KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.5

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.5 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.5.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.5

Question 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) \(\left(\mathrm{y}^{2}+\frac{3}{2}\right)\left(\mathrm{y}^{2}-\frac{3}{2}\right)\)
(v) (3 – 2x) (3 + 2x)
Solution:
(i) (x + 4) (x + 10)
Suitable identity is (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)
Suitable identity is (x + a) (x + b) = x² + (a + b)x + ab
(x + 8) (x – 10)= (x)² + (8 – 10)x + (8 × -10)
= x² + (-2)x + (-80)
= x² – 2x – 80

(iii) (3x + 4) (3x – 5)
Suitable identity is (x + a) (x + b) = x² + (a + b)x + ab
(3x + 4) (3x – 5) = (3x)² + (+4 – 5)3x + (4 x -5)
= 9x² + (-1)3x + (-20)
= 9x² – 3x – 20

(iv) \(\left(\mathrm{y}^{2}+\frac{3}{2}\right)\left(\mathrm{y}^{2}-\frac{3}{2}\right)\)
Suitable identity is (a + b) (a – b) = a² – b²
\(\begin{aligned}\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right) &=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2} \\ &=y^{4}-\frac{9}{4} \end{aligned}\)

(v) (3 – 2x) (3 + 2x)
Suitable identity is (a + b) (a – b) = a² – b²
(3 – 2x) (3 + 2x) = (3)² – (2x)²
= 9 – 4x².

Question 2.
Evaluate the following products without mulitplying directly :
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution:
(i) 103 × 107
= (100 + 3) (100 – 3)
As per Identity,
(a + b)(a – b) = a² – b²
(100 + 3)(100 – 3) = (100)² – (3)²
= 10000 – 9
= 9991

(ii) 95 × 96
= (100 – 5) (100 + 6)
As per Identity.
(x + a) (x + b) = x² + (a + b)x + ab
(100 – 5) (100 + 6) = (100)² + (-5 + 6)100 + (-5 × 6)
= 10000 + (1)100 + (-30)
= 10000 + 100 – 30
= 10100 – 30
= 10070

(iii) 104 × 96
= (100 + 4) (100 – 4)
As per Identity,
(a + b) (a – b) = a² – b²
(100 + 4) (100 – 4) = (100)² – (4)²
= 10000 – 16
= 9984

Question 3.
Factorise the following using appropriate identities :
(i) 9x² + 6xy + y²
(ii) 4y² – 4y + 1
(iii) \(x^{2}-\frac{y^{2}}{100}\)
Solution:
(i) 9x² + 6xy + y²
= (3x)² + 2(3x)(y) + (y)²
= (3x + y)² [∵ x² + 2xy + y² = (x + y)²]

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

(iii) \(x^{2}-\frac{y^{2}}{100}\)

Question 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) \(\left(\frac{1}{4} a-\frac{1}{2} b+1\right)^{2}\)
Solution:
(i) (x + 2y + 4z)²
(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
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx (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
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx (-2x + 3y + 2z)²
= (-2x)² + (3y)²2 + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x² + 9x² + 4z² – 12xy + 12yz – 8zx

(iv) (3a – 7b – c)²
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca (3a – 7b – c)² =(3a)² + (-7b)² + (-c)² + 2(3a)(-7b)
+ 2(-7b)(-c) + 2(-c)(3a) = 9a² + 49b² + c² – 42ab – 14bc – 6ca.

(v) (-2x + 5y – 3z)²
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx (-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

(vi) \(\left(\frac{1}{4} a-\frac{1}{2} b+1\right)^{2}\)
(a + b + c)² = a² + b² + c² +2ab + 2bc + 2ca

Question 5.
Factorise :
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz- 8xz
Solution:
(i) 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.

(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz- 8xz

Question 6.
Write the following cubes in expanded form:
(i) (2x + 1)³
(ii) (2a – 3b)³
(iii) \(\left[\frac{3}{2} x+1\right]^{3}\)
(iv) \(\left[\mathrm{x}-\frac{2}{3} \mathrm{y}\right]^{3}\)
Solution:
(i) (2x + 1)³
As per Identity,
(a + b)³ = a³ + b³ + 3ab(a + b)
(2x + 1)³ = (2x)³ + (1)³ + 3(2x)(1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 1 + 12x² + 6x
(2x + 1)³ = 8x³ + 12x² + 6x + 1.

(ii) (2a – 3b)³
As per 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)
(2a – 3b)³= 8a³ – 27b³ – 36a²b + 54ab²

(iii) \(\left[\frac{3}{2} x+1\right]^{3}\)
As per Identity,
(a + b)³ = a³ – b³ – 3ab(a + b)

(iv) \(\left[\mathrm{x}-\frac{2}{3} \mathrm{y}\right]^{3}\)
As per Identity,
(a – b)³ = a³ – b³ – 3ab(a – b)

Question 7.
Evaluate the following using suitable identities :
(i) (99)³
(ii) (102)³
(iii) (998)³
Solution:
(i) (99)³ = (100 – 1)³
As per Identity,
(a – b)³ = a³ – b³ – 3ab(a – b)
(100 – 1)³ = (100)³ – (1)³ – 3(100)(1)(100- 1)
(99)³ = 1000000 – 1 – 300(100 – 1)
= 1000000 – 1 – 30000 + 300
= 970299
∴ (99)³ = 970299

(ii) (102)³ = (100 + 2)³
As per Identity,
(a + b)³ = a³ + b³ + 3ab(a + b)
(100 + 2)³ = (100)³ + (2)³ + 3( 100)(2)( 100 + 2)
= 1000000 + 8 + 600( 100 + 2)
= 1000000 + 8 + 60000 + 1200
(100 + 2)³ = 1061208
∴ (102)³= 1061208

(iii) (998)³ = (1000 – 2)³
As per Identity,
(a – b)³ = a³ – b³ – 3ab(a – b)
(1000 – 2)³ = (1000)³ – (2)3 – 3( 1000) (2)(1000 – 2)
= 1000000000 – 8 – 6000(1000 – 1)
= 1000000000 – 8 – 6000000 + 6000
= 994005992
∴ (998)³ = 994005992

Question 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) \(27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\)
Solution:
(i) 8a³ + b³ + 12a²b + 6ab²
= (2a)³ + (b)³ + 3(2a)(b)(2a + b)
= (2a + b)³ ∵ a³ + b³ + 3ab(a + b) = (a + b)³

(ii) 8a³ – b³ – 12ab² + 6ab³
= (2a)³ + (-b)³ + 3(2a)(-b)(2a – b)
= (2a – b)³ ∵ a³ – b³ – 3ab(a – b) = (a – b)³

(iii) 27 – 125a³ – 135a + 225a²
OR 125a³ – 225a² + 135a – 27
= (5a)³ + (-3)³ + 3(5a)(-3)(5a – 3)
= (5a – 3)³ ∵ a³ – b³ – 3ab(a – b) = (a – b)³

(iv) 64a³ – 27b³ – 144a²b + 108ab²
= (4a)³ + (-3b)³ + 3(4a)(-3b)(4a – 3b)
= (4a-3b)³ ∵ a³ – b³ – 3ab(a – b) = (a – b)³

Question 9.
Verify :
(i) x³ + y³ = (x + y) (x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution:
(i) 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
R.H.S. = x³ + y³
∴ L.H.S. = R.H.S.
∴ x³ + y³ = (x + y)(x² – xy + y²)

(ii) 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 – xy³ – y³
R.H.S. = x³ – y³
∴L.H.S. = R.H.S.
∴ x³ – y³ = (x – y)(x² + xy + y²)

Question 10.
Factorise each of the following :
(i) 27y³ + 125z³
(ii) 64m³ – 343n³
[Hint: See Question 9).
Solution:
(i) 27y³ + 125z³
= (3y)³ + (5z)³
= (3y + 5z) {(3y)² – (3y)(5z) + (5z)²}
= (3y + 5z) (9y² – 15yz + 25z²)

(ii) 64m³ – 343n³
= (4m)³ – (7n)³
= (4m – 7n) {(4m)² + (4m) (7n) + (7n)²}
= (4m – 7n) (17m² + 28mn + 49n²)

Question 11.
Factorise : 27x³ + y³ + z³ – 9xyz
Solution:
27x³ + y³ + z³ – 9xyz
= (3x)³ + (y)3³ + (z)³ – 3(3x)(y)(z)
= (3x + y + z) {(3x)² + (y)² + (z)² – (3x)(y) – (y)(z) – (z)(3x)}
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3zx)

Question 12.
Verify that x³ + y³ + z³ – 3xyz = \(\frac{1}{2}\) (x + y + z)[(x – y)² + (y – z)² + (z – x)²]
Solution:
L.H.S.= x³ + y³+ z³ – 3xyz
R.H.S. = \(\frac{1}{2}\) (x+y+z) [(x – y)² + (y – z)² + (z – x)²]
= \(\frac{1}{2}\) (x+y+z) [x² 2xy + y²+ y² – 2yz + z² + z² -2zx + x²]
= \(\frac{1}{2}\) (x+y+z) [2x²+ 2y²+ 2z²– 2xy – 2yz – 2zx]
= \(\frac{1}{2}\) (x+y+z) × (x² + y² + z² – xy – yz – zx)
= (x + y + z)(x² + y² + z² – xy – yz – zx)
R.H.S.= x³ + y³ + z³ – 3xyz
∴ L.H.S. = R.H.S.
∴ x³ + y³ + z³ – 3xyz = \(\frac{1}{2}\) (x + y + z)[(x – y)² + (y – z)² + (z – x)³]

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Solution:
x + y + z = 0
x + y = -z
by cubing on both sides,
(x + y)³ = (-z)3
x³+ y³ + 3xy(x + y) = -z³
x³ + y³ + 3xy (-z) = -z³
(∵ x + y =-z data given)
x³ + y³ – 3xyz = -z³
∴ x³ + y³ + z³ = 3xyz.

Question 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
Solution:
(i) (-12)³ + (7)³ + (5)³
Let -12 = x, 7 = y, 5 = z, then
x³ + y³ + z³ = 3xyz.
∴ (-12)³ + (7)³ + (5)³
= 3(-12)(7)(5)
= -36 × 35
= -1260.

(ii) (28)³ + (-15)³ + (-13)³
Let x = 28, y = -15, z = -13, then
x³ + y³ + z³ = 3xyz
∴ (28)³ + (-15)³ + (-13)³
= 3(28)(- 15)(-13)
= 84 × 195
= 16380

Question 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 : 35y² + 13y – 12
Solution:
(i) Area: 25a² – 35a + 12
= 25a² – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
= (5a -4) (5a – 3)
length × breadth = Area of rectangle
(5a – 4) (5a – 3) = 25a² – 35a + 12
∴ Length = (5a – 4)
Breadth = (5a – 3)

(ii) Area : 35y² + 13y – 12
Area of Rectangle – Length × Breadth
35y² + 13y – 12
35y² + 28y – 15y
7y(5y + 4) – 3(5y + 4)
(7y – 3) (5y + 4)
∴ Length = (7y – 3)
Breadth = (5y + 4)

Question 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
Solution:
(i) length × breadth × height = Volume of Cuboid
= 3x² – 12x
= 3x² – 12x
= 3x(x – 4)
= 3 × x × (x – 4)
l × b × h
∴ Length = 3
Breadth = x
Height = (x – 4)

(ii) length × breadth × height = Volume of Cuboid
= 12ky² + 8ky – 20k
12ky² + 8ky – 20k
= 4k(3y² + 2y – 5)
= 4k (3y² + 5y – 3y – 5)
= 4k (y(3y + 5) – 1 (3y + 5))
= 4k × (3y + 5) × (y – 1)
= l × b × h
∴ Length = 4k
Breadth = (3y + 5)
Height = (y – 1)

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