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2. Polynomials Mathematics Exercise - 2.5 class 9 Maths in English - CBSE Study

2. Polynomials Mathematics Exercise - 2.5 class 9 Maths in English - CBSE Study

Exercise 2.5 class 9 Maths all chapters ncert books solutions 2. Polynomials Mathematics Exercise - 2.5 class 9 Maths in English - CBSE Study. 2. Polynomials Mathematics Class 9 exercise - 2.5 class 9 Maths cbse board school study materials like cbse notes in English medium, all chapters and exercises are covered the ncert latest syllabus 2024 - 25.

NCERT Solutions For Class 9 in Hindi and English Medium

Exercise 2.5 class 9 Maths

In this chapter-2. Polynomials of NCERT Solution class 9 Mathematics in English we cover all ncert book exercises and examples also quick revision notes or point which will help the student in exams. 2. Polynomials Mathematics Class 9 exercise - 2.5 class 9 Maths cbse board school study materials like cbse notes in English medium, all chapters and exercises are covered the ncert latest syllabus 2024 - 25.

Exercise 2.5 class 9 Maths here we are going to discuss our class subjects and chapters and exercise with solution also questions answer for 2. Polynomials प्रश्न उत्तर (questions and answers) 2. Polynomials के प्रश्न उत्तर अध्याय 2. Polynomials प्रश्न उत्तर

NCERT Solution Class 9 Mathematics

  • Quick revision chapter - review short Notes class 9 Mathematics
  • Latest 2024 - 25 Intext (textual) questions answers for class 9
  • Solved NCERT Book Solutions [UPADATED] 2024 - 25 class 9 in English medium
  • Extra Additional and Important questions for upcoming exams 2024 - 25
  • 2. Polynomials class 9 प्रश्न उत्तर and class 9 2. Polynomials question answer.
  • Exercise 2.5 कक्षा 9 and similarily you can find 2. Polynomials कक्षा 9 and so mmany chapter included in this ncert books solutions 2. Polynomials के प्रश्न उत्तर

    • Exercise 2.5 class 9 Maths

    कक्षा 9 विज्ञान पाठ 2. Polynomials के प्रश्न उत्तर पाठ 2. Polynomials class 9 question answer chapter 2. Polynomials class 9

    If you revise these questions and answers for class 9 Mathematics Exercise 2.5 in English also class 9 2. Polynomials

    2. Polynomials

    Exercise 2.5

    Algebraic Identities: 

     ∵ ∴

    (1) (x + y)2 = x2 + 2xy + y2

    (2)  (x - y)2 = x2 - 2xy + y2

    (3)  x2 - y2 = (x + y) (x - y) 

    (4)  (x + a) (x + b) = x2 + (a + b)x + ab 

    (5)  (x + y)3 = x3 + 3x2y + 3xy2 + y3

    (6)  (x - y)3 = x3 - 3x2y + 3xy2 - y3

    (7)  x3 + y3 = (x + y) (x2 - xy + y2)

    (8)  x3 - y3 = (x - y) (x2 + xy + y2)

    (9)  (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

    (10) x3 + y3 + z3 - 3xyz = ( x + y + z) (x2 + y2 + z2 - xy - yz - zx)

     

    Exercise 2.5 

    Q1. Use suitable identities to find the following products:

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

    (ii) (x + 8) (x – 10)

    (iii) (3x + 4) (3x – 5)

    (v) (3 – 2x) (3 + 2x)

    Solution: 

    (i) (x + 4) (x + 10) 

    Using identity;  (x + a) (x + b) = x2 + (a + b)x + ab 

    (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  

    (x + 8) (x – 10) = x2 + [8 + (-10)]x + (8)(-10) 

    = x2 - 2x - 80  

    (iii) (3x + 4) (3x – 5)

    Using identity;  (x + a) (x + b) = x2 + (a + b)x + ab  

    (3x + 4) (3x – 5) = (3x)2 + [4 + (-5)]3x + (4)(-5) 

    = 9x2 - 3x - 20  

    Using identity;  (x + y) (x - y) = x2 - y2 

    (v) (3 – 2x) (3 + 2x)

    Using identity; (x + y) (x - y) = x2 - y2 

    (3 – 2x) (3 + 2x) = (3)2 - (2x)2

    = 9 - 4x2

    Q2. Evaluate the following products without multiplying directly:

    (i) 103 × 107

    (ii) 95 × 96

    (iii) 104 × 96

    Solution: 

    (i) 103 × 107 = (100 + 3) (100 + 7) 

    Using identity;  (x + a) (x + b) = x2 + (a + b)x + ab  

    (100 + 3) (100 + 7) = (100)2​ + (3 + 7)100 + 3×7 

    =10000 + 1000 + 21     

    = 11021

    (ii) 95 × 96 = (90 + 5) (90 + 6) 

    Using identity;  (x + a) (x + b) = x2 + (a + b)x + ab  

    (90 + 5) (90 + 6) = (90)2​ + (5 + 6)90 + 5×6 

    =8100 + 990 + 30     

    = 9120

    (iii)  104 × 96 = (100 + 4) (100 - 4) 

    Using identity; (x + y) (x - y) = x2 - y2  

    (100)2 - (4)2

    =10000 - 16      

    = 9984

    3. Factorise the following using appropriate identities:

    (i) 9x2 + 6xy + y2 

    (ii) 4y2 – 4y + 1

    Solution:

    (i) 9x2 + 6xy + y2 

    = (3x)2 + 2.3x.y + (y)2     [ ∵ x2 + 2xy + y2 = (x + y)2]

    ∴ = (3x + y)2 

    =  (3x + y)  (3x + y)

    (ii) 4y2 - 4y + 1 

    = (2y)2 - 2.2y.1 + (1)2     [ ∵ x2 - 2xy + y2 = (x - y)2]

    ∴ = (2y - 1)2 

    =  (2y - 1)  (2y - 1)

      

     

    [ ∵ x2 - y2 = (x + y) (x - y) ​]

    Q4. 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

    Solution:

    (i) (x + 2y + 4z)2  

    Here let as a = x, b = 2y, c = 4z and putting the values of a, b and c in the

    Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

    ∴ (x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

       = x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx                   

    (ii) (2x – y + z)2 

    Here let as a = 2x, b = - y, c = z and putting the values of a, b and c in the

    Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

    ∴ (2x – y + z)2 = (2x)2 + (- y)2 + (z)2 + 2(2x)(- y) + 2(- y)(z) + 2(z)(2x)

       = 4x2 + y2 + z2 - 4xy - 2yz + 4zx           

    (iii) (–2x + 3y + 2z)2

    Here let as a = - 2x, b = 3y, c = 2z and putting the values of a, b and c in the

    Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

    ∴ (–2x + 3y + 2z)2 

       = (– 2x)2 + (3y)2 + (2z)2 + 2(–2x)(3y) + 2(3y)(2z) + 2(2z)(–2x)

       = 4x2 + 9y2 + 4z2 – 12xy  + 12yz – 8zx  

    (iv) (3a – 7b – c)2 

    Here let as x = 3a, y = – 7b, z = – c and putting the values of x, y and z in the

    Identity (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

    ∴ (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

    Here let as a = - 2x, b = 5y, c = –3z and putting the values of a, b and c in the

    Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

    ∴ (–2x + 5y – 3z)2

       = (– 2x)2 + (5y)2 + (– 3z)2 + 2(–2x)(5y) + 2(5y)(– 3z) + 2(– 3z)(–2x)

       = 4x2 + 25y2 + 9z2 – 20xy  – 30yz + 12zx  

    Q5. Factorise:
    (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

    Solution:

    (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

    = (2x)2 + (3y)2 + (4z)2 + 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x)

     [∵ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2 ]

    = (2x + 3y + 4z)2 

    = (2x + 3y + 4z) (2x + 3y + 4z)


    Q6. Write the following cubes in expanded form:
    (i) (2x + 1)3 

    (ii) (2a – 3b)3

    Solution:

    (i) (2x + 1)3 

    [using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]

    (2x + 1)3 = (2x)3 + 3(2x)2(1) + 3(2x)(1)2 + (1)3

                   = 8x3 + 12x2 + 6x + 1

    (ii) (2a – 3b)3

    [Using identity  (x - y)3 = x3 - 3x2y + 3xy2 - y3]

    (2a - 3b)3 = (2a)3 - 3(2a)2(3b) + 3(2a)(3b)2 - (3b)3

                   = 8a3 - 36a2b + 54ab2 - 27b3

    algebraic identities

    [using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]

    algebraic identities

    algebraic identities

    [using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]

    Q7. Evaluate the following using suitable identities:

    (i) (99)3 

    (ii) (102)3 

    (iii) (998)3

    Solution: 

    (i) (99)3 

    = (100 - 1)3

    [using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]

    (100 - 1)3 = (100)3 - 3(100)2(1) + 3(100)(1)2 - (1)3

                    = 1000000 - 30000 + 300 - 1

                    = 1000300 - 30001

                    = 970299

    (ii) (102)3 

    = (100 + 2)3

    [using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]

    (100 + 2)3 = (100)3 + 3(100)2(2)+ 3(100)(2)2 + (2)3

                    = 1000000 + 60000 + 1200 + 8

                    = 1061208

    (iii) (998)3

    = (1000 - 2)3

    [using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]

    (1000 - 2)3 = (1000)3 - 3(1000)2(2)+ 3(1000)(2)2 - (2)3

                    = 1000000000 - 6000000 + 12000 - 8

                    = 1000012000 - 6000008

                    = 994011992

    Q8. Factorise each of the following:

    (i) 8a3 + b3 + 12a2b + 6ab2 

    (ii) 8a2 – b2 – 12a2b + 6ab2

    (iii) 27 – 125a3 – 135a + 225a2

    (iv) 64a3 – 27b3 – 144a2b + 108ab2  

    Solution:

    (i) 8a3 + b3 + 12a2b + 6ab2 

    = (2a)3 +(b)3 + 3(2a)2(b) + 3(2a)(b)2

    [Using identity  x3  + y3 + 3x2y + 3xy2 = (x + y)3 ]

    = (2a)3 +(b)3 + 3(2a)2(b) + 3(2a)(b)2 = (2a + b)3

    = (2a + b)(2a + b)(2a + b)

    (ii) 8a2 – b2 – 12a2b + 6ab2

    = (2a)3 - (b)3 - 3(2a)2(b) + 3(2a)(b)2

    [Using identity  x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]

    = (2a)3 - (b)3 - 3(2a)2(b) + 3(2a)(b)2 = (2a - b)3​

    = (2a - b)(2a - b)(2a - b)

    (iii) 27 – 125a3 – 135a + 225a2

    = (3)3 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2

    [Using identity  x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]

    = (3)3 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2= (3 - 5a)3​

    = (3 - 5a)(3 - 5a)(3 - 5a)

    (iv) 64a3– 27b3 – 144a2b + 108ab2  

    = (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2

    [Using identity  x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]

    = (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2 = (4a - 3b)3​

    = (4a - 3b)(4a - 3b)(4a - 3b)

    [Using identity  x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]

    Q9. Verify:

    (i) x3 + y3 = (x + y) (x2 – xy + y2)

    Solution:

    RHS = (x + y) (x2 – xy + y2)

    = x(x2 – xy + y2) + y (x2 – xy + y2)

    = x3 – x2y + xy2 + x2y – xy2 + y3 

    = x3 + y3

     ∵ LHS = RHS Verified 

    (ii) x3 – y3 = (x – y) (x2 + xy + y2)

    Solution:

    RHS = (x - y) (x2 + xy + y2)

    x(x2 + xy + y2) - y(x2 + xy + y2)

    = x3 + x2y + xy2 – x2y – xy2 – y3 

    = x3 – y3

    ∵ LHS = RHS Verified 

    Q10. Factorise each of the following:

    (i) 27y3 + 125z3 

    (ii) 64m3 – 343n3

    Solution: 

    (i) 27y3 + 125z3 

    = (3y)3 + (5z)3

    [Using identity x3 + y3 = (x + y) (x2 – xy + y2) ]

    (3y)3 + (5z)3​ = (3y + 5y) [(3y)2 - (3y)(5z) + (5z)2]

    = (3y + 5y) (9y2 - 15yz + 25z2)

    (ii) 64m3 – 343n3

    Solution: 

    (ii) 64m3 – 343n3

    = (4m)3 – (7n)3

    [Using identity x3 – y3 = (x – y) (x2 + xy + y2) ]

    (4m)3 – (7n)3​ = (4m – 7n) [(4m)2 + (4m)(7n) + (7n)2]

    = (4m – 7n) (16m2 + 28mn + 49n​2)

    Q11. Factorise : 27x3 + y3 + z3 – 9xyz

    Solution: 

    = (3x)3 + (y)3 + (z)3 - 9xyz 

    ∵ x3 + y3 + z3 - 3xyz =  (x + y + z) (x2 + y2 + z​2 - xy - yz - zx)

    Using identity: 

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

    = (3x + y + z) (9x2 + y2 + z2 - 3xy - yz - 3zx)

    Q12. Verify that:

    x3 + y3 + z3 - 3xyz =½ (x + y + z) [(x -y)2 + (y - z)2 + (z - x)2]

    LHS =  ½(x + y + z) [x2 - 2xy + y2 + y2 - 2yz + z2 + z2 - 2xz + x2]

            = ½(x + y + z) (2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz)

            = ½ × 2(x + y + z)(x2 + y2 + z2 - xy - yz - xz)

            = (x + y + z)(x2 + y2 + z2 - xy - yz - xz) 

            = x3 + y3 + z3 - 3xyz                 [Using Identity]

    LHS = RHS 

    ​

     

    Topic Lists:

    2. Polynomials अध्याय के अन्य दुसरे विषय भी देखे :

  • Exercise 2.1
  • Exercise 2.2
  • Exercise 2.3
  • Exercise 2.4
  • Exercise 2.5

    • 2. Polynomials Mathematics Exercise - 2.5 class 9 Maths in English - CBSE Study

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    Chapter-Lists: Mathematics

  • Chapter 1. Number Systems
  • Chapter 2. Polynomials
  • Chapter 3. Coordinate Geometry
  • Chapter 4. Linear Equation In Two Variables
  • Chapter 5. Introduction To Euclid’s Geometry
  • Chapter 6. Lines and Angles
  • Chapter 7. Triangles
  • Chapter 8. Quadrilaterals
  • Chapter 9. Area Parallelograms and Triangles
  • Chapter 10. Circles
  • Chapter 11. Constructions
  • Chapter 12. Herons Formula
  • Chapter 13. Surface Areas and Volumes
  • Chapter 14. Statistics
  • Chapter 15. Probability

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