# Extra Questions: Class 9 Maths Chapter 2 Polynomials

NCERT Class 9 Maths Chapter 2 Polynomials :

- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.1
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.2
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.3
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.4
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.5
- Extra Questions Class 9 Maths Chapter 2 Polynomials

**Question 1.**** **Which of the following expressions are polynomials?

**(i)** x^{5} - 2x^{3} + + x + 7

**Solution :**

We have x^{5} - 2x^{3} + + x + 7

The highest degree of the variable = 5

**(ii)** \(\displaystyle \frac{1}{{\sqrt{2}}}{{x}^{2}}\,-\,\sqrt{2}x\,+\,2\)

**Solution :**

We have \(\displaystyle \frac{1}{{\sqrt{2}}}{{x}^{2}}\,-\,\sqrt{2}x\,+\,2\)

The highest degree of the variable = 2

**Question 2. **Coefficient of x in √3 - 2√2x + 4x^{2}

**Solution :**

We have √3 - 2√2x + 4x^{2}

The coefficient of x is 2√2x

**Question 3. **Classify the following as linear, quadratic, and cubic polynomials:

**(i)** 2x^{2 } + 4x

**Solution :**

We have ** **2x^{2 } + 4x

It is a quadratic polynomial.

**Question 4. **If p(x) = 5 - 4x + 2x^{2} , find **(i) **p(0) ** (ii) **p(3) **(iii) **p(-2)

**Solution :**

Given 5 - 4x + 2x^{2}

[ After put x = 0 ]

5 - 4(0) + 2(0)^{2}

= 5 - 0 + 2 × 0

= 5 - 0 + 0

= 5

[ After put x = 3 ]

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

= 5 - 12 + 2 × 9

= 5 - 12 + 18

= 23 - 12

= 11

[ After put x = -2 ]

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

= 5 + 8 + 2 × 4

= 5 + 8 + 8

= 21

After put p(0), p(3) and p(-2) we get 5, 11 and 21.

**Question 5. **Find the zero of the polynomial:

**(i) **p(x) = x - 5

**Solution :**

Given x - 5

x - 5 = 0

= x = 5

**(ii) **g(x) = 5 - 4x

Given 5 - 4x

5 - 4x = 0

= - 4x = - 5

= x = \(\displaystyle \frac{{-\,5}}{{-\,4}}\)

= x =

= x = \(\displaystyle \frac{{\,5}}{4}\)

**Question 6. **Using remainder theorem, find the remainder when.

**(i) **( x^{3} - 6x^{2} + 9x + 3 ) is divided by ( x - 1 )

**Solution :**

Let p(x) = x^{3} - 6x^{2} + 9x + 3

So, x - 1 = 0

x = 1

Using remainder theorem,

when p(x) = x^{3} - 6x^{2} + 9x + 3

than p ( 1 ) = ( 1 )^{3} - 6( 1 )^{2} + 9 × 1 + 3

= 1 - 6 × 1 + 9 + 3

= 1 - 6 + 9 + 3

= 13 - 6

= 7

So, answer = 7

**Question 7.** Using factor theorem, show that:

**(i) **( x - 2 ) is a factor of ( x^{3} - 8 )

**Solution :**

Let p(x) = x^{3} - 8

So, x - 2 = 0

x = 2

Using remainder theorem,

when p(x) = x^{3} - 8

than p ( 2 ) = ( 2 )^{3} - 8

= 8 - 8

= 0

So, p(x) is a factor of ( x^{3} - 8 )

**Question 8.** Find the value of k for which ( x - 1 ) is a factor of ( 2x^{3} + 9x^{2} + x + k )

**Solution :**

Let p(x) = 2x^{3} + 9x^{2} + x + k

Given x - 1 is factor of p(x)

So, x - 1 = 0

x = 1

Using the factor theorem,

when p( x ) = 2x^{3} + 9x^{2} + x + k

than p(1) = 2(1)^{3} + 9(1)^{2} + 1 + k

= 2 × 1 + 9 × 1 + 1 + k = 0

= 2 + 9 + 1 + k = 0

= 12k = 0

= k = -12

The value of k = -12

**Question 9. **Factorize :

**(i)** 18x^{2}y - 24xyz

**Solution :**

Given 18x^{2}y - 24xyz

18x^{2}y - 24xyz [ find the common ]

= 6xy ( 3x - 4z )

So, answer = 6xy ( 3x - 4z )

**(ii) **x^{3} - 2x^{2}y + 3xy^{2} - 6y^{3}

**Solution :**

Given x^{3} - 2x^{2}y + 3xy^{2} - 6y^{3}

x^{3} - 2x^{2}y + 3xy^{2} - 6y^{3 } [ find the common ]

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

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

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

**Question 10. **Factorize :

**(i) **25x^{2} - 64y^{2}

**Solution :**

Given 25x^{2} - 64y^{2}

[ Using identity = ( a + b ) ( a - b ) = a^{2} - b^{2} ]

25x^{2} - 64y^{2}

= ( 5x + 8y ) ( 5x - 8y )

∴ 25x^{2} - 64y^{2} = ( 5x + 8y ) ( 5x - 8y )

**(ii) **( 4x - 2y )^{3}

**Solution :**

We have ( 4x - 2y )^{3}

[ using identity = ( x - y )^{3} = x^{3} - y^{3} - 3xy ( x - y ) ]

( 4x - 2y )^{3}

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

= 64x^{3} - 8y^{3} - 24xy ( 4x - 2y )

= 64x^{3} - 8y^{3} - 96x^{2}y + 48xy^{2}

So, answer = 64x^{3} - 8y^{3} - 96x^{2}y + 48xy^{2}

**Question 11. **Factorize :

**(i) **x^{2} + 18x + 32

**Solution :**

Given x^{2} + 18x + 32

[ Using the splitting the middle term method ]

So, x^{2} + 18x + 32

= x^{2} + 16x + 2x + 32 [ find the common ]

= x ( x + 16 ) + 2 ( x + 16 )

= ( x + 2 ) ( x + 16 )

**(ii)** 7x + 49x + 84

**Solution :**

Given 7x + 49x + 84

[ Using the splitting the middle term method ]

So, 7x + 49x + 84

= 7x + 28x + 21x + 84

= 7x ( 1 + 4 ) + 1x ( 1 + 4 )

= ( 7x + x ) ( 1 + 4 )

**(iii) **√2x^{2} + 3x + √2

**Solution :**

Given √2x^{2} + 3x + √2

[ Using the splitting the middle term method ]

√2x^{2} + 2x + x + √2

= √2x ( x + √2 ) + 1 ( x + √2 )

= ( x + √2 ) ( √2x + 1 )

**Question 12. **Expand :

**(i) **( a + 2b + 5c )^{2}

**Solution :**

We have ( a + 2b + 5c )^{2}

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

( a + 2b + 5c )^{2}

= ( a )^{2} + ( 2b )^{2} + ( 5c )^{2} + 2 × a × 2b + 2 × 2b × 5c + 2 × 5c × a

= a^{2} + 4b^{2} + 25c^{2} + 4ab + 20bc + 10ac

So, answer = a^{2} + 4b^{2} + 25c^{2} + 4ab + 20bc + 10ac

**(ii) **( a - 2b - 3c )^{2}

**Solution :**

We have** **( a - 2b - 3c )^{2}

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

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

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

= a^{2} + 4b^{2} + 9c^{2} - 4ab + 12bc - 6ac

So, answer = a^{2} + 4b^{2} + 9c^{2} - 4ab + 12bc - 6ac

**Question 13. **Factorize :

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

**Solution :**

We have 9x^{2} + 16y^{2} + 4z^{2} - 24xy + 16yz - 12xz

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

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

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

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

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

So, answer = ( -3x + 4y + 2z ) ( -3x + 4y + 2z )

**(ii) **4x^{2} + 9y^{2} + 16z^{2} + 12xy - 24yz - 16xz

**Solution :**

We have 4x^{2} + 9y^{2} + 16z^{2} + 12xy - 24yz - 16xz

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

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}

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

So, answer = ( 2x + 3y - 4z ) ( 2x + 3y - 4z )

**Question 14. **Evaluate :

**(i) **( 99 )^{2}

**Solution :**

We have ( 99 )^{2}

( 100 - 1 )^{2}

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

( 100 - 1 )^{2}

= ( 100 )^{2} - 2 × 100 × 1 + ( 1 )^{2}

= 10000 - 200 + 1

= 10001 - 200

= 9801

So, answer = 9801

**(ii) **991 × 1009

**Solution :**

We have 991 × 1009

( 1000 - 9 ) × ( 1000 + 9 )

[ using identity = x^{2} - y^{2} = ( x + y ) ( x - y) ]

( 1000 - 9 ) × ( 1000 + 9 )

= ( 1000 )^{2} - ( 9 )^{2}

= 1000000 - 81

= 999919

So, answer = 999919

**(iii) **117 × 83

**Solution :**

We have ** **117 × 83

( 100 + 17 ) × ( 100 - 17 )

[ using identity = x^{2} - y^{2} = ( x + y ) ( x - y) ]

( 100 + 17 ) × ( 100 - 17 )

= ( 100 )^{2} - ( 17 )^{2}

= 10000 - 289

= 9711

So, answer = 9711

**Question 15. **Expand :

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

**Solution :**

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

[ using identity = ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y ) ]

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

= \(\displaystyle {{\left( {\frac{2}{3}x} \right)}^{3}}\) + ( 1 )^{3} + 3 × \(\displaystyle \left( {\frac{2}{3}x} \right)\) × 1 ( \(\displaystyle \left( {\frac{2}{3}x} \right)\) + 1 )

=

= \(\displaystyle \frac{8}{{27}}{{x}^{3}}\,+\,1+\,2x\,\times \,\left( {\frac{2}{3}x\,+\,1} \right)\)

= \(\displaystyle \frac{8}{{27}}{{x}^{3}}\,+\,1+\,\frac{4}{3}{{x}^{2}}\,+\,2x\)

So, answer = \(\displaystyle \frac{8}{{27}}{{x}^{3}}\,+\,1+\,\frac{4}{3}{{x}^{2}}\,+\,2x\)

**(ii) **( 3a - 2b )^{3}

**Solution :**

We have ( 3a - 2b )^{3}

[ using identity = ( x - y )^{3} = x^{3} - y^{3} - 3xy ( x - y ) ]

( 3a - 2b )^{3}

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

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

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

So, answer = 27a^{3} - 8b^{3} - 54a^{2}b + 36ab^{2}

**Question 16. **Evaluate :

**(i) **( 95 )^{3}

**Solution :**

We have ( 95 )^{3}

( 100 - 5 )^{3}

[ using identity = ( x - y )^{3} = x^{3} - y^{3} - 3xy ( x - y ) ]

( 100 - 5 )^{3}

= ( 100 )^{3} - ( 5 )^{3} - 3 × 100 × 5 ( 100 - 5 )

= 1000000 - 125 - 1500 ( 100 - 5 )

= 1000000 - 125 - 150000 + 7500

= 1007500 - 150125

= 857375

So, answer = 857375

**(ii) **( 0.2 )^{3} - ( 0. 3 )^{3} + ( 0.1 )^{3}

**Solution :**

We have ( 0.2 )^{3} - ( 0. 3 )^{3} + ( 0.1 )^{3}

= 0.008 - 0.0027 + 0.001

= 0.009 - 0.027

= 0.018

So, answer = 0.018

**Question 17. **Factorize :

**(i) **125a^{3} + b^{3} + 64c^{3} - 60abc

**Solution :**

We have 125a^{3} + b^{3} + 64c^{3} - 60abc

[ using identity = x^{3} + y^{3} + z^{3} - 3xyz = ( x + y + z ) ( x^{2} + y^{2} + z^{2} - xy - yz - zx ) ]

125a^{3} + b^{3} + 64c^{3} - 60abc

= ( 5a + b + 4c ) ( 5a^{2} + b^{2} + 4c^{2} - 5a × b + b × -4c - 4c × 5a )

= ( 5a + b + 4c ) ( 25a^{2} + b^{2} - 5ab - 4bc - 20ca )

So, answer = ( 5a + b + 4c ) ( 25a^{2} + b^{2} - 5ab - 4bc - 20ca )

**(ii) **216 + 27b^{3} + 8c^{3} - 108bc

We have 216 + 27b^{3} + 8c^{3} - 108bc

[ using identity = x^{3} + y^{3} + z^{3} - 3xyz = ( x + y + z ) ( x^{2} + y^{2} + z^{2} - xy - yz - zx ) ]

216 + 27b^{3} + 8c^{3} - 108bc

= ( 6 + 3b + 2c ) ( 6^{2} + 3b2 + 2c^{2} - 6 × 3b - 3b × 2c - 2c × 6 )

= ( 6 + 3b + 2c ) ( 36 + 9b^{2} + 4c^{2} - 18b - 6bc - 12c )

So, answer = ( 6 + 3b + 2c ) ( 36 + 9b^{2} + 4c^{2} - 18b - 6bc - 12c )

**Question 18.** Find the product :

**(i) **( x - 2y + 3 ) ( x2 + 4y2 + 2xy - 3x + 6y + 9 )

We have ( x - 2y + 3 ) ( x2 + 4y2 + 2xy - 3x + 6y + 9 )

Let, First bracket = ( x - 2y + 3 )

Second bracket = ( x2 + 4y2 + 2xy - 3x + 6y + 9 )

[ we have multiply to the first bracket by the second bracket ]

( x - 2y + 3 ) ( x2 + 4y2 + 2xy - 3x + 6y + 9 )

= x ( x2 + 4y2 + 2xy - 3x + 6y + 9 ) - 2y ( x2 + 4y2 + 2xy - 3x + 6y + 9 ) + 3 ( x2 + 4y2 + 2xy - 3x + 6y + 9 )

= x3 + 4xy2 + 2x2y - 3x2 + 6xy + 9x - 2x2y - 8y3 - 4xy2 + 6xy - 12y2 - 18y + 3x2 + 12y2 + 6xy - 9x + 18y + 27

= x3 + 6xy + 6xy + 6xy - 8y3 + 27

= x3 + 18xy - 8y3 + 27

= x3 - 8y3 + 27 + 18xy

So, answer = x3 - 8y3 + 27 + 18xy

**Question 19. **Write the following in the expanded form :

**(i) **( 2a - 3b - c )2

**Solution :**

We have ( 2a - 3b - c )2

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

( 2a - 3b - c )2

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

= 4a^{2} + 9b^{2} + c^{2} - 12ab + 6bc - 4ca

So, answer = 4a^{2} + 9b^{2} + c^{2} - 12ab + 6bc - 4ca

**Question 20. **Simplify each of the following :

**(i) **( 4x + 2y )3 + ( 4x - 2y )3

**Solution :**

We have ( 4x + 2y )3 + ( 4x - 2y )3

[ using identity = ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y ) and ( x - y )^{3} = x^{3} - y^{3} - 3xy ( x - y ) ]

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

= [ { ( 4x )3 + ( 2y )3 + 3 × 4x × 2y ( 4x + 2y ) } + { ( 4x )3 - ( 2y )3 - 3 × 4x × 2y ( 4x - 2y ) } ]

= 64x3 + 8y3 + 24xy ( 4x + 2y ) + 64x3 - 8y3 - 24xy ( 4x - 2y )

= 64x3 + 8y3 + 96x2y + 48xy2 + 64x3 - 8y3 - 96x2y + 48xy2

= 64x3 + 64x3+ 48xy2 + 48xy2

= 128x3 + 96xy2

So, answer = 128x3 + 96xy2

**(ii) **175 × 175 + 2 × 174 × 25 + 25 × 25

**Solution :**

Given 175 × 175 + 2 × 175 × 25 + 25 × 25

Multiply all these

= 30625 + 8750 + 625

= 40000

S0, answer = 40000 .

NCERT Class 9 Maths Chapter 2 Polynomials :

- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.1
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.2
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.3
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.4
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.5
- Extra Questions Class 9 Maths Chapter 2 Polynomials