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AP Board Class 9 Maths Chapter 2 Polynomials and Factorisation Ex 2.5 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.5 Book Answers

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Andhra Pradesh State Board Class 9th Maths Chapter 2 Polynomials and Factorisation Ex 2.5 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths
Chapters	Maths Chapter 2 Polynomials and Factorisation Ex 2.5
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AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.5

Question 1.
Use suitable identities to find the following products.
i) (x + 5) (x + 2)
Solution:
(x + 5) (x + 2)
= x² + (5 + 2)x + 5 x 2
[ ∵ (x + a) (x + b) = x² + (a + b) x + ab]
= x² + 7x + 10

ii) (x – 5) (x – 5)
Solution:
(x – 5) (x – 5)
= (x – 5)² = x² – 2(x) (5) + 5²
[ ∵(x – y)² = x² – 2xy + y²]
= x² – 10x + 25

iii) (3x + 2) (3x – 2)
Solution:
(3x + 2) (3x – 2) = (3x)² – (2)²
[∵ (x + y) (x – y) =x²– y²]
= 9x² – 4

iv) (𝑥2+1𝑥2)(𝑥2−1𝑥2)
Solution:

v) (1 + x) (1 + x)
Solution:
(1 + x) (1 + x)
= (1 + x)² = 1² + 2 (1) (x) + x²
[∵(x + y)² = x² + 2xy + y²]
= 1 + 2x + x²

Question 2.
Evaluate the following products with¬out actual multiplication.
i) 101 x 99
Solution:
101 x 99
= (100 + 1) (100 – 1)
= 100² – 1²
= 10000 – 1
= 9999

ii) 999 x 999
Solution:
999 x 999
= 999²
= (1000 – 1)²
= 1000² – 2 x (1000) x 1 + 1²
= 1000000-2000 + 1
= 998001

iii) 5012×4912
Solution:

iv) 501 x 501
Solution:
501 x 501
= (500 + 1) (500 + 1)
= (500 + 1)²
= 500² + 2 x (500) x 1 + 1²
= 250000 + 1000 + 1 = 251001

v) 30.5 x 29.5 = (30 + 0.5) (30 – 0.5)
= 30² – (0.5)²
= 900 – 0.25
= 899.75

Question 3.
Factorise the following using appro-priate identities.
i) 16x² + 24xy + 9y²
Solution:
16x² + 24xy + 9y2
= (4x)² + 2 (4x) (3y) + (3y)²
= (4x + 3y)² = (4x + 3y) (4x + 3y)
[ ∵ (x + y)² = x² + 2xy + y²]

ii) 4y² – 4y + 1
Solution:
4y² – 4y + 1
= (2y)² – 2 (2y) (1) + (1)²
[ ∵ (x -y)² = x² – 2xy + y²]
= (2y -1)² = (2y – 1) (2y-1)

iii) 4𝑥2−𝑦225
Solution:

iv) 18a² – 50
Solution:
18a² – 50 = 2 (9a² – 25)
= 2[(3a)² – (5)²]
[ ∵ x² – y² = (x + y) (x – y)]
= 2 (3a + 5) (3a – 5)

v) x² + 5x + 6
Solution:
x² + 5x + 6 = x² + (3 + 2) x + 3 x 2
[ ∵ (x + a) (x + b) = x² + (a + b) x + a . b]
= (x + 3) (x + 2)

vi) 3p² – 24p + 36
Solution:
3p² – 24p + 36
= 3[p² – 8p + 12]
= 3[p² + (- 6 – 2)p + (- 6) (- 2)]
[ ∵ (x + a) (x + b) = x² + (a + b) x + ab]
= 3 (p – 6) (p – 2)

Question 4.
Expand each of the following, using suitable identities.
i) (x + 2y + 4z)2
(x + 2y + 4z)² = (x)² + (2y)² + (4z)² + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x)
[ ∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]
= x² + 4y² + 16z² + 4xy + 16yz + 8zx

ii) (2a – 3b)³
Solution:
(2a – 3b)³ = (2a)³ – 3 (2a)² (3b) + 3 (2a) (3b)² – (3b)³
[ ∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8a³ – 3(4a²) (3b) + 3 (2a) (9b²) – 27b³
= 8a³ – 36a²b + 54ab²-27b³
(or)
∵ (a – b)³ = a³ – b³– 3ab (a – b)]
= (2a)³ – (3b)³ – 3(2a) (3b) (2a – 3b)
= 8a³ – 27b³ – 18ab (2a – 3b)

iii) (- 2a + 5b – 3c)²
Solution:
(- 2a + 5b – 3c)²
= (- 2a)² + (5b)² + (- 3c)² + 2 (- 2a) (5b) + 2 (5b) (- 3c) + 2 (- 3c) (- 2a)
= 4a² + 25b² + 9c² – 20ab – 30bc + 12ca
[ ∵ (x + y + z)² = x² + y² + z² + 2xy +2yz +2za]

iv) [𝑎4−𝑏2+1]2
Solution:

v) (p + 1)³
Solution:
(p + 1)³
= (P)³ + 3 (p)² (1) + 3 (p) (1)² + (1)³
[ ∵ (x + y)³ = x³ + 3x²y + 3xy² + y³]
= p³ + 3p² + 3p + 1

vi) (𝑥−23𝑦)3
Solution:

Question 5.
Factorise
i) 25x² + 16y² + 4z² – 40xy + 16yz – 20xz
Solution:
25x² + 16y² + 4z² – 40xy + 16yz – 20xz
= (5x)² + (- 4y)² + (- 2z)² + 2(5x) (- 4y) + 2 (- 4y) (- 2z) + 2 (- 2z) (5x)
= (5x – 4y – 2z)² = (- 5x + 4y +, 2z)²

ii) 9a² + 4b² + 16c² + 12ab – 16bc – 24ca
Solution:
9a² + 4b² + 16c² + 12ab – 16bc -24ca
= (3a)² + (2b)² + (- 4c)²+ 2 (3a) (2b) + 2 (2b) (- 4c) + 2(- 4c) (3a)
= (3a + 2b – 4c)²

Question 6.
If a + b + c = 9 and ab + be + ca = 26, find a² + b² + c².
Solution:
Given that a + b + c = 9
Squaring on both sides,
(a + b + c)² = 9²
⇒ a²+ b² + c²+ 2 (ab + be + ca) = 81 ⇒ a² + b² + c² = 81 – 2 (ab + be + ca)
(by problem)
= 81 – 2 x 26
= 81 – 52 = 29

Question 7.
Evaluate the following by using suit¬able identities. m EachgM)
i) (99)³
Solution:
(99)² = (100 – 1)³
= 100³ – 3 (100)² (1) + 3 (100) (1)² – 1³
[ ∵ (x – y)³ = x³ – 3x²y + 3xy² + y³]
= 1000000 – 30000 + 300 – 1
= 970299

ii) (102)³
Solution:
(102)³ = (100 + 2)³
= 100³ + 3 (100)² (2) + 3 (100) (2)² + 2³
[ ∵ (x + y)³ = x³ + 3x²y + 3xy² + y³]
= 1000000 + 60000 + 1200 + 8
= 1061208

iii) (998)³
Solution:
(998)³ =(1000 – 2)³
[ ∵ (x – y)³ = x³ – 3x²y + 3xy² – y³] = 1000³– 3(1000)²(2) + 3(1000)(2)²– 2³
= 1000000000 – 6000000 + 12000 – 8
= 994011992

iv) (1001)³
Solution:
(1001)³ = (1000 + 1)³ .
[ ∵ (x + y)³ = x³ + 3x²y + 3xy² + y³] = 1000³ + 3(1000)²(1) + 3(1000) (1)² + 1³
= 1000000000 + 3000000 + 3000 + 1
= 1003003001

Question 8.
Factorise each of the following.
i) 8a³ + b³ + 12a² b + 6ab²
Solution:
8a³ + b³ + 12a² b + 6ab²
= (2a)³ + (b)³ + 3 (2a)² (b) + 3 (2a) (b)²
= (2a + b)³

ii) 8a³ – b³ – 12a² b + 6ab²
Solution:
8a³ – b³ – 12a² b + 6ab²
= (2a)³ – (b)³ – 3 (2a)² (b) + 3 (2a) (b)²
= (2a – b)³

iii) 1 – 64a³ -12a + 48a²
Solution:
1 – 64a³ – 12a + 48a²
= (1)³ – (4a)³ – 3(1)² (4a) + 3(1) (4a)²
= (1 – 4a)³

iv) 8𝑝3−125𝑝2+625𝑝−1125
Solution:

Question 9.
Verify i) x³ + y³ = (x + y) (x² – xy + y²);
ii) x³ – y³ = (x – y) (x² + xy + y²)
Using some non-zero positive integers and check by actual multiplication. Can you
call these as identities ?
i) x³ + y³ = (x + y) (x² – xy + y²)
Solution:
Given x³ + y³ = (x + y) (x² – xy + y²)
L.H.S = x³ + y³
R.H.S = (x + y) (x² – xy + y²)
= x (x² – xy + y²) + y (x² – xy + y²)
= x³ -x²y + xy² + x²y – xy² + y³
= x³ + y³
= L.H.S
∴ L.H.S = R.H.S

Take x = 3, y = 2
L.H.S = 3³ + 2³ = 27 + 8 = 35
R.H.S = (3 + 2) (3² – 3 x 2 + 2²)
= 5 x (9 – 6 + 4)
= 5 x 7 = 35
∴ L.H.S = R.H.S

ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution:
Given that x³ – y³ = (x – y) (x² + xy + y²)
L.H.S = x³ – y³
R.H.S = (x – y) (x² + xy + y²)
= x (x² + xy + y²) – y (x² + xy + y²)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³= L.H.S

L.H.S = 3³ – 2³ = 27 – 8 = 19
R.H.S = (3 – 2) (3² + 3 x 2 + 2²)
= 1 x (9 + 6 + 4)
= 1 x 19 = 19
∴ L.H.S = R.H.S
We can call the above two expressions as identities

Question 10.
Factorise by using the above results (identities).
i) 27a³ + 64b³
Solution:
27a³+ 64b³ = (3a)³ + (4b)³
= (3a + 4b) {(3a)² – (3a) (4b) + (4b)²}
= (3a + 4b) (9a² – 12ab + 16b²)

ii) 343y³ – 1000
Solution:
343y³ – 1000 = (7y)³ – (10)³
= (7y – 10) [(7y)² + (7y) (10) + (10)²]
= (7y – 10) (49y² + 70y + 100)

Question 11.
Factorise 27x³ + y³ + z³ – 9xyz using identity.
Solution:
Given 27x³ + y³ + z³ – 9xyz
= (3x)³ + (y)³ + (z)³ – 3 (3x) (y) (z)
= (3x + y + z)
[(3x)² + y² + z² – (3x) (y) – (y) (z) – (z) (3x)]
[ ∵ (x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z²– xy – yz – zx)
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3xz)

Question 12.
Verify that x³+ y³ + z³ – 3xyz = 1/2 (x + y + z) [(x – y)² + (y – z)² + (z – x)² ]
(OR)
Verify that
p³ + q³ + r³ – 3pqr = 1/2 (p + q + r)
[(p – q)² + (q – r)² + (r – p)²]
Solution:
Given x³+ y³ + z³ – 3xyz = 1/2 (x + y + z) [(x – y)² + (y – z)² + (z – x)² ]
R-H.S = 1/2 (x + y + z) [(x – y)² + (y – z)²+ (z – x)²]
= 1/2 (x + y + z) [x² + y² – 2xy + y² + z² – 2yz + z² + x² – 2xz]
= 1/2 (x + y + z) [2x² + 2y² + 2z² – 2xy – 2yz – 2zx]
= 1/2 (x + y + z) (2) [x² + y² + z² – xy – yz – zx]
= (x + y + z) (x² + y² + z² – xy – yz – zx)
= L.H.S
Hence proved.

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3xyz
Solution:
Given x + y + z = 0
To prove x³ + y³ + z³ = 3xyz
We have an identity
(x + y + z) (x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz
Substituting x + y + z = 0in the above equation, we get
0 x (x² + y² + z² -xy-yz-zx)
= x³ + y³ + z³ – 3xyz
⇒ x³ + y³ + z³ – 3xyz = 0
⇒ x³ + y³ + z³ = 3xyz

Question 14.
Without actual calculating the cubes, find the value of each of the following.
i) (- 10)³ + 7³ + 3³
Solution:
Given (-10)³ + 7³ + 3³
Sum of the bases = -10 + 7 + 3 = = 0
∴ (- 10)³ + 7³ + 3³
= 3 (- 10) x (7) x 3
= -630
[ ∵ x + y + z = 0 then x³ + y³ + z³ = 3xyz]

ii) (28)³ + (- 15)³ + (- 13)³
Solution:
Given (28)³ + (- 15)³+ (- 13)³
Sum of the bases = 28 + (- 15) + (- 13) = 0
∴ (28)³ + (- 15)³ + (- 13)³
= 3 x 28 x (- 15) x (- 13)
= 16380

iii) (12)3+(13)3−(56)3 read it as (12)3+(13)3+(−56)3
Solution:

iv) (0.2)³ – (0.3)³ + (0.1)³
Solution:
Given that (0.2)³ – (0.3)³ + (0.1)³
= (0.2)³ + (- 0.3)³ + (0.1)³
Sum of the bases = 0.2 – 0.3 + 0.1 = 0
∴ (0.2)³ + (-0.3)³ + (0.1)³
= 3 x (0.2) (- 0.3) (0.1)
= -0.018

Question 15.
Give possible expressions for the length and breadth of the rectangle whose area is given by
i) 4a² + 4a – 3
Given that area = 4a² + 4a – 3
= 4a² + 6a – 2a – 3
= 2a (2a + 3) – 1 (2a + 3)
= (2a – 1) (2a + 3)
∴ Length = (2a + 3); breadth = (2a – 1).

ii) 25a²– 35a + 12
Solution:
Given that area = 25a² – 35a +12
= 25a² – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
∴ (5a – 4) (5a – 3) are the length and breadth.

Question 16.
What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below ?
i) 3x³ – 12x
Solution:
Volume = 3x³– 12x
= 3x (x² – 4)
= 3x (x + 2) (x – 2) are the dimensions.

ii) 12y² + 8y – 20
Solution:
Given that volume = 12y² + 8y – 20
= 4 (3y² + 2y – 5)
= 4 [3y² + 5y – 3y – 5]
= 4 [y (3y + 5) – 1 (3y + 5)]
= 4 (3y + 5) (y – 1)
Hence 4, (3y + 5) and (y – 1) are the dimensions.

Question 17.
Show that if 2 (a² + b² ) = (a + b)² then a = b
Solution:
Given that 2 (a² + b² ) = (a + b)²
To prove a = b
As 2 (a² + b² ) = (a + b)²
We have
2a² + 2b² = a² + 2ab + b²
2a² – a² + 2b² – b² = 2ab
a² + b² = 2ab
This is possible only when a = b
∴ a = b

AP Board Textbook Solutions PDF for Class 9th Maths

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