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Class 9 Exam  >  Class 9 Notes  >  NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths

NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths PDF Download

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) (y2 + 3/2) (y2 - 3/2)             

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

Answer

(i) Using identity, (x + 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, (x + 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, (x + a) (x + b) = x2 + (a + b) x + ab
Here, x = 3x , a = 4 and b = -5
(3x + 4) (3x - 5) = (3x) 2 + {4 + (-5)}3x + {4×(-5)}
= 9x2 + 3x(4 - 5) - 20
= 9x2 - 3x - 20

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

(v) (3 - 2x) (3 + 2x)
Using identity, (x + 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, (x + 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, (x + 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, (x + y) (x -y) = x2 - y2
Here, x = 100 and y = 4
104 × 96 = (100 + 4) (100 - 4) = (100)2 - (4)2 = 10000 - 16 = 9984 

3. Factorise the following using appropriate identities:

(i) 9x2 + 6xy + y2                 

(ii) 4y2 - 4y + 1              

(iii) x2 - y2/100

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) + y2 = (3x + y)2 = (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) + 12 = (2y - 1)2 = (2y - 1) (2y - 1)

(iii) x2 - y2/100 = x2 - (y/10)2
Using identity, a2 - b2 = (a + b) (a - b)
Here, a = x and b = (y/10)
x2 - y2/100 = x2 - (y/10)2 = (x - y/10) (x + y/10)

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

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

(ii) (2x - y + z)2                    

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

(iv) (3a - 7b - c)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)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = x, b = 2y and c = 4z
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + (2×x×2y) + (2×2y×4z) + (2×4z×x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

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

(iv) (3a - 7b - c)2
Using identity, (a + b + c)2 = 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)2 = 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)2 = 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)2 = 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)2 = 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)3 = a3 + b3 + 3ab(a + b)
(2x + 1)3 = (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)3 = a3 - b3 - 3ab(a - b) 
(2a - 3b)3 = (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)3 = a3 + b3 + 3ab(a + b)
[3/2 x + 1]3 = (3/2 x)3 + 13 + (3×3/2 x×1)(3/2 x + 1)
= 27/8 x3 + 1 + 9/2 x(3/2 x + 1)
= 27/8 x3 + 1 + 27/4 x2 + 9/2 x
= 27/8 x3 + 27/4 x2 + 9/2 x + 1

(iv) [x - 2/3 y]3
Using identity, (a - b)3 = 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)3 = a3 - b3 - 3ab(a - b) 
(100 - 1)3 = (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)3 = a3 + b3 + 3ab(a + b)
(100 + 2)3 = (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)3 = a3 - b3 - 3ab(a - b) 
(1000 - 2)3 = (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)3 = 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)3 = 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)3 = 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)3 = 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)3 = 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) x3 + y3 = (x + y) (x2 - xy + y2)             

(ii) x3 - y3 = (x - y) (x2 + xy + y2)

Answer

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

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


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

(ii) 64m3 - 343n3

Answer

(i) 27y3 + 125z3
Using identity, x3 + y3 = (x + y) (x2 - xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) {(3y)2 - (3y)(5z) + (5z)2}
= (3y + 5z) (9y2 - 15yz + 25z)2 

(ii) 64m3 - 343n3 
Using identity, x3 - y3 = (x - y) (x2 + xy + y2 ) 
64m3 - 343n3 = (4m)3 - (7n)3
= (4m + 7n) {(4m)2 + (4m)(7n) + (7n)2}
= (4m + 7n) (16m2 + 28mn + 49n)2 


11. Factorise : 27x3 + y3 + z3 - 9xyz

Answer

27x3 + y3 + z3 - 9xyz = (3x)3 + y3 + z3 - 3×3xyz
Using identity, x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
27x3 + y3 + z3 - 9xyz
= (3x + y + z) {(3x)2 + y2 + z2 - 3xy - yz - 3xz}
= (3x + y + z) (9x2 + y2 + z2 - 3xy - yz - 3xz)

12. Verify that: x3 + y3 + z3 - 3xyz = 1/2(x + y + z) [(x - y)2 + (y - z)2 + (z - x)2]

Answer

We know that,
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
⇒ x3 + y3 + z3 - 3xyz = 1/2×(x + y + z) 2(x2 + y2 + z2 - xy - yz - xz)
= 1/2(x + y + z) (2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz) 
= 1/2(x + y + z) [(x2 + y2 -2xy) + (y2 + z2 - 2yz) + (x2 + z2 - 2xz)] 
= 1/2(x + y + z) [(x - y)2 + (y - z)2 + (z - x)2]


13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

 Answer

We know that,
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
Now put (x + y + z) = 0,  
x3 + y3 + z3 - 3xyz = (0)(x2 + y2 + z2 - xy - yz - xz) 
⇒ x3 + y3 + z3 - 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 + y3 + z3 = 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 + y3 + z3 = 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 : 3x2 - 12x
 (ii) Volume : 12ky2 + 8ky - 20k

Answer

(i) Volume : 3x2 - 12x 
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 : 12ky2 + 8ky - 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)

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Ans. A polynomial is a mathematical expression consisting of variables and coefficients that are combined by mathematical operations such as addition, subtraction, and multiplication, but not division, and raised to non-negative integer powers.
2. How to identify the degree of a polynomial?
Ans. The degree of a polynomial is the highest power of the variable present in the polynomial. To identify the degree of a polynomial, we need to find the highest power of the variable in the polynomial expression.
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Ans. The zero of a polynomial is also known as the root or solution of a polynomial equation. It is a value of the variable for which the polynomial expression becomes zero.
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Ans. The factor theorem is a fundamental theorem of algebra that states that if a polynomial P(x) is divided by (x-a), then the remainder is zero if and only if (x-a) is a factor of P(x).
5. What is the difference between a monomial and a polynomial?
Ans. A monomial is a polynomial expression with only one term, whereas a polynomial is an expression with more than one term. In other words, a monomial is a special case of a polynomial where the polynomial degree is one.
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NCERT Solutions Chapter 2 - Polynomials (III)

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NCERT Solutions Chapter 2 - Polynomials (III)

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NCERT Solutions Chapter 2 - Polynomials (III)

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Additional Information about NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths for Class 9 Preparation

NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths Free PDF Download

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

NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths Notes on EduRev are your ultimate resource for success.

NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths Class 9 Questions

The "NCERT Solutions Chapter 2 - Polynomials (III), Class 9, Maths Class 9 Questions" guide is a valuable resource for all aspiring students preparing for the Class 9 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

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