Class 9 2. Polynomials Exercise 2.5: NCERT Book Solutions
Class 9 chapter 2. Polynomials ncert solution for board exams and term 1 and term 2 exams.
NCERT Solutions
All chapters of ncert books Mathematics 2. Polynomials Exercise 2.5 is solved by exercise and chapterwise for class 9 with questions answers also with chapter review sections which helps the students who preparing for UPSC and other competitive exams and entrance exams.
Class 9 chapter 2. Polynomials ncert solution for board exams and term 1 and term 2 exams. - 2. Polynomials - Exercise 2.5: NCERT Book Solutions for class 9th. All solutions and extra or additional solved questions for 2. Polynomials : Exercise 2.5 Mathematics class 9th:English Medium NCERT Book Solutions. Class 9 chapter 2. Polynomials ncert solution for board exams and term 1 and term 2 exams.
2. Polynomials : Exercise 2.5 Mathematics class 9th:English Medium NCERT Book Solutions
Class 9 chapter 2. Polynomials ncert solution for board exams and term 1 and term 2 exams. - 2. Polynomials - Exercise 2.5: NCERT Book Solutions for class 9th. All solutions and extra or additional solved questions for 2. Polynomials : Exercise 2.5 Mathematics class 9th:English Medium NCERT Book Solutions.
Class 9 2. Polynomials Exercise 2.5: NCERT Book Solutions
NCERT Books Subjects for class 9th Hindi Medium
2. Polynomials
Class 9 chapter 2. Polynomials ncert solution for board exams and term 1 and term 2 exams.
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
[using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]
[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 + 49n2)
Q11. Factorise : 27x3 + y3 + z3 – 9xyz
Solution:
= (3x)3 + (y)3 + (z)3 - 9xyz
∵ x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - 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
Class 9 chapter 2. Polynomials ncert solution for board exams and term 1 and term 2 exams.
See other sub-topics of this chapter:
Class 9 chapter 2. Polynomials ncert solution for board exams and term 1 and term 2 exams. - 2. Polynomials - Exercise 2.5: NCERT Book Solutions for class 9th. All solutions and extra or additional solved questions for 2. Polynomials : Exercise 2.5 Mathematics class 9th:English Medium NCERT Book Solutions. Class 9 chapter 2. Polynomials ncert solution for board exams and term 1 and term 2 exams.
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Mathematics Chapter List
1. Number Systems
2. Polynomials
3. Coordinate Geometry
4. Linear Equation In Two Variables
5. Introduction To Euclid’s Geometry
6. Lines and Angles
7. Triangles
8. Quadrilaterals
9. Area Parallelograms and Triangles
10. Circles
11. Constructions
12. Herons Formula
13. Surface Areas and Volumes
14. Statistics
15. Probability
