# Class 9 NCERT Solutions of Chapter 2-Polynomial

**The NCERT solutions of class 9 maths chapter 2- Polynomial** is a basic chapter of algebra. The **NCERT** **solutions of the chapter 2-Polynomial** will help you in understanding the questions based on the different structures of **Polynomial.** The **NCERT solutions of the chapter 2-Polynomial** provide the use of all algebraic identities which are used in higher classes also in another branch of **mathematics**. All the **NCERT** **solutions of the chapter 2-Polynomial** are solved by an expert teacher of maths as per the CBSE norms.

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### Exercise 2.1

**Q1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.**

**(i) 4x² –3x + 7**

**(ii) y² + √2**

**(iii) 3√t + t√2**

Ans. (i) 4x² –3x + 7

Hint: Polynomial is the algebraic expression in which the power of the variable is a whole number

The given expression is a polynomial in one variable x since each power of x (2,1,0) is a whole number

(ii) y² + √2

The given expression is polynomial in one variable y since each power of y (2,0) is a whole number

(iii) 3√t + t√2

The given expression is not a polynomial since the power of variable t is (1/2, 1) in which 1/2 is not a whole number

The given expression is not a polynomial since the power of variable y is (1, –1) in which –1 is not a whole number.

The given expression is a polynomial in three variables x,y, and t since each exponent of the variable is 10, 3, and 50 in which all are whole numbers.

**Q2.Write the coefficient of x² in each of the following.**

**(i) 2 + x² + x**

**(ii) 2 – x² + x³**

Ans.

(i) 2 + x² + x

In the given expression the term containing x² has the coefficient 1

(ii) 2 – x² + x³

In the given expression the term containing x² has the coefficient –1

In the given expression the term containing x² has the coefficient π/2.

In the given expression there is no any term x², so the coefficient of x² is 0.

**Q3. Give one example each of a binomial of degree 35, and a monomial of degree 100.**

Ans. Binomial of degree 35 is following

Monomial of degree 100 is following

**Q4. Write the degree of the following polynomial**

**(i) 5x³ + 4x² + 7x**

**(ii) 4 – y²**

**(iii)5t –√7**

**(iv) 3**

Ans.

(i) 5x³ + 4x² + 7x

The highest exponent of the variable x is 3, so the degree of the given polynomial is 3

(ii) 4 – y²

The highest exponent of the variable y in the expression is 2, so the degree of the polynomial is 2

(iii)5t –√7

The highest exponent of the variable t in the expression is 1, so the degree of the polynomial is 1

(iv) 3

There is no variable in the given term, so the degree of the polynomial is 0

**Q5. Classify the following as linear, quadratic, and cubic polynomial.**

**(i) x² + x**

**(ii) x – x³**

**(iii) y + y² + 4**

**(iv) 1 + x**

**(v) 3t**

**(vi) r²**

**(vii) 7x³**

Ans.

(i) x² + x

The given polynomial is of the degree 2, so it is a quadratic polynomial

(ii) x – x³

The given polynomial is of the degree 3, so it is a cubic polynomial

(iii) y + y² + 4

The given polynomial is of the degree 2, so the given polynomial is a quadratic polynomial

(iv) 1 + x

The given polynomial is of the degree 1, so the given polynomial is a linear polynomial

(v) 3t

The given polynomial is of the degree 1, so the given polynomial is a linear polynomial

(vi) r²

The given polynomial is of the degree 2, so the given polynomial is a quadratic polynomial

(vii) 7x³

The given polynomial is of the degree 3, so the given polynomial is a cubic polynomial

**See the video for exercise 2.1 solutions**

**Exercise 2.2**

**Q1. Find the value of polynomial 5x –4x² +3 at**

**(i) x = 0 (ii)x = –1 (iii) x = 2**

Ans. Let the given polynomial is expressed as p(x)

(i) x = 0

p(x) = 5x –4x² +3

p(0) = 5× 0 –4 × 0 + 3 = 3

(ii)x = –1

p(x) = 5x –4x² +3

p(–1) = 5 ×–1 – 4 (–1) ² + 3 = –5 –4 + 3 = –9 + 3 = –6

(iii) x = 2

p(x) = 5x –4x² +3

p(2) = 5 × 2 – 4 ×2² + 3 = 10 – 16 + 3 = –3

**Q2.Find p(0), p(1) and p(2) for each of the following polynomials**

**(i) p(y) = y² –y + 1**

**(ii) p(t) = 2 +t +2t² – t³**

**(iii) p(x) = x³**

**(iv) p(x) = (x +1)(x – 1)**

Ans.

(i) p(y) = y² –y + 1

p(0) = 0 – 0 +1 = 1

p(1) = 1² –1 + 1 = 1

p(2) = 2² –2 + 1 = 3

(ii) p(t) = 2 +t +2t² – t³

p(0) = 2 +0 +2×0² – 0³= 2

p(1) = 2 +1 +2×1² – 1³ = 3 +2 – 1 = 4

p(2)=2 +2 +2×2² – 2³ = 4 + 8 – 8 = 4

(iii) p(x) = x³

p(0) = 0³ = 0

p(1) = 1³ = 1

p(2) = 2³ = 8

(iv) p(x) = (x +1)(x – 1)

p(0) = (0+1)(0 – 1) = –1

p(1) = (1 +1)(1 – 1) = 0

p(2) = (2 +1)(2 – 1) = 3

**Q3. Verify whether the following are zeroes of the polynomial , indicated against them.**

Ans.

Since the value of the polynomial at x = – 1/3 is zero, so x = – 1/3 is the zero of the given polynomial

The value of the polynomial is not zero at the value ,x = 4/5, so x = 4/5 is not the zero of the given polynomial

p(1) = 1² – 1 = 0

The value of the polynomial at x = 1, is zero, so x = 1 is the zero of the given polynomial.

The value of the polynomial at x = –1/2, is not zero, so x = –1/2 is not the zero of the given polynomial.

p(0) = 0

The value of the polynomial at x = 0, is zero, so x = 0 is the zero of the given polynomial

The value of the polynomial at x = –m/l is zero, so x = –m/l is the zero of the given polynomial

The value of the polynomial at x = –1/l√3 is zero, so x = –1/√3 is the zero of the given polynomial

The value of the polynomial at x = 2/√3 is not zero, so x = 2/√3 is not the zero of the given polynomial

The value of the polynomial at x = 1/2 is not zero, so x = 1/2 is not the zero of the given polynomial

**Q4. Find the zero of the polynomial in each of the given cases.**

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

**(ii) p(x) = x – 5**

**(iii) p(x) = 2 x + 5**

**(iv) p(x) = 3x – 2**

**(v) p(x) = 3x**

**(vi) p(x) = ax, a ≠ 0**

**(vii) p(x) = cx + d, c ≠ 0 where c and d are real numbers**

Ans.

(i) p(x) = x +5

x +5 = 0

x = – 5

Hence the zero of the given polynomial is x =– 5

(ii) p(x) = x – 5

x – 5 = 0

x = 5

Hence the zero of the given polynomial is x = 5

(iii) p(x) = 2 x + 5

2 x + 5 = 0

Hence the zero of the given polynomial is – 5/2

(iv) p(x) = 3x – 2

3x – 2 = 0

Hence the zero of the given polynomial is 2/3

(v) p(x) = 3x

3x = 0

x = 0

Hence the zero of the given polynomial is 0

(vi) p(x) = ax, a ≠ 0

ax = 0

x = 0

Hence the zero of the given polynomial is 0

vii) p(x) = cx + d

cx + d = 0

Hence zero of the given polynomial is – d/c

**EXERCISE-2.3**

**Q1.Find the remainder when is divided by**

**(i) x**

**(iii) x**

**(iv) x+π**

**(v) 5+2x**

Ans.

(i) We have to divide the polynomial ** **by the polynomial x + 1

Let’s find the zero of the x = 0

x = – 1

Putting the value of x = – 1, in

= – 1 + 3 – 3 + 1 = 0

(ii) Dividing the polynomial ** **by **, **we can get the reminder in the following way

Putting the value in

(iii) Dividing the polynomial ** **by x, we can get the reminder by the following way

Putting x = 0 the zero of the polynomial x, in the given polynomial

0 + 0 + 0 +1 = 1

(iv) Dividing the polynomial ** **by x + π, we can get the reminder by the following way

x + π = 0

x = – π , is the zero of the polynomial x + π

Putting the value x = – π, in the given polynomial

The reminder is

(v) Dividing the polynomial ** **by 5 + 2x, we can get the reminder by the following way

5+2x = 0

, is the zero of the polynomial 5+2x

Putting the value of in the polynomial **, **we get the reminder

**Q2.Find the remainder when is divided by .**

The zero the polynomial x – a, is

x – a = 0

x = a

Putting this value in

a³ – a × a² + 6 × a – a

= a³ – a³ +6a – a = 5a

Therefore the required reminder is 5a

**Q3.Check whether is a factor of .**

The zero of the polynomial 7 + 3x is as follows

7 + 3x = 0

Putting this value of x in , we shall get the required reminder

**EXERCISE-2.4**

**Q1.Determine which of the following polynomials has a factor.**

**(i)**

**(ii)**

**(iii)**

**(iv)**

Ans.

(i)

The zero of the polynomial x + 1 is as follows

x + 1 = 0

x = – 1

Putting this value of x = – 1, in the given polynomial

(–1)³ + 1² + 1 + 1

– 1 + 1 + 1 + 1 = 2

Since value of the polynomial at x = – 1, is not zero so (x +1) is not the factor of the given polynomial

(ii)

The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1

Putting x = – 1, in the given polynomial

= 1 – 1 + 1 – 1 + 1 = 1

Since value of the given polynomial at x = 1, is not zero,therefore (x +1) is not the factor of the given polynomial

(iii)

The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1

Putting x = – 1, in the given polynomial

= 1 – 3 + 3 –1 +1 = 1

Since value of the given polynomial at x = 1, is not zero,therefore (x +1) is not the factor of the given polynomial

(iv)

The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1

Putting x = – 1, in the given polynomial

The value of the polynomial at x = – 1 is not zero, so (x + 1) is not the factor of the given polynomial

**Q2. Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases**

**(i) p (x)= 2x ^{3} + x^{2} – 2x – 1, g (x) = x + 1**

**(ii) p(x)= x**

^{3}+ 3x^{2}+ 3x + 1, g (x) = x + 2**(iii) p (x) = x**

^{3}– 4x^{2}+ x + 6, g (x) = x – 3Ans. (i) p (x)= 2x^{3} + x^{2} – 2x – 1, g (x) = x + 1

x + 1 = 0

x = – 1

Putting the value of x = – 1 in the given polynomial

p(x) = 2x^{3} + x^{2} – 2x – 1

= 2× – 1 + 1 + 2 – 1

p(–1) = – 2 + 3 – 1 = 0

Since value of p(–1) is zero ,therefore g(x) is the factor of the given polynomial

(ii) p(x)= x^{3} + 3x^{2} + 3x + 1, g (x) = x + 2

x + 2 = 0

x = – 2

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

p(– 2) = (– 2)^{3} + 3(–2)² + 3 (– 2) +1 = -8 +12 -6 +1 = -1

Since the value of p(– 2) is not zero, therefore the given polynomial is not the factor of (x + 2)

(iii) p (x) = x^{3} – 4x^{2} + x + 6, g (x) = x – 3

x – 3 = 0

x = 3

Putting the value of x in the given polynomial

p(3) = 3³ –4× 3² + 3 + 6

= 27– 36 + 9

= 0

Since the value of p(3) is zero, so g(x) is the factor of p(x)

**Q3.Find the value of k, if x – 1 is a factor of p (x) in each of the following cases**

**(i) p (x) = x ^{2} + x + k**

**(ii) p (x) = 2x**

^{2}+ kx + √2**(iii) p (x) = kx**

^{2}– √2 x + 1**(iv) p (x) = kx**

^{2}– 3x + kAns.

For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x^{2} + x + k

Since, p(1) = (1)^{2} +1 + k

⇒ p(1) = k + 2 = 0

⇒ k = -2.

(ii) Here, p (x) = 2x^{2} + kx + √2

Since, p(1) = 2(1)^{2} + k(1) + √2

= 2 + k + √2 =0

k = -2 – √2 = -(2 + √2)

(iii) Here, p (x) = kx^{2} – √2 x + 1

Since, p(1) = k(1)^{2} – (1) + 1

= k – √2 + 1 = 0

⇒ k = √2 -1

(iv) Here, p(x) = kx^{2} – 3x + k

p(1) = k(1)^{2} – 3(1) + k

= k – 3 + k

= 2k – 3 = 0

⇒ k =

**Q4. Factorise**

**(i) 12x ^{2} – 7x +1**

**(ii) 2x**

^{2}+ 7x + 3**(iii) 6x**

^{2}+ 5x – 6**(iv) 3x**

^{2}– x – 4Ans.

(i) We have,

12x

^{2}– 7x + 1 = 12x

^{2}– 4x- 3x + 1

= 4x (3x – 1 ) -1 (3x – 1)

= (3x -1) (4x -1)

Thus, 12x

^{2}-7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x^{2} + 7x + 3 = 2x^{2} + x + 6x + 3

= x(2x + 1) + 3(2x + 1)

= (2x + 1)(x + 3)

Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

(iii) We have, 6x^{2} + 5x – 6 = 6x^{2} + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (2x + 3)(3x – 2)

Thus, 6x^{2} + 5x – 6 = (2x + 3)(3x – 2)

(iv) We have, 3x^{2} – x – 4 = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)

Thus, 3x^{2} – x – 4 = (3x – 4)(x + 1)

**Q5. Factorise**

**(i) x ^{3} – 2x^{2} – x + 2**

**(ii) x**

^{3}– 3x^{2}– 9x – 5**(iii) x**

^{3}+ 13x^{2}+ 32x + 20**(iv) 2y**

^{3}+ y^{2}– 2y – 1Ans.

(i) We have, x

^{3}– 2x

^{2}– x + 2

Rearranging the terms, we have x

^{3}– x – 2x

^{2}+ 2

= x(x

^{2}– 1) – 2(x

^{2}-1) = (x

^{2}– 1)(x – 2)

= [(x)

^{2}– (1)

^{2}](x – 2)

= (x – 1)(x + 1)(x – 2)

[∵ (a

^{2}– b

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

Therefore , x

^{3}– 2x

^{2}– x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x^{3} – 3x^{2} – 9x – 5

= x^{3} + x^{2} – 4x^{2} – 4x – 5x – 5 ,

= x^{2} (x + 1) – 4x(x + 1) – 5(x + 1)

= (x + 1)(x^{2} – 4x – 5)

= (x + 1)(x^{2} – 5x + x – 5)

= (x + 1)[x(x – 5) + 1(x – 5)]

= (x + 1)(x – 5)(x + 1)

Therefore, x^{3} – 3x^{2} – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x^{3} + 13x^{2} + 32x + 20

= x^{3} + x^{2} + 12x^{2} + 12x + 20x + 20

= x^{2}(x + 1) + 12x(x +1) + 20(x + 1)

= (x + 1)(x^{2} + 12x + 20)

= (x + 1)(x^{2} + 2x + 10x + 20)

= (x + 1)[x(x + 2) + 10(x + 2)]

= (x + 1)(x + 2)(x + 10)

Therefore, x^{3} + 13x^{2} + 32x + 20

= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y^{3} + y^{2} – 2y – 1

= 2y^{3} – 2y^{2} + 3y^{2} – 3y + y – 1

= 2y^{2}(y – 1) + 3y(y – 1) + 1(y – 1)

= (y – 1)(2y^{2} + 3y + 1)

= (y – 1)(2y^{2} + 2y + y + 1)

= (y – 1)[2y(y + 1) + 1(y + 1)]

= (y – 1)(y + 1)(2y + 1)

Therefore, 2y^{3} + y^{2} – 2y – 1

= (y – 1)(y + 1)(2y +1)

**Exercise 2.5**

**Q1.Use suitable identities to find the following products**

**(i) **

**(ii) **

**(iii)**

**(iv) **

Ans.

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

**Q2.Evaluate the following products without multiplying directly:**

**(i)103×107**

**(ii)95×96**

**(iii)104× 96**

Ans.

(i)103×107

Modifying the given expression in form of (x + a)(x +b) as following

(103)×(107) = (100 +3)(100 +7)

(100 +3)(100 +7) = 100² +(3 +7)100 + 3×7 = 10000 +1000 + 21 = 110021

(ii) 95×96

95×96 = (100 – 5)(100 – 4)

(100 – 5)(100 – 4) = 100² +( – 5 – 4)100 +( – 4)( – 5)

10000 – 900 + 20 = 9120

(iii) 104× 96

(100 +4)(100 – 4)

(100 +4)(100 – 4) = 100² – 4² = 10000 – 16 = 9984

**Q3.Factorize the following using appropriate identities**

(i)

(ii)

(iii)

Ans.

(i)

Using the following identity

(ii)

(iii)

**Q4.Expand each of the following, using a suitable identity**

(i)

(ii)

(iii)

(iv)

(v)

(iv)

Ans.

(i)

(ii)

(iii)

(iv)

(v)

(iv)

**Q5.Factorise:**

(i)

(ii)

Ans.

(i)

Since the terms –24yz and –16xz, both are having negative sign, so the term z which is common between them must be negative

(ii)

**Q6.Write the following cubes in expanded form**

Ans.

Applying the following identity

=

=

Applying the following identity

Using the following identity

**Q7.Evaluate the following using suitable identities**

**(i)99³ (ii)102³ (iii) 998³ **

Ans.

(i)99³

Modifying 99 into (100– 1) and using the following identity

(x +y)³ = x³ + 3x²y + 3y²x + y³

= 1000000 – 30000 + 300 – 1

= 970299

** (ii)102³**

Modifying 102 into ( 100 +2) and using the following identity

(x +y)³ = x³ + 3x²y + 3y²x + y³

(100 +2)³ = 100³ + 3× 100² ×2 + 3×2²× 100 + 2³

= 1000000 + 60000 +1200 +8

=1061208

**(iii) 998³ **

Modifying the number 998 into (1000 – 2) and using the following identity

(x +y)³ = x³ + 3x²y + 3y²x + y³

(1000 –2)³ = 1000³ – 3× 1000² ×2 + 3×2²× 1000 – 2³

1000000000 –6000000 +12000 – 8

=994011992

**Q8.Factorise each of the following**

(1) 8a³ + b³ +12a²b + 6ab²

(ii) 8a³ – b³ – 12a²b + 6ab²

(iii) 27 – 125a³ – 135a + 225a²

(iv) 64a³ – 27b³ – 144a²b + 108b²a

Ans.

(1) 8a³ + b³ +12a²b + 6ab²

Using the following identity

(x + y)³ = x³ + y³ +3x²y + 3y²x

Modifying the given expression as following

(2a)³ + b³ +3×(2a)²×b + 3×b²×2a

= (2a + b)³

(ii) 8a³ – b³ – 12a²b + 6ab²

Using the following identity

(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x

Modifying the given expression as following

(2a)³ – b³ –3×(2a)²×b + 3×b²×2a

= (2a – b)³

(iii) 27 – 125a³ – 135a + 225a²

Using the following identity

(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x

27 – 125a³ – 135a + 225a²

= 3³ – (5a)³ –3×(3)²×5a + 3×(5a)²×5a

= (3 – 5a)³

(iv) 64a³ – 27b³ – 144a²b + 108b²a

(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x

Modifying the given expression as following

(4a)³ – (3b)³ – 3×(4a)²×3b + 3×(3b)²× 4a

(4a – 3b)³

(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x

Modifying the given expression as following

**Q9.Verify**

Ans.

Using the following identity

Using the following identity

Hence Proved

**Q10.Factorise each of the following**

Ans.

Modifying the given expression according to the identity

Modifying the given expression according to the identity

**Q11.Factorise**

Using the following identity

Modifying the given expression according to the identity

**Q12.Verify that**

Applying the following identity

Dividing and multiplying the RHS by 2

**Q13. If x + y + z = 0, show that x ^{3} + y^{3} + z^{3} = 3 xyz.**

Ans.

Using the following identity

Since, x + y + z = 0

x^{3} + y^{3} + z^{3} – 3 xyz = 0

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

Hence if x + y + z = 0, then x^{3} + y^{3} + z^{3} = 3 xyz

**Q14.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}**

Ans.

(i) We have, (– 12)^{3} + (7)^{3} + (5)^{3}

Let x = – 12, y = 7 and z = 5

Then, x + y + z = – 12 + 7 + 5 = 0

We know that if x + y + z = 0, then, x^{3} + y^{3} + z^{3} = 3xyz

∴ (-12)^{3} + (7)^{3} + (5)^{3} = 3×– 12×7×5

= – 1260

(ii) We have, (28)^{3} + (–15)^{3} + (–13)^{3}

Let x = 28, y = –15 and z = – 13

Then, x + y + z = 28 – 15 – 13 = 0

We know that if x + y + z = 0, then x^{3} + y^{3} + z^{3} = 3xyz

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

= 16380

**Q15.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 – 12Ans.

Area of a rectangle = Length x Breadth

(i) 25a^{2} – 35a + 12 = 25a^{2} – 20a – 15a + 12

= 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)

Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y^{2}+ 13y -12 = 35y^{2} + 28y – 15y -12

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

Thus, the possible length and breadth are (7y – 3) and (5y + 4).

**Q16.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 – 20kAns.

Volume of a cuboid = Length x Breadth x Height

(i) We have, 3x

^{2}– 12x = 3(x

^{2}– 4x)

= 3 x x x (x – 4)

∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) We have, 12ky^{2} + 8ky – 20k

= 4[3ky^{2} + 2ky – 5k] = 4[k(3y^{2} + 2y – 5)]

= 4 x k x (3y^{2} + 2y – 5)

= 4k[3y^{2} – 3y + 5y – 5]

= 4k[3y(y – 1) + 5(y – 1)]

= 4k[(3y + 5) x (y – 1)]

= 4k x (3y + 5) x (y – 1)

Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

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