**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

**KSEEB Solutions For Class 9 Maths Polynomials 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) = x^{2} + (a + b)x + ab

(x + 4) (x + 10) = x^{2} + (4 + 10)x + (4 × 10)

= x^{2} + 14x + 40

(ii) (x + 8) (x- 10)

Suitable identity is (x + a) (x + b) = x^{2} + (a + b)x + ab

(x + 8) (x – 10)= (x)^{2} + (8 – 10)x + (8 × -10)

= x^{2} + (-2)x + (-80)

= x^{2} – 2x – 80

(iii) (3x + 4) (3x – 5)

Suitable identity is (x + a) (x + b) = x^{2} + (a + b)x + ab

(3x + 4) (3x – 5) = (3x)^{2} + (+4 – 5)3x + (4 x -5)

= 9x^{2} + (-1)3x + (-20)

= 9x^{2} – 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) = a^{2} – b^{2}

\(\begin{aligned}\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right) &=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2} \\ &=y^{4}-\frac{9}{4} \end{aligned}\)

(v) (3 – 2x) (3 + 2x)

Suitable identity is (a + b) (a – b) = a^{2} – b^{2}

(3 – 2x) (3 + 2x) = (3)^{2} – (2x)^{2}

= 9 – 4x^{2}.

**KSEEB Solutions For Class 9 Maths 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) = a^{2} – b^{2}

(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) = x^{2} + (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) = a^{2} – b^{2}

(100 + 4) (100 – 4) = (100)^{2} – (4)^{2}

= 10000 – 16

= 9984

**Exercise 4.5 Class 9 Polynomials Question 3.**

Factorise the following using appropriate identities :

(i) 9x^{2} + 6xy + y^{2}

(ii) 4y^{2} – 4y + 1

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

Solution:

(i) 9x^{2} + 6xy + y^{2}

= (3x)^{2} + 2(3x)(y) + (y)^{2}

= (3x + y)^{2} [∵ x^{2} + 2xy + y^{2} = (x + y)^{2}]

(ii) 4y^{2} – 4y + 1

= (2y)^{2} – 2(2y)(1) + (1)^{2}

= (2y – 1)^{2} [∵ x^{2} – 2xy + y^{2} = (x – y)^{2}]

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

**9th Maths Polynomials Exercise 4.5 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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

(x + 2y + 4z)^{2} =(x)^{2} + (2y)^{2} + (4z)^{2} + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

= x^{2} + 4y^{2} + 16z^{2} + 4xy + 16yz + 8zx

(ii) (2x – y + z)^{2}

(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx (2x – y + z)^{2}= (2x)^{2} + (-y)^{2} + (z)^{2} + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)

= 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4zx.

(iii) (-2x + 3y + 2z)^{2}

(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx (-2x + 3y + 2z)^{2
}= (-2x)^{2} + (3y)^{2}2 + (2z)^{2} + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)

= 4x^{2} + 9x^{2} + 4z^{2} – 12xy + 12yz – 8zx

(iv) (3a – 7b – c)^{2}

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca (3a – 7b – c)^{2} =(3a)^{2} + (-7b)^{2} + (-c)^{2} + 2(3a)(-7b)

+ 2(-7b)(-c) + 2(-c)(3a) = 9a^{2} + 49b^{2} + c^{2} – 42ab – 14bc – 6ca.

(v) (-2x + 5y – 3z)^{2}

(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx (-2x + 5y – 3z)^{2} = (-2x)^{2} + (5y)^{2} + (-3z)^{2} + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x) = 4x^{2} + 25y^{2} + 9z^{2} – 20xy – 30yz + 12zx

(vi) \(\left(\frac{1}{4} a-\frac{1}{2} b+1\right)^{2}\)

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} +2ab + 2bc + 2ca

**Class 9 Maths Chapter 4 Exercise 4.5 Solutions Question 5.**

Factorise :

(i) 4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

(ii) 2x^{2} + y^{2} + 8z^{2} – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz- 8xz

Solution:

(i) 4x^{2} + 9y^{2} + 16z^{2}+ 12xy – 24yz – 16xz

= (2x)^{2}+ (3y)^{2} + (-4z)^{2} + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)

= (2x + 3y – 4z)^{2}.

(ii) 2x^{2} + y^{2} + 8z^{2} – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz- 8xz

**KSEEB Solutions For Class 9th Maths Polynomials 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} = a^{3} + b^{3} + 3ab(a + b)

(2x + 1)^{3} = (2x)^{3} + (1)^{3} + 3(2x)(1)(2x + 1)

= 8x^{3} + 1 + 6x(2x + 1)

= 8x^{3} + 1 + 12x^{2} + 6x

(2x + 1)^{3} = 8x^{3} + 12x^{2} + 6x + 1.

(ii) (2a – 3b)^{3}

As per Identity,

(a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

(2a – 3b)^{3}= (2a)^{3}– (3b)^{3} – 3(2a)(3b)(2a-3b)

= 8a^{3} – 27b^{3} – 18ab(2a – 3b)

(2a – 3b)^{3}= 8a^{3} – 27b^{3} – 36a^{2}b + 54ab^{2}

(iii) \(\left[\frac{3}{2} x+1\right]^{3}\)

As per Identity,

(a + b)^{3} = a^{3} – b^{3} – 3ab(a + b)

(iv) \(\left[\mathrm{x}-\frac{2}{3} \mathrm{y}\right]^{3}\)

As per Identity,

(a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

**KSEEB Solutions For Class 9 Maths Polynomials Exercise 4.5 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} = a^{3} – b^{3} – 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} = a^{3} + b^{3} + 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} = a^{3} – b^{3} – 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

**KSEEB Solutions For Class 9 Maths Chapter 4 Question 8.**

Factorise each of the following :

(i) 8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}

(ii) 8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}

(iii) 27- 125a^{3}– 135a + 225a^{2}

(iv) 64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}

(v) \(27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\)

Solution:

(i) 8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}

= (2a)^{3} + (b)^{3} + 3(2a)(b)(2a + b)

= (2a + b)^{3} ∵ a^{3} + b^{3} + 3ab(a + b) = (a + b)^{3}

(ii) 8a^{3} – b^{3} – 12ab^{2} + 6ab^{3}

= (2a)^{3} + (-b)^{3} + 3(2a)(-b)(2a – b)

= (2a – b)^{3} ∵ a^{3} – b^{3} – 3ab(a – b) = (a – b)^{3}

(iii) 27 – 125a^{3} – 135a + 225a^{2}

OR 125a^{3} – 225a^{2} + 135a – 27

= (5a)^{3} + (-3)^{3} + 3(5a)(-3)(5a – 3)

= (5a – 3)^{3} ∵ a^{3} – b^{3} – 3ab(a – b) = (a – b)^{3}

(iv) 64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}

= (4a)^{3} + (-3b)^{3} + 3(4a)(-3b)(4a – 3b)

= (4a-3b)^{3} ∵ a^{3} – b^{3} – 3ab(a – b) = (a – b)^{3}

**Maths Exercise 4.5 Class 9 Question 9.**

Verify :

(i) x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

(ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

Solution:

(i) x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2})

L.H.S. = x^{3} + y^{3}

R.H.S. = (x + y)(x^{2} – xy + y^{2})

= x(x^{2} – xy + y^{2}) + y(x^{2} – xy + y^{2})

= x^{3} – x^{2}y + xy^{2} + x^{2}y – xy^{2} + y^{3}

R.H.S. = x^{3} + y^{3}

∴ L.H.S. = R.H.S.

∴ x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2})

(ii) x^{3} – y^{3} = (x – y)(x^{2} + xy + y^{2})

L.H.S. = x^{3} – y^{3}

R.H.S. = (x – y)(x^{2} + xy + y^{2})

= x(x^{2} + xy + y^{2}) – y(x^{2} + xy + y^{2})

= x^{3} + x^{2}y + xy^{2} – x^{2}y – xy^{3} – y^{3}

R.H.S. = x^{3} – y^{3}

∴L.H.S. = R.H.S.

∴ x^{3} – y^{3} = (x – y)(x^{2} + xy + y^{2})

**Polynomials Class 9 KSEEB Solutions Question 10.**

Factorise each of the following :

(i) 27y^{3} + 125z^{3}

(ii) 64m^{3} – 343n^{3}

[Hint: See Question 9).

Solution:

(i) 27y^{3} + 125z^{3}

= (3y)^{3} + (5z)^{3}

= (3y + 5z) {(3y)^{2} – (3y)(5z) + (5z)^{2}}

= (3y + 5z) (9y^{2} – 15yz + 25z^{2})

(ii) 64m^{3} – 343n^{3}

= (4m)^{3} – (7n)^{3}

= (4m – 7n) {(4m)^{2} + (4m) (7n) + (7n)^{2}}

= (4m – 7n) (17m^{2} + 28mn + 49n^{2})

Question 11.

Factorise : 27x^{3} + y^{3} + z^{3} – 9xyz

Solution:

27x^{3} + y^{3} + z^{3} – 9xyz

= (3x)^{3} + (y)3^{3} + (z)^{3} – 3(3x)(y)(z)

= (3x + y + z) {(3x)^{2} + (y)^{2} + (z)^{2} – (3x)(y) – (y)(z) – (z)(3x)}

= (3x + y + z) (9x^{2} + y^{2} + z^{2} – 3xy – yz – 3zx)

**KSEEB Maths Solutions For Class 9 Question 12.**

Verify that x^{3} + y^{3} + z^{3} – 3xyz = \(\frac{1}{2}\) (x + y + z)[(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]

Solution:

L.H.S.= x^{3} + y^{3}+ z^{3} – 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) [x^{2} 2xy + y^{2}+ y^{2} – 2yz + z^{2} + z^{2} -2zx + x^{2}]

= \(\frac{1}{2}\) (x+y+z) [2x^{2}+ 2y^{2}+ 2z^{2}– 2xy – 2yz – 2zx]

= \(\frac{1}{2}\) (x+y+z) × (x^{2} + y^{2} + z^{2} – xy – yz – zx)

= (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)

R.H.S.= x^{3} + y^{3} + z^{3} – 3xyz

∴ L.H.S. = R.H.S.

∴ x^{3} + y^{3} + z^{3} – 3xyz = \(\frac{1}{2}\) (x + y + z)[(x – y)^{2} + (y – z)^{2} + (z – x)^{3}]

**9th Standard Maths Exercise 4.5 Question 13.**

If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3xyz.

Solution:

x + y + z = 0

x + y = -z

by cubing on both sides,

(x + y)^{3} = (-z)3

x^{3}+ y^{3} + 3xy(x + y) = -z^{3}

x^{3} + y^{3} + 3xy (-z) = -z^{3}

(∵ x + y =-z data given)

x^{3} + y^{3} – 3xyz = -z^{3}

∴ x^{3} + y^{3} + z^{3} = 3xyz.

**KSEEB Solutions For Class 9th Maths 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

x^{3} + y^{3} + z^{3} = 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

x^{3} + y^{3} + z^{3} = 3xyz

∴ (28)^{3} + (-15)^{3} + (-13)^{3
}= 3(28)(- 15)(-13)

= 84 × 195

= 16380

**KSEEB 9th Maths Solutions Question 15.**

Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :

(i) Area: 25a^{2} – 35a +12

(ii) Area : 35y^{2} + 13y – 12

Solution:

(i) Area: 25a^{2} – 35a + 12

= 25a^{2} – 20a – 15a + 12

= 5a(5a – 4) – 3(5a – 4)

= (5a -4) (5a – 3)

length × breadth = Area of rectangle

(5a – 4) (5a – 3) = 25a^{2} – 35a + 12

∴ Length = (5a – 4)

Breadth = (5a – 3)

(ii) Area : 35y^{2} + 13y – 12

Area of Rectangle – Length × Breadth

35y^{2} + 13y – 12

35y^{2} + 28y – 15y

7y(5y + 4) – 3(5y + 4)

(7y – 3) (5y + 4)

∴ Length = (7y – 3)

Breadth = (5y + 4)

**KSEEB Solutions Class 9 Maths Question 16.**

What are the possible expressions for the dimensions of the cuboids whose volumes are given below :

(i) Volume: 3x^{2} – 12x

(ii) Volume: 12ky^{2} + 8ky – 20k

Solution:

(i) length × breadth × height = Volume of Cuboid

= 3x^{2} – 12x

= 3x^{2} – 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

= 12ky^{2} + 8ky – 20k

12ky^{2} + 8ky – 20k

= 4k(3y^{2} + 2y – 5)

= 4k (3y^{2} + 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)

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.5 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.5, drop a comment below and we will get back to you at the earliest.