Do I Need to Practice All Questions in NCERT Solutions for Class 9 Maths Polynomials?

Yes, practising all the questions is beneficial as they cover various algebraic concepts related to polynomials. Start by solving the examples to understand the steps involved in solving polynomial problems.

Do you need help learning the NCERT Solutions for Class 9 Maths Chapter 2?

Using ALLEN's NCERT solutions for Class 9 Maths Chapter 2 on a regular basis will help you achieve high exam scores. You can learn how to format your exam answers from the solutions. For students to easily understand the concepts, every little detail is covered in an easy-to-understand manner.

What are the Key Formulas in NCERT Solutions for Class 9 Maths Chapter 2?

The chapter includes important formulas such as the factor theorem, remainder theorem, and key algebraic identities. These are essential for understanding and solving polynomial problems effectively. These are essential for understanding and solving polynomial problems effectively. These are essential for understanding and solving polynomial problems effectively.

NCERT Solutions Class 9 Maths Chapter 2 Polynomial

Use suitable identities to find the following products : (i)

(x + 4) (x + 10)

(ii)

(x + 8) (x - 10)

(iii)

(3 x + 4) (3 x - 5)

(iv)

(y^{2} + \frac{3}{2}) (y^{2} - \frac{3}{2})

(v)

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

Sol. (i)

(x + 4) (x + 10)

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

= x^{2} + 14 x + 40

(ii)

(x + 8) (x - 10)

= (x + 8) {x + (- 10)}

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

= x^{2} - 2 x - 80

(iii)

(3 x + 4) (3 x - 5)

= (3 x + 4) (3 x - 5) = (3 x + 4) (3 x + (- 5))

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

= 9 x^{2} - 3 x - 20

(iv)

(y^{2} + \frac{3}{2}) (y^{2} - \frac{3}{2})

Let,

y^{2} = x

\Rightarrow (y^{2} + \frac{3}{2}) (y^{2} - \frac{3}{2}) = (x + \frac{3}{2}) (x - \frac{3}{2}) = x^{2} - \frac{9}{4}

[using identity

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

] Substituting

x = y^{2}

, we get

= (y^{2})^{2} - \frac{9}{4}

= y^{4} - \frac{9}{4}

(v)

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

(3)^{2} - (2 x)^{2} = 9 - 4 x^{2}

[using identity

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

]

Evaluate the following products without multiplying directly: (i)

103 \times 107

(ii)

95 \times 96

(iii)

104 \times 96

Sol. (i)

103 \times 107 = (100 + 3) \times (100 + 7)

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

= 10000 + 1000 + 21 = 11021

Alternate solution:

103 \times 107 = (105 - 2) \times (105 + 2)

= (105)^{2} - (2)^{2} = (100 + 5)^{2} - 4

= (100)^{2} + 2 (100) (5) + (5)^{2} - 4

= 10000 + 1000 + 25 - 4

= 11021

. (ii)

95 \times 96

= (90 + 5) \times (90 + 6)

= (90)^{2} + (5 + 6) 90 + (5) (6)

= 8100 + 990 + 30 = 9120

(iii)

104 \times 96

= (100 + 4) \times (100 - 4)

[using identity

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

]

= (100)^{2} - (4)^{2} = 10000 - 16

= 9984

Factorise the following using appropriate identities: (i)

9 x^{2} + 6 x y + y^{2}

(ii)

4 y^{2} - 4 y + 1

(iii)

x^{2} - \frac{y ^{2}}{100}

Sol. (i)

9 x^{2} + 6 x y + y^{2}

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

= (3 x + y)^{2}

= (3 x + y) (3 x + y)

(ii)

4 y^{2} - 4 y + 1

= (2 y)^{2} - 2 (2 y) (1) + (1)^{2}

= (2 y - 1)^{2} = (2 y - 1) (2 y - 1)

(iii)

x^{2} - \frac{y ^{2}}{100}

[using identity

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

]

x^{2} - (\frac{y}{10})^{2} = (x + \frac{y}{10}) (x - \frac{y}{10})

Expand each of the following using suitable identities : (i)

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

(ii)

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

(iii)

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

(iv)

(3 a - 7 b - c)^{2}

(v)

(- 2 x + 5 y - 3 z)^{2}

(vi)

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

Sol. (i)

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

2 (x) (2 y) + 2 (2 y) (4 z) + 2 (4 z) (x)

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

(ii)

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

= (2 x - y + z) (2 x - y + z)

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

y)

(z) + 2 (z) (2 x)

= 4 x^{2} + y^{2} + z^{2} - 4 x y - 2 yz + 4 z x

(iii)

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

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

2 (- 2 x) (2 z) + 2 (3 y) (2 z)

= 4 x^{2} + 9 y^{2} + 4 z^{2} - 12 x y - 8 x z + 12 yz

(iv)

(3 a - 7 b - c)^{2} = (3 a - 7 b - c) (3 a - 7 b - c)

= (3 a)^{2} + (- 7 b)^{2} + (- c)^{2} + 2 (3 a) (- 7 b) +

2 (3 a) (- c) + 2 (- 7 b) (- c)

= 9 a^{2} + 49 b^{2} + c^{2} - 42 ab - 6 a c + 14 b c

(v)

(- 2 x + 5 y - 3 z)^{2}

= (- 2 x + 5 y - 3 z) (- 2 x + 5 y - 3 z)

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

2 (- 2 x) (- 3 z) + 2 (- 3 z) (5 y)

= 4 x^{2} + 25 y^{2} + 9 z^{2} - 20 x y + 12 x z - 30 yz

(vi)

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

= (\frac{1}{4} a - \frac{1}{2} b + 1) (\frac{1}{4} a - \frac{1}{2} b + 1)

= (\frac{1}{4} a)^{2} + (- \frac{1}{2} b)^{2} + (1)^{2} + 2 (\frac{1}{4} a) (- \frac{1}{2} b)

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

= \frac{1}{16} a^{2} + \frac{1}{4} b^{2} + 1 - \frac{1}{4} ab - b + \frac{1}{2} a

Factorise : (i)

4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 yz - 16 x z

(ii)

2 x^{2} + y^{2} + 8 z^{2} - 22 x y + 42 yz - 8 x z

Sol. (i)

4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 yz - 16 x z

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

2 (3 y) (- 4 z) + 2 (- 4 z) (2 x)

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

= (2 x + 3 y - 4 z) (2 x + 3 y - 4 z)

(ii)

2 x^{2} + y^{2} + 8 z^{2} - 22 x y + 42 yz - 8 x z

= (- 2 x)^{2} + y^{2} + (22 z)^{2} + 2 (- 2 x) y +

2 y (22 z) + 2 (22 z) (- 2 x)

= (- 2 x + y + 22 z)^{2}

Write the following cubes in expanded form : (i)

(2 x + 1)^{3}

(ii)

(2 a - 3 b)^{3}

(iii)

(\frac{3}{2} x + 1)^{3}

(iv)

(x - \frac{2}{3} y)^{3}

Sol. (i)

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

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

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

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

(ii)

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

= 8 a^{3} - 27 b^{3} - 18 ab (2 a - 3 b)

= 8 a^{3} - 27 b^{3} - 36 a^{2} b + 54 a b^{2}

(iii)

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

= \frac{27}{8} x^{3} + 1 + \frac{27}{4} x^{2} + \frac{9}{2} x

= \frac{27}{8} x^{3} + \frac{27}{4} x^{2} + \frac{9}{2} x + 1

(iv)

(x - \frac{2}{3} y)^{3} = x^{3} - (\frac{2}{3} y)^{3} - 3 x (\frac{2}{3} y) (x - \frac{2}{3} y)

= x^{3} - \frac{8}{27} y^{3} - 2 xy (x - \frac{2}{3} y)

= x^{3} - \frac{8}{27} y^{3} - 2 x^{2} y + \frac{4}{3} x^{2}

Evaluate the following using suitable identities : (i)

(99)^{3}

(ii)

(102)^{3}

(iii)

(998)^{3}

Sol. (i)

(99)^{3} = (100 - 1)^{3}

= (100)^{3} - (1)^{3} - 3 (100) (1) (100 - 1) = 1000000 - 1 - 300 (100 - 1) = 1000000 - 1 - 30000 + 300 = 970299

(ii)

(102)^{3} = (100 + 2)^{3}

= (100)^{3} + (2)^{3} + 3 (100) (2) (100 + 2) = 1000000 + 8 + 600 (100 + 2) = 1000000 + 8 + 60000 + 1200 = 1061208

(iii)

(998)^{3} = (1000 - 2)^{3}

= (1000)^{3} - (2)^{3} - 3 (1000) (2) (1000 - 2) = 1000000000 - 8 - 6000 (1000 - 2) = 994011992

Factorise each of the following : (i)

8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2}

(ii)

8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}

(iii)

27 - 125 a^{3} - 135 a + 225 a^{2}

(iv)

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a^{2}

(v)

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

Sol. (i)

8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2}

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

(ii)

8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}

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

= (2 a - b)^{3}

(iii)

27 - 125 a^{3} - 135 a + 225 a^{2}

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

= (3 - 5 a)^{3}

(iv)

64 a^{3} - 27 b^{3} - 144 a^{2} b + 180 a b^{2}

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

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

(v)

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

= (3 p)^{3} - (\frac{1}{6})^{3} - 3 (3 p) (\frac{1}{6}) (3 p - \frac{1}{6})

= (3 p - \frac{1}{6})^{3} = (3 p - \frac{1}{6}) (3 p - \frac{1}{6}) (3 p - \frac{1}{6})

Verify: (i)

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

(ii)

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

Sol. (i)

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

\Rightarrow x^{3} + y^{3} = (x + y)^{3} - 3 x y (x + y)

\Rightarrow x^{3} + y^{3} = (x + y) {(x + y)^{2} - 3 x y}

\Rightarrow x^{3} + y^{3} = (x + y) (x^{2} + 2 x y + y^{2} - 3 x y)

\Rightarrow x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(ii)

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

\Rightarrow x^{3} - y^{3} = (x - y)^{3} + 3 x y (x - y)

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

\Rightarrow x^{3} - y^{3} = (x - y) (x^{2} + y^{2} - 2 x y + 3 x y)

\Rightarrow x^{3} - y^{3} = (x - y) (x^{2} + y^{2} + x y)

Factorise each of the following : (i)

27 y^{3} + 125 z^{3}

(ii)

64 m^{3} - 343 n^{3}

Sol. (i)

27 y^{3} + 125 z^{3} = (3 y)^{3} + (5 z)^{3}

= (3 y + 5 z) {(3 y)^{2} - (3 y) (5 z) + (5 z)^{2}} = (3 y + 5 z) (9 y^{2} - 15 yz + 25 z^{2})

(ii)

64 m^{3} - 343 n^{3}

= (4 m)^{3} - (7 n)^{3}

= (4 m - 7 n) {16 m^{2} + 4 m .7 n + (7 n)^{2}}

= (4 m - 7 n) (16 m^{2} + 28 mn + 49 n^{2})

Factorise:

27 x^{3} + y^{3} + z^{3} - 9 x yz

Sol.

27 x^{3} + y^{3} + z^{3} - 9 x yz

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

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

(y)

(z) - (z) (3 x))

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

Verify that

x^{3} + y^{3} + z^{3} - 3 x yz = \frac{1}{2} (x + y + z)

[(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

Sol.

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

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

+ z^{2}) + (z^{2} - 2 z x + x^{2})]

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

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

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

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

If

x + y + z = 0

, show that

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

Sol. We know

x^{3} + y^{3} + z^{3} - 3 x yz

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

Given:

x + y + z = 0

= (0) (x^{2} + y^{2} + z^{2} - x y - yz - z x) = 0

or

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

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}

Sol. (i)

(- 12)^{3} + (7)^{3} + (5)^{3}

- 12 + 7 + 5 = 0

(- 12)^{3} + (7)^{3} + (5)^{3}

= 3 (- 12) (7) (5) = - 1260

[using identity] if

a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 abc

(ii)

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

28 - 15 - 13 = 0

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

= 3 (28) (- 15) (- 13) = 16380

[using identity] if

a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 abc

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

= 25 a^{2} - 35 a + 12

(ii) Area

= 35 y^{2} + 13 y - 12

Sol. (i) Area

= 25 a^{2} - 35 a + 12

= 25 a^{2} - 20 a - 15 a + 12 = 5 a (5 a - 4) - 3 (5 a - 4) = (5 a - 3) (5 a - 4)

Here, Length = 5 a - 3, Breadth = 5 a - 4

(ii)

35 y^{2} + 13 y - 12

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

= 7 y (5 y + 4) - 3 (5 y + 4)

= (5 y + 4) (7 y - 3)

Here, Length

= 5 y + 4

, Breadth

= 7 y - 3

.

What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume:

3 x^{2} - 12 x

(ii) Volume:

12 ky^{2} + 8 xy - 20 k

Sol. (i) Volume

= 3 x^{2} - 12 x

= 3 x (x - 4) = 3 \times x \times (x - 4)

∴

Dimensions are 3 units, x -units and (

x - 4

) units (ii)

12 ky^{2} + 8 ky - 20 k

= 4 k (3 y^{2} + 2 y - 5) = 4 k (3 y^{2} + 5 y - 3 y - 5)

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

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

∴

Dimensions of cuboid are

4 k, 3 y + 5, y - 1

NCERT Solutions Class 9 Maths Chapter 2 : All Exercises	Total Questions
Class 9 Maths Chapter 2 Exercise 2.1 Solutions	5 Question
Class 9 Maths Chapter 2 Exercise 2.2 Solutions	4 Questions
Class 9 Maths Chapter 2 Exercise 2.3 Solutions	6 Questions
Class 9 Maths Chapter 2 Exercise 2.4 Solutions	12 Questions
Total	27

Frequently Asked Questions:

What is a polynomial in Class 9 Maths Solutions?

Do I Need to Practice All Questions in NCERT Solutions for Class 9 Maths Polynomials?

Do you need help learning the NCERT Solutions for Class 9 Maths Chapter 2?

What are the Key Formulas in NCERT Solutions for Class 9 Maths Chapter 2?

How do you identify whether an expression is a polynomial or not?

What are the types of polynomials in Class 9?

What is an example of a polynomial in Class 9?

What is the degree of a polynomial?

What is a zero polynomial and its degree?

What are the zeroes of a polynomial?

How many zeroes can a polynomial have?

What is factorization in polynomials?

Join ALLEN!

Related Articles:-

NCERT Solutions Class 9 Maths Chapter 1 Number Systems

NCERT Solutions Class 9 Maths Chapter 7 Geometry of Triangles

NCERT Solutions Class 9 Maths Chapter 9 Circles

NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume

NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions Class 9 Maths Chapter 12 Statistics

Frequently Asked Questions:

What is a polynomial in Class 9 Maths Solutions?

Do I Need to Practice All Questions in NCERT Solutions for Class 9 Maths Polynomials?

Do you need help learning the NCERT Solutions for Class 9 Maths Chapter 2?

What are the Key Formulas in NCERT Solutions for Class 9 Maths Chapter 2?

How do you identify whether an expression is a polynomial or not?

What are the types of polynomials in Class 9?

What is an example of a polynomial in Class 9?

What is the degree of a polynomial?

What is a zero polynomial and its degree?

What are the zeroes of a polynomial?

How many zeroes can a polynomial have?

What is factorization in polynomials?

Join ALLEN!

Related Articles:-

NCERT Solutions Class 9 Maths Chapter 1 Number Systems

NCERT Solutions Class 9 Maths Chapter 7 Geometry of Triangles

NCERT Solutions Class 9 Maths Chapter 9 Circles

NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume

NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions Class 9 Maths Chapter 12 Statistics

NCERT Solutions Class 9 Maths Chapter 2 Polynomial

1.0Download NCERT Solutions Class 9 Maths Chapter 2 PDF Online

2.0NCERT Solutions for Class 9 Maths Chapter 2 Polynomials: Overview

3.0Chapter 2 Polynomials Subtopics of NCERT Class 9 Maths

4.0Polynomials in NCERT Solutions for Class 9 Maths, Chapter 2: All Exercises

5.0NCERT Questions with Solutions for Class 9 Maths Chapter 2 - Detailed Solutions

Exercise : 2.1

Exercise : 2.2

Exercise : 2.3

Exercise : 2.4

Exercise : 2.5

6.0Advantages of NCERT Solutions for Class 9 Maths Chapter 2