NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5.

- Polynomials Class 9 Ex 2.1
- Polynomials Class 9 Ex 2.2
- Polynomials Class 9 Ex 2.3
- Polynomials Class 9 Ex 2.4
- Polynomials Class 9 Ex 2.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Polynomials |

Exercise |
Ex 2.5 |

Number of Questions Solved |
16 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

**Ex 2.5 Class 9 Maths Question 1.**

**Use suitable identities to find the following products:**

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

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

(iii) (3x+4)(3x+2x)

(iv) (\( { y }^{ 2 }+\cfrac { 3 }{ 2 }\))( \( { y }^{ 2 }-\cfrac { 3 }{ 2 }\))

(v) (3-2x)(3+2x)

**Solution:
**

**Ex 2.5 Class 9 Maths Question 2.**

**Evaluate the following products without multiplying directly:**

(i) 103 x 107

(ii) 95 x 96

(iii) 104 x 96

**Solution:**

**(i)**103 x 107 = (100 + 3) (100 + 7)

= (100)

^{2}+ (3 + 7) (100) + 3 x 7

= 100 x 100 + (10)(100) + 21

= 10000 +1000 + 21 = 11021

**(ii)** 95 x 96 = (100 – 5) (100 – 4)

= (100)^{2} + (-5 – 4)(100) + (-5)(-4) – 100 x 100 + (-9)(100) + 20

= 10000 – 900 + 20 = 9120

**(iii)** 104 x 96 = (100 + 4)(100 – 4)

= (100)^{2} – (4)^{2
}= 10000 -16

= 9984

**Ex 2.5 Class 9 Maths Question 3.**

**Factorise the following using appropriate identities :
**(i) 9x

^{2}+ 6xy + y

^{2}

(ii) 4y

^{2}– 4y +1

(iii) x

^{2}–\(\cfrac { { y }^{ 2 } }{ 100 }\)

**Solution:**

**Ex 2.5 Class 9 Maths 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) \((\cfrac { 1 }{ 4 } a-\cfrac { 1 }{ 2 } b+1)\)

^{2 }

**Solution:**

**Ex 2.5 Class 9 Maths Question 5.**

**Factorise:**

(i) 4x^{2} + 9y^{2} + I62^{2} + 12xy – 24yz -16xz

(ii) 2x^{2}+ y^{2} + 82^{2} – 2\( \sqrt { 2 }\)xy + 4\( \sqrt { 2 }\)yz – 8×2

**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

= (-a\( \sqrt { 2 }\)x)^{2} + (y)^{2} + (2\( \sqrt { 2 }\)z)^{2} + (2 – \( \sqrt { 2 }\)x) (y) + 2(y) (2\( \sqrt { 2 }\)z) + 2(2\( \sqrt { 2 }\)z)(-\( \sqrt { 2 }\)x)

= (-\( \sqrt { 2 }\)x + y + 2\( \sqrt { 2 }\)z)^{2}

**Ex 2.5 Class 9 Maths Question 6.
Write the following cubes in expanded form :
**(i) (2x+1)

^{3}

**(ii) (2a-3b)**

^{ }^{3 }(iii) \((\cfrac { 3 }{ 2 } x+1)\)

^{3}

(iv) \((x-\cfrac { 2 }{ 3 } y)\)

^{3 }

**Solution:**

**Ex 2.5 Class 9 Maths Question 7.**

**Evaluate the following using suitable identities :**

(i) (99)

^{3}

(ii) (102)

^{3}

(iii) (998)

^{3 }

**Solution:**

**Ex 2.5 Class 9 Maths Question 8.**

**Factorise each of the following:**

(i) 8a^{3} + b^{3} + 12a^{2}6 + 6ab^{2
}(ii) 8a^{3} – b^{3} -12a^{2}6 + 6a6^{2
}(iii) 27 – 125a^{3} – 135 a + 225 a^{2 }64a^{3}

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

**Solution:
**

**Ex 2.5 Class 9 Maths Question 9.**

**Verify:**

(1) x

^{3}+ y

^{3}= Or + y)(x

^{2}-xy + y

^{2})

(ii) x

^{3}-y

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

^{2}+ xy + y

^{2})

**Solution:**

**(i)**We know that,

(x + y)

^{3}= x

^{3}+ y

^{3}+ 3xy(x + y)

=> x

^{3}+ y

^{3}= (x + y)

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

= (x + y)[(x + y)

^{2}-3xy]

= (x + y) [x

^{2}+ y

^{2}+ 2xy – 3x] = (x + y)[x

^{2}+ y

^{2}– xy]

= RHS

**Hence proved.**

**(ii)** We know that, (x – y)^{3} = x^{3} – y^{3} -3xy(x – y)

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

= (x-y)[(x – y)^{2} +3xy]

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

= (x — y)[x^{2} + y^{2} + xy]

= RHS **Hence proved.**

**Ex 2.5 Class 9 Maths Question 10.**

**Factorise each of the following:
**(i) 27y

^{3}+ 125z

^{3}

(ii) 64m

^{3}-343n

^{3 }[Hint: See question 9]

**Solution:**

**(i)**27 y

^{3}+125z

^{3}= (3y)

^{3}+ (5z)

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

^{2}– (3y)(5z) + (5z)

^{2}]

= (3y + 5z) (9 y

^{2}– 15yz + 25z

^{2})

**(ii)** 64m^{3} -343n^{3} = (4m)^{3} -(7n)^{3}

= (4m-7n)[(4m)^{2} + (4m)(7n) + (7n)^{2}]

= (4m – 7n)[16m^{2} + 28mn + 49n^{2}]

**Ex 2.5 Class 9 Maths Question 11.**

**Factorise :**

27x^{3} + y^{3} + z^{3} – 9xyz

**Solution:**

27x^{3} +y^{3} +z^{3} -9xyz = (3x)^{3} + y^{3} +z^{3} -3(3x)(y)(z)

= (3x + y + z)[(3x)^{2} + y^{2} + z^{2} – (3x)y – yz -z(3x)]

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

**Ex 2.5 Class 9 Maths Question 12.**

Verify that

x^{3} + y^{3} +z^{3} -3xyz = \(\cfrac { 1 }{ 2 } \) (x + y + z)[(x -y)^{2} +(y-z)^{2} +(z-x)^{2}]

**Solution:**

We have, x^{3} + y^{3} + z^{3} – 3xyz

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

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

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

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

= \(\frac { 1 }{ 2 } \)(x + y + z)[(x-y)^{2} + (y-z)^{2} +(z-x)^{2}]

**Ex 2.5 Class 9 Maths Question 13.**

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

**Solution:**

We know that,

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

= 0(x^{2} + y^{2} + z^{2} – xy- yz-zx) (∵ x + y + z = 0 given)

= 0

=> x^{3} + y^{3} + z^{3} = 3xyz ** Hence proved.**

**Ex 2.5 Class 9 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)** We know that, x^{3} + y^{3} + z^{3} – 3xyz

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

Also, we know that, if

x + y + z = 0

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

Given expression is (-12)^{3} + (7)^{3} + (5)^{3}.

Here, -12 + 7 + 5=0

∴ (-12)^{3} + (7)^{3} + (5)^{3} = 3 x (-12) x 7 x 5 = -1260

**
(ii)** Given expression is (28)

^{3}+ (-15)

^{3}+ (-13)

^{3 }Here, 28 + (-15) + (-13) = 28 -15-13 = 0

∴ (28)

^{3}+ (-15)

^{3}+ (-13)

^{3}= 3 x (28) x (-15) x (-13) = 16380

**Ex 2.5 Class 9 Maths Question 15.**

**Give possible expressions for the length and the 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)**We have,

Area of rectangle = 25a

^{2}– 35a +12 [by splitting the middle term]

= 25a

^{2}-20a-15a+12

= 5a(5a – 4) – 3(5a – 4)

= (5a-4)(5a-3)

Hence, possible expression for length = (5a – 3) and possible expression for breadth = (5a – 4)

**(ii)** We have,

Area of rectangle = 35y+ 13y -12

= 35y^{2} + (28 – 15)y -12

= 35y^{2} + 28y-15y-12

= (35y^{2} +28y)-(15y + 12)

= 7y(5y + 4) – 3(5y + 4)

= (7y – 3) (5y + 4)

**Ex 2.5 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)** We have,

Volume of cuboid = 3x^{2} -12x = 3x(x – 4)

So, the possible expressions for the dimensions of the cuboids are 3, x and x – 4.

[∵ volume of cuboid = length x breadth x height]

**(ii)** We have,

Volume of cuboid = 12ky^{2} + 8ky – 20k = 12ky^{2} + (20 -12)ky – 20k [by splitting the middle term]

= 12ky^{2} + 20 ky – 12ky – 20 k = 4ky(3y + 5) – 4k(3y + 5) = (3y + 5)(4ky – 4k)

= (3y + 5)4k(y -1) = 4k(3y + 5)(y -1)

So, the possible expressions for the dimensions of the cuboid are 4 k, 3y + 5 and y -1.

We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5, drop a comment below and we will get back to you at the earliest.