1/6)(3p - 1/6)

9. Verify : 

(i) x3y3 = (xy) (x2xy + y2)             

(ii) x3y3 = (xy) (x2xy + y2)

Answer

(i) x3y3 = (xy) (x2xy + y2)
We know that, 
(x + y)= x3 + y3 + 3xy(x + y) 
⇒ x3 + y= (x + y)- 3xy(x + y)
⇒ x3 + y= (x + y)[(x + y)2 - 3xy]   {Taking (x+y) common}
⇒ x3 + y= (x + y)[(x2 + y+ 2xy) - 3xy] 
⇒ x3 + y= (x + y)(x2 + y- xy) 

(ii) x3y3 = (xy) (x2xy + y2 )
We know that, 
(x - y)= x3 - y3 - 3xy(x - y) 
⇒ x3 - y= (x - y)+ 3xy(x - y)
⇒ x3 + y= (x - y)[(x - y)2 + 3xy]    {Taking (x-y) common}
⇒ x3 + y= (x - y)[(x2 + y- 2xy) + 3xy] 
⇒ x3 + y= (x + y)(x2 + y+ xy)


10. Factorise each of the following:
 (i) 27y3 + 125z3                     

(ii) 64m3 - 343n3

Answer

(i) 27y3 + 125z3
Using identity, x3y3 = (xy) (x2xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
(3y5z) {(3y)2(3y)(5z) + (5z)2}
= (3y5z) (9y2 - 15yz + 25z)2 

(ii) 64m3 - 343n3 
Using identity, x3y3 = (xy) (x2xy + y2
64m3 - 343n3 = (4m)3 - (7n)3
(4m7n) {(4m)2(4m)(7n) + (7n)2}
= (4m7n) (16m2 + 28mn + 49n)2 


11. Factorise : 27x3y3z3 - 9xyz

Answer

27x3 + y3 + z3 - 9xyz = (3x)3y3z3 - 3×3xyz
Using identity, x3y3z3 - 3xyz = (x + y + z)(x2y2z2xy - yz - xz)
27x3 + y3 + z3 - 9xyz
(3x + y + z) {(3x)2y2z23xy - yz - 3xz}
(3x + y + z) (9x2y2z23xy - yz - 3xz)

12. Verify that: x3 + y3 + z33xyz = 1/2(x + y + z) [(xy)+ (y - z)+ (z - x)2]

Answer

We know that,
x3y3z3 - 3xyz = (x + y + z)(x2y2z2xy - yz - xz)
 x3y3z3 - 3xyz = 1/2×(x + y + z) 2(x2y2z2xy - yz - xz)
= 1/2(x + y + z) (2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz) 
1/2(x + y + z) [(x2y2 -2xy) + (y+ z2 - 2yz) + (x2z- 2xz)] 
= 1/2(x + y + z) [(x - y)+ (y - z)+ (z - x)2]


13. If x + y + z = 0, show that x3 + y3z3 = 3xyz.

 Answer


We know that,
x3y3z3 - 3xyz = (x + y + z)(x2y2z2xy - yz - xz)
Now put (x + y + z) = 0,  
x3y3z3 - 3xyz = (0)(x2y2z2xy - yz - xz) 
⇒ x3y3z3 - 3xyz = 0


14. Without actually calculating the cubes, find the value of each of the following:
 (i) (-12)3 + (7)3 + (5)3
 (ii) (28)3 + (-15)3 + (-13)3


Answer

(i) (-12)3 + (7)3 + (5)3
 Let x = -12, y = 7 and z = 5
We observed that, x + y + z = -12 + 7 + 5 = 0

We know that if,
x + y + z = 0, then x3 + y3z3 = 3xyz
(-12)3 + (7)3 + (5)3 = 3(-12)(7)(5) = -1260

(ii) (28)3 + (-15)3 + (-13)3
 Let x = 28, y = -15 and z = -13
We observed that, x + y + z = 28 - 15 - 13 = 0

We know that if,
x + y + z = 0, then x3 + y3z3 = 3xyz
(28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) = 16380

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 
 (i) Area : 25a2 - 35a + 12
 (ii) Area : 35 y2 + 13y - 12


Answer
(i) Area : 25a2 - 35a + 12
Since, area is product of length and breadth therefore by factorizing the given area, we can know the length and breadth of rectangle.
25a2 - 35a + 12
25a2 - 15a -20a + 12
= 5a(5a - 3) - 4(5a - 3)
(5a - 4)(5a - 3)
Possible expression for length = 5a - 4
Possible expression for breadth = 5a - 3

(ii) Area : 35 y2 + 13y - 12
35 y2 + 13y - 12
35y2 - 15y + 28y - 12
= 5y(7y - 3) + 4(7y - 3)
(5y + 4)(7y - 3)
Possible expression for length = (5y + 4)
Possible expression for breadth (7y - 3)

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? 

(i) Volume : 3x212x
 (ii) Volume : 12ky28ky - 20k


Answer

(i) Volume : 3x2 - 12
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid. 
3x2 - 12x
= 3x(x - 4)
Possible expression for length = 3
Possible expression for breadth x
Possible expression for height (x - 4)

(ii) Volume : 12ky28ky - 20k 
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid. 
12ky2 + 8ky - 20k
= 4k(3y2 + 2y - 5)
= 4k(3y2 +5y - 3y - 5)
4k[y(3y +5) - 1(3y + 5)]
4k (3y +5) (y - 1)
Possible expression for length = 4k
Possible expression for breadth (3y +5)
Possible expression for height (y - 1)

Share with a friend

Complete Syllabus of Class 9

Content Category

Related Searches

Ex 2.1 NCERT Solutions - Polynomials

,

video lectures

,

Maths Class 9 Notes | EduRev

,

Ex 2.2 NCERT Solutions - Polynomials

,

Exam

,

NCERT Solutions Chapter 2 - Polynomials (III)

,

Important questions

,

Summary

,

Free

,

ppt

,

Previous Year Questions with Solutions

,

NCERT Solutions Chapter 2 - Polynomials (III)

,

Polynomials - Chapter Notes, Class 9 Mathematics

,

mock tests for examination

,

NCERT Solutions, Chapter 2: Polynomials, Class 10 (Mathematics)

,

Objective type Questions

,

NCERT Solutions Chapter 2 - Polynomials (III)

,

Viva Questions

,

pdf

,

NCERT Solutions - Polynomials, Class 10, Maths

,

Class 9

,

Sample Paper

,

practice quizzes

,

Maths Class 9 Notes | EduRev

,

MCQs

,

past year papers

,

Class 9

,

study material

,

Class 9

,

Semester Notes

,

Extra Questions

,

Maths Class 9 Notes | EduRev

,

shortcuts and tricks

;