# NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

## NCERT Solutions Class 9 Maths Chapter 2 โ Free PDF Download

**NCERT Solutions Class 9 Maths Chapter 2 Polynomials** are provided here. These NCERT Solutions are created by BYJUโS expert faculties to help students in the preparation of their board exams. These expert faculties solve and provide the NCERT Solutions for Class 9 so that it would help students to solve the problems comfortably. They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT textbook for Class 9.

In **NCERT Solutions for Class 9,** students are introduced to a lot of important topics which will be helpful for those who wish to pursue Mathematics as a subject in further classes. Based on these NCERT Solutions. These solutions help students to prepare for their upcoming Board Exams by covering the whole syllabus which follows NCERT guidelines.Chapter 2- Polynomials

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## Exercise 2.1 Page: 32

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

**(i) 4x ^{2}โ3x+7**

Solution:

The equation 4x^{2}โ3x+7 can be written as 4x^{2}โ3x^{1}+7x^{0}

Since *x* is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers, we can say that the expression 4x^{2}โ3x+7 is a polynomial in one variable.

**(ii) y ^{2}+โ2**

Solution:

The equation y^{2}+**โ2** can be written as y^{2}+**โ**2y^{0}

Since y is the only variable in the given equation and the powers of y (i.e., 2 and 0) are whole numbers, we can say that the expression y^{2}+**โ**2 is a polynomial in one variable.

**(iii) 3โt+tโ2**

Solution:

The equation 3โt+tโ2 can be written as 3t^{1/2}+โ2t

Though, *t* is the only variable in the given equation, the powers of *t* (i.e.,1/2) is not a whole number. Hence, we can say that the expression 3โt+tโ2 is **not **a polynomial in one variable.

**(iv) y+2/y**

Solution:

The equation y+2/y an be written as y+2y^{-1}

Though, *y *is the only variable in the given equation, the powers of *y* (i.e.,-1) is not a whole number. Hence, we can say that the expression y+2/y is **not **a polynomial in one variable.

**(v) x ^{10}+y^{3}+t^{50}**

Solution:

Here, in the equation x^{10}+y^{3}+t^{50}

Though, the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression

x^{10}+y^{3}+t^{50}. Hence, it is **not **a polynomial in one variable.

**2. Write the coefficients of x ^{2} in each of the following:**

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

Solution:

The equation 2+x^{2}+x can be written as 2+(1)x^{2}+x

We know that, coefficient is the number which multiplies the variable.

Here, the number that multiplies the variable x^{2} is 1

, the coefficients of x^{2 }in 2+x^{2}+x is 1.

**(ii) 2โx ^{2}+x^{3}**

Solution:

The equation 2โx^{2}+x^{3 }can be written as 2+(โ1)x^{2}+x^{3}

We know that, coefficient is the number (along with its sign, i.e., โ or +) which multiplies the variable.

Here, the number that multiplies the variable x^{2} is -1

the coefficients of x^{2 }in 2โx^{2}+x^{3 }is -1.

**(iii) (/2)x ^{2}+x**

Solution:

The equation (/2)x^{2 }+x can be written as (/2)x^{2} + x

We know that, coefficient is the number (along with its sign, i.e., โ or +) which multiplies the variable.

Here, the number that multiplies the variable x^{2} is /2.

the coefficients of x^{2 }in (/2)x^{2 }+x is /2.

**(iii)โ2x-1**

Solution:

The equation โ2x-1 can be written as 0x^{2}+โ2x-1 [Since 0x^{2} is 0]

We know that, coefficient is the number (along with its sign, i.e., โ or +) which multiplies the variable.

Here, the number that multiplies the variable x^{2}is 0

, the coefficients of x^{2 }in โ2x-1 is 0.

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

Solution:

Binomial of degree 35: A polynomial having two terms and the highest degree 35 is called a binomial of degree 35

Eg., 3x^{35}+5

Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of degree 100

Eg., 4x^{100}

**4. Write the degree of each of the following polynomials:**

**(i) 5x ^{3}+4x^{2}+7x**

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, 5x^{3}+4x^{2}+7x = 5x^{3}+4x^{2}+7x^{1}

The powers of the variable x are: 3, 2, 1

the degree of 5x^{3}+4x^{2}+7x is 3 as 3 is the highest power of x in the equation.

**(ii) 4โy ^{2}**

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, in 4โy^{2},

The power of the variable y is 2

the degree of 4โy^{2} is 2 as 2 is the highest power of y in the equation.

**(iii) 5tโโ7**

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, in 5t**โโ7 ,**

The power of the variable t is: 1

the degree of 5t**โโ7 **is 1 as 1 is the highest power of y in the equation.

**(iv) 3**

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, 3 = 3ร1 = 3ร x^{0}

The power of the variable here is: 0

the degree of 3 is 0.

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

Solution:

We know that,

Linear polynomial: A polynomial of degree one is called a linear polynomial.

Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial.

Cubic polynomial: A polynomial of degree three is called a cubic polynomial.

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

Solution:

The highest power of x^{2}+x is 2

the degree is 2

Hence, x^{2}+x is a quadratic polynomial

**(ii) xโx ^{3}**

Solution:

The highest power of xโx^{3 }is 3

the degree is 3

Hence, xโx^{3} is a cubic polynomial

**(iii) y+y ^{2}+4**

Solution:

The highest power of y+y^{2}+4 is 2

the degree is 2

Hence, y+y^{2}+4is a quadratic polynomial

**(iv) 1+x**

Solution:

The highest power of 1+x is 1

the degree is 1

Hence, 1+x is a linear polynomial.

**(v) 3t**

Solution:

The highest power of 3t is 1

the degree is 1

Hence, 3t is a linear polynomial.

**(vi) r ^{2}**

Solution:

The highest power of r^{2 }is 2

the degree is 2

Hence, r^{2}is a quadratic polynomial.

**(vii) 7x ^{3}**

Solution:

The highest power of 7x^{3 }is 3

the degree is 3

Hence, 7x^{3} is a cubic polynomial.

## Exercise 2.2 Page: 34

**1. Find the value of the polynomial (x)=5xโ4x ^{2}+3 **

**(i) x = 0**

**(ii) x = โ 1**

**(iii) x = 2**

Solution:

Let f(x) = 5xโ4x^{2}+3

(i) When x = 0

f(0) = 5(0)-4(0)^{2}+3

= 3

(ii) When x = -1

f(x) = 5xโ4x^{2}+3

f(โ1) = 5(โ1)โ4(โ1)^{2}+3

= โ5โ4+3

= โ6

(iii) When x = 2

f(x) = 5xโ4x^{2}+3

f(2) = 5(2)โ4(2)^{2}+3

= 10โ16+3

= โ3

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

**(i) p(y)=y ^{2}โy+1**

Solution:

p(y) = y^{2}โy+1

โดp(0) = (0)^{2}โ(0)+1=1

p(1) = (1)^{2}โ(1)+1=1

p(2) = (2)^{2}โ(2)+1=3

**(ii) p(t)=2+t+2t ^{2}โt^{3}**

Solution:

p(t) = 2+t+2t^{2}โt^{3}

โดp(0) = 2+0+2(0)^{2}โ(0)^{3}=2

p(1) = 2+1+2(1)^{2}โ(1)^{3}=2+1+2โ1=4

p(2) = 2+2+2(2)^{2}โ(2)^{3}=2+2+8โ8=4

**(iii) p(x)=x ^{3}**

Solution:

p(x) = x^{3}

โดp(0) = (0)^{3 }= 0

p(1) = (1)^{3 }= 1

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

**(iv) P(x) = (xโ1)(x+1)**

Solution:

p(x) = (xโ1)(x+1)

โดp(0) = (0โ1)(0+1) = (โ1)(1) = โ1

p(1) = (1โ1)(1+1) = 0(2) = 0

p(2) = (2โ1)(2+1) = 1(3) = 3

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

**(i) p(x)=3x+1, x=โ1/3**

Solution:

For, x = -1/3, p(x) = 3x+1

โดp(โ1/3) = 3(-1/3)+1 = โ1+1 = 0

โด -1/3 is a zero of p(x).

**(ii) p(x)=5xโฯ, x = 4/5**

Solution:

For, x = 4/5, p(x) = 5xโฯ

โด p(4/5) = 5(4/5)- = 4-

โด 4/5 is not a zero of p(x).

**(iii) p(x)=x ^{2}โ1, x=1, โ1**

Solution:

For, x = 1, โ1;

p(x) = x^{2}โ1

โดp(1)=1^{2}โ1=1โ1 = 0

p(โ1)=(-1)^{2}โ1 = 1โ1 = 0

โด1, โ1 are zeros of p(x).

**(iv) p(x) = (x+1)(xโ2), x =โ1, 2**

Solution:

For, x = โ1,2;

p(x) = (x+1)(xโ2)

โดp(โ1) = (โ1+1)(โ1โ2)

= (0)(โ3) = 0

p(2) = (2+1)(2โ2) = (3)(0) = 0

โดโ1,2 are zeros of p(x).

**(v) p(x) = x ^{2}, x = 0**

Solution:

For, x = 0 p(x) = x^{2}

p(0) = 0^{2 }= 0

โด 0 is a zero of p(x).

**(vi) p(x) = lx+m, x = โm/l**

Solution:

For, x = -m/*l *; p(x) = *l*x+m

โด p(-m/*l)*= *l*(-m/*l*)+m = โm+m = 0

โด-m/*l* is a zero of p(x).

**(vii) p(x) = 3x ^{2}โ1, x = -1/โ3 , 2/โ3**

Solution:

For, x = -1/โ3 , 2/โ3 ; p(x) = 3x^{2}โ1

โดp(-1/โ3) = 3(-1/โ3)^{2}-1 = 3(1/3)-1 = 1-1 = 0

โดp(2/โ3 ) = 3(2/โ3)^{2}-1 = 3(4/3)-1 = 4โ1=3 โ 0

โด-1/โ3 is a zero of p(x) but 2/โ3 is not a zero of p(x).

**(viii) p(x) =2x+1, x = 1/2**

Solution:

For, x = 1/2 p(x) = 2x+1

โด p(1/2)=2(1/2)+1 = 1+1 = 2โ 0

โด1/2 is not a zero of p(x).

**4. Find the zero of the polynomials in each of the following cases:**

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

Solution:

p(x) = x+5

โ x+5 = 0

โ x = โ5

โด -5 is a zero polynomial of the polynomial p(x).

**(ii) p(x) = xโ5**

Solution:

p(x) = xโ5

โ xโ5 = 0

โ x = 5

โด 5 is a zero polynomial of the polynomial p(x).

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

Solution:

p(x) = 2x+5

โ 2x+5 = 0

โ 2x = โ5

โ x = -5/2

โดx = -5/2 is a zero polynomial of the polynomial p(x).

**(iv) p(x) = 3xโ2 **

Solution:

p(x) = 3xโ2

โ 3xโ2 = 0

โ 3x = 2

โx = 2/3

โดx = 2/3 is a zero polynomial of the polynomial p(x).

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

Solution:

p(x) = 3x

โ 3x = 0

โ x = 0

โด0 is a zero polynomial of the polynomial p(x).

**(vi) p(x) = ax, a0**

Solution:

p(x) = ax

โ ax = 0

โ x = 0

โดx = 0 is a zero polynomial of the polynomial p(x).

**(vii)p(x) = cx+d, c โ 0, c, d are real numbers.**

Solution:

p(x) = cx + d

โ cx+d =0

โ x = -d/c

โด x = -d/c is a zero polynomial of the polynomial p(x).

## Exercise 2.3 Page: 40

**1. Find the remainder when x ^{3}+3x^{2}+3x+1 is divided by**

**(i) x+1**

Solution:

x+1= 0

โx = โ1

โดRemainder:

p(โ1) = (โ1)^{3}+3(โ1)^{2}+3(โ1)+1

= โ1+3โ3+1

= 0

**(ii) xโ1/2**

Solution:

x-1/2 = 0

โ x = 1/2

โดRemainder:

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

= (1/8)+(3/4)+(3/2)+1

= 27/8

**(iii) x**

Solution:

x = 0

โดRemainder:

p(0) = (0)^{3}+3(0)^{2}+3(0)+1

= 1

**(iv) x+ฯ**

Solution:

x+ฯ = 0

โ x = โฯ

โดRemainder:

p(0) = (โฯ)^{3 }+3(โฯ)^{2}+3(โฯ)+1

= โฯ^{3}+3ฯ^{2}โ3ฯ+1

**(v) 5+2x**

Solution:

5+2x=0

โ 2x = โ5

โ x = -5/2

โดRemainder:

(-5/2)^{3}+3(-5/2)^{2}+3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1

= -27/8

**2. Find the remainder when x ^{3}โax^{2}+6xโa is divided by x-a.**

Solution:

Let p(x) = x^{3}โax^{2}+6xโa

xโa = 0

โดx = a

Remainder:

p(a) = (a)^{3}โa(a^{2})+6(a)โa

= a^{3}โa^{3}+6aโa = 5a

**3. Check whether 7+3x is a factor of 3x ^{3}+7x.**

Solution:

7+3x = 0

โ 3x = โ7

โ x = -7/3

โดRemainder:

3(-7/3)^{3}+7(-7/3) = -(343/9)+(-49/3)

= (-343-(49)3)/9

= (-343-147)/9

= -490/9 โ 0

โด7+3x is not a factor of 3x^{3}+7x

## Exercise 2.4 Page: 43

**1. Determine which of the following polynomials has (x + 1) a factor:**

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

Solution:

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

The zero of x+1 is -1. [x+1 = 0 means x = -1]

p(โ1) = (โ1)^{3}+(โ1)^{2}+(โ1)+1

= โ1+1โ1+1

= 0

โดBy factor theorem, x+1 is a factor of x^{3}+x^{2}+x+1

**(ii) x ^{4}+x^{3}+x^{2}+x+1**

Solution:

Let p(x)= x^{4}+x^{3}+x^{2}+x+1

The zero of x+1 is -1. . [x+1= 0 means x = -1]

p(โ1) = (โ1)^{4}+(โ1)^{3}+(โ1)^{2}+(โ1)+1

= 1โ1+1โ1+1

= 1 โ 0

โดBy factor theorem, x+1 is not a factor of x^{4} + x^{3} + x^{2} + x + 1

**(iii) x ^{4}+3x^{3}+3x^{2}+x+1 **

Solution:

Let p(x)= x^{4}+3x^{3}+3x^{2}+x+1

The zero of x+1 is -1.

p(โ1)=(โ1)^{4}+3(โ1)^{3}+3(โ1)^{2}+(โ1)+1

=1โ3+3โ1+1

=1 โ 0

โดBy factor theorem, x+1 is not a factor of x^{4}+3x^{3}+3x^{2}+x+1

**(iv) x ^{3 }โ x^{2}โ (2+โ2)x +โ2**

Solution:

Let p(x) = x^{3}โx^{2}โ(2+โ2)x +โ2

The zero of x+1 is -1.

p(โ1) = (-1)^{3}โ(-1)^{2}โ(2+โ2)(-1) + โ2 = โ1โ1+2+โ2+โ2

= 2โ2 โ 0

โดBy factor theorem, x+1 is not a factor of x^{3}โx^{2}โ(2+โ2)x +โ2

**2. 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}โ2xโ1, g(x) = x+1**

Solution:

p(x) = 2x^{3}+x^{2}โ2xโ1, g(x) = x+1

g(x) = 0

โ x+1 = 0

โ x = โ1

โดZero of g(x) is -1.

Now,

p(โ1) = 2(โ1)^{3}+(โ1)^{2}โ2(โ1)โ1

= โ2+1+2โ1

= 0

โดBy factor theorem, g(x) is a factor of p(x).

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

Solution:

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

g(x) = 0

โ x+2 = 0

โ x = โ2

โด Zero of g(x) is -2.

Now,

p(โ2) = (โ2)^{3}+3(โ2)^{2}+3(โ2)+1

= โ8+12โ6+1

= โ1 โ 0

โดBy factor theorem, g(x) is not a factor of p(x).

**(iii) p(x)=x ^{3}โ4x^{2}+x+6, g(x) = xโ3**

Solution:

p(x) = x^{3}โ4x^{2}+x+6, g(x) = x -3

g(x) = 0

โ xโ3 = 0

โ x = 3

โด Zero of g(x) is 3.

Now,

p(3) = (3)^{3}โ4(3)^{2}+(3)+6

= 27โ36+3+6

= 0

โดBy factor theorem, g(x) is a factor of p(x).

**3. Find the value of k, if xโ1 is a factor of p(x) in each of the following cases:**

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

Solution:

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

โ (1)^{2}+(1)+k = 0

โ 1+1+k = 0

โ 2+k = 0

โ k = โ2

**(ii) p(x) = 2x ^{2}+kx+**โ2

Solution:

If x-1 is a factor of p(x), then p(1)=0

โ 2(1)^{2}+k(1)+โ2 = 0

โ 2+k+โ2 = 0

โ k = โ(2+โ2)

**(iii) p(x) = kx ^{2}โ**โ

**2x+1**

Solution:

If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

โ k(1)^{2}-โ2(1)+1=0

โ k = โ2-1

**(iv) p(x)=kx ^{2}โ3x+k**

Solution:

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

โ k(1)^{2}โ3(1)+k = 0

โ kโ3+k = 0

โ 2kโ3 = 0

โ k= 3/2

**4. Factorize:**

**(i) 12x ^{2}โ7x+1**

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -7 and product =1ร12 = 12

We get -3 and -4 as the numbers [-3+-4=-7 and -3ร-4 = 12]

12x^{2}โ7x+1= 12x^{2}-4x-3x+1

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

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

**(ii) 2x ^{2}+7x+3**

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 7 and product = 2ร3 = 6

We get 6 and 1 as the numbers [6+1 = 7 and 6ร1 = 6]

2x^{2}+7x+3 = 2x^{2}+6x+1x+3

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

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

**(iii) 6x ^{2}+5x-6 **

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 5 and product = 6ร-6 = -36

We get -4 and 9 as the numbers [-4+9 = 5 and -4ร9 = -36]

6x^{2}+5x-6 = 6x^{2}+9xโ4xโ6

= 3x(2x+3)โ2(2x+3)

= (2x+3)(3xโ2)

**(iv) 3x ^{2}โxโ4 **

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -1 and product = 3ร-4 = -12

We get -4 and 3 as the numbers [-4+3 = -1 and -4ร3 = -12]

3x^{2}โxโ4 = 3x^{2}โxโ4

= 3x^{2}โ4x+3xโ4

= x(3xโ4)+1(3xโ4)

= (3xโ4)(x+1)

**5. Factorize:**

**(i) x ^{3}โ2x^{2}โx+2**

Solution:

Let p(x) = x^{3}โ2x^{2}โx+2

Factors of 2 are ยฑ1 and ยฑ 2

Now,

p(x) = x^{3}โ2x^{2}โx+2

p(โ1) = (โ1)^{3}โ2(โ1)^{2}โ(โ1)+2

= โ1โ2+1+2

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor ร Quotient + Remainder

(x+1)(x^{2}โ3x+2) = (x+1)(x^{2}โxโ2x+2)

= (x+1)(x(xโ1)โ2(xโ1))

= (x+1)(xโ1)(x-2)

**(ii) x ^{3}โ3x^{2}โ9xโ5**

Solution:

Let p(x) = x^{3}โ3x^{2}โ9xโ5

Factors of 5 are ยฑ1 and ยฑ5

By trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

Now,

p(x) = x^{3}โ3x^{2}โ9xโ5

p(5) = (5)^{3}โ3(5)^{2}โ9(5)โ5

= 125โ75โ45โ5

= 0

Therefore, (x-5) is the factor of p(x)

Now, Dividend = Divisor ร Quotient + Remainder

(xโ5)(x^{2}+2x+1) = (xโ5)(x^{2}+x+x+1)

= (xโ5)(x(x+1)+1(x+1))

= (xโ5)(x+1)(x+1)

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

Solution:

Let p(x) = x^{3}+13x^{2}+32x+20

Factors of 20 are ยฑ1, ยฑ2, ยฑ4, ยฑ5, ยฑ10 and ยฑ20

By trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

p(x)= x^{3}+13x^{2}+32x+20

p(-1) = (โ1)^{3}+13(โ1)^{2}+32(โ1)+20

= โ1+13โ32+20

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor ร Quotient +Remainder

(x+1)(x^{2}+12x+20) = (x+1)(x^{2}+2x+10x+20)

= (xโ5)x(x+2)+10(x+2)

= (xโ5)(x+2)(x+10)

**(iv) 2y ^{3}+y^{2}โ2yโ1**

Solution:

Let p(y) = 2y^{3}+y^{2}โ2yโ1

Factors = 2ร(โ1)= -2 are ยฑ1 and ยฑ2

By trial method, we find that

p(1) = 0

So, (y-1) is factor of p(y)

Now,

p(y) = 2y^{3}+y^{2}โ2yโ1

p(1) = 2(1)^{3}+(1)^{2}โ2(1)โ1

= 2+1โ2

= 0

Therefore, (y-1) is the factor of p(y)

Now, Dividend = Divisor ร Quotient + Remainder

(yโ1)(2y^{2}+3y+1) = (yโ1)(2y^{2}+2y+y+1)

= (yโ1)(2y(y+1)+1(y+1))

= (yโ1)(2y+1)(y+1)

## Exercise 2.5 Page: 48

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

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

Solution:

Using the identity, (x+a)(x+b) = x ^{2}+(a+b)x+ab[Here, a = 4 and b = 10]

We get,

(x+4)(x+10) = x^{2}+(4+10)x+(4ร10)

= x^{2}+14x+40

**(ii) (x+8)(x โ10) **

Solution:

Using the identity, (x+a)(x+b) = x ^{2}+(a+b)x+ab[Here, a = 8 and b = โ10]

We get,

(x+8)(xโ10) = x^{2}+(8+(โ10))x+(8ร(โ10))

= x^{2}+(8โ10)xโ80

= x^{2}โ2xโ80

**(iii) (3x+4)(3xโ5)**

Solution:

Using the identity, (x+a)(x+b) = x ^{2}+(a+b)x+ab[Here, x = 3x, a = 4 and b = โ5]

We get,

(3x+4)(3xโ5) = (3x)^{2}+[4+(โ5)]3x+4ร(โ5)

= 9x^{2}+3x(4โ5)โ20

= 9x^{2}โ3xโ20

**(iv) (y ^{2}+3/2)(y^{2}-3/2)**

Solution:

Using the identity, (x+y)(xโy) = x^{2}โy^{ 2}[Here, x = y^{2}and y = 3/2]

We get,

(y^{2}+3/2)(y^{2}โ3/2) = (y^{2})^{2}โ(3/2)^{2}

= y^{4}โ9/4

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

**(i) 103ร107**

Solution:

103ร107= (100+3)ร(100+7)

Using identity, [(x+a)(x+b) = x^{2}+(a+b)x+ab

Here, x = 100

a = 3

b = 7

We get, 103ร107 = (100+3)ร(100+7)

= (100)^{2}+(3+7)100+(3ร7))

= 10000+1000+21

= 11021

**(ii) 95ร96 **

Solution:

95ร96 = (100-5)ร(100-4)

Using identity, [(x-a)(x-b) = x^{2}-(a+b)x+ab

Here, x = 100

a = -5

b = -4

We get, 95ร96 = (100-5)ร(100-4)

= (100)^{2}+100(-5+(-4))+(-5ร-4)

= 10000-900+20

= 9120

**(iii) 104ร96**

Solution:

104ร96 = (100+4)ร(100โ4)

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

Here, a = 100

b = 4

We get, 104ร96 = (100+4)ร(100โ4)

= (100)^{2}โ(4)^{2}

= 10000โ16

= 9984

**3. Factorize the following using appropriate identities:**

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

Solution:

9x^{2}+6xy+y^{2 }= (3x)^{2}+(2ร3xรy)+y^{2}

Using identity, x^{2}+2xy+y^{2 }= (x+y)^{2}

Here, x = 3x

y = y

9x^{2}+6xy+y^{2 }= (3x)^{2}+(2ร3xรy)+y^{2}

= (3x+y)^{2}

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

**(ii) 4y ^{2}โ4y+1**

Solution:

4y^{2}โ4y+1 = (2y)^{2}โ(2ร2yร1)+1

Using identity, x^{2} โ 2xy + y^{2 }= (x โ y)^{2}

Here, x = 2y

y = 1

4y^{2}โ4y+1 = (2y)^{2}โ(2ร2yร1)+1^{2}

= (2yโ1)^{2}

= (2yโ1)(2yโ1)

**(iii) x ^{2}โy^{2}/100**

Solution:

x^{2}โy^{2}/100 = x^{2}โ(y/10)^{2}

Using identity, x^{2}-y^{2 }= (x-y)(x+y)

Here, x = x

y = y/10

x^{2}โy^{2}/100 = x^{2}โ(y/10)^{2}

= (xโy/10)(x+y/10)

**4. Expand each of the following, using suitable identities:**

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

**(ii) (2xโy+z) ^{2}**

**(iii) (โ2x+3y+2z) ^{2}**

**(iv) (3a โ7bโc) ^{2}**

**(v) (โ2x+5yโ3z) ^{2}**

**((1/4)a-(1/2)b +1) ^{2}**

Solution:

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

Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx

Here, x = x

y = 2y

z = 4z

(x+2y+4z)^{2 }= x^{2}+(2y)^{2}+(4z)^{2}+(2รxร2y)+(2ร2yร4z)+(2ร4zรx)

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

**(ii) (2xโy+z) ^{2} **

Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx

Here, x = 2x

y = โy

z = z

(2xโy+z)^{2 }= (2x)^{2}+(โy)^{2}+z^{2}+(2ร2xรโy)+(2รโyรz)+(2รzร2x)

= 4x^{2}+y^{2}+z^{2}โ4xyโ2yz+4xz

**(iii) (โ2x+3y+2z) ^{2}**

Solution:

Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx

Here, x = โ2x

y = 3y

z = 2z

(โ2x+3y+2z)^{2 }= (โ2x)^{2}+(3y)^{2}+(2z)^{2}+(2รโ2xร3y)+(2ร3yร2z)+(2ร2zรโ2x)

= 4x^{2}+9y^{2}+4z^{2}โ12xy+12yzโ8xz

**(iv) (3a โ7bโc) ^{2}**

Solution:

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

Here, x = 3a

y = โ 7b

z = โ c

(3a โ7bโ c)^{2 }= (3a)^{2}+(โ 7b)^{2}+(โ c)^{2}+(2ร3a รโ 7b)+(2รโ 7b รโ c)+(2รโ c ร3a)

= 9a^{2} + 49b^{2 }+ c^{2}โ 42ab+14bcโ6ca

**(v) (โ2x+5yโ3z) ^{2}**

Solution:

Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx

Here, x = โ2x

y = 5y

z = โ 3z

(โ2x+5yโ3z)^{2 }= (โ2x)^{2}+(5y)^{2}+(โ3z)^{2}+(2รโ2x ร 5y)+(2ร 5yรโ 3z)+(2รโ3z รโ2x)

= 4x^{2}+25y^{2 }+9z^{2}โ 20xyโ30yz+12zx

**(vi) ((1/4)a-(1/2)b+1) ^{2}**

Solution:

Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx

Here, x = (1/4)a

y = (-1/2)b

z = 1

**5. Factorize:**

**(i) 4x ^{2}+9y^{2}+16z^{2}+12xyโ24yzโ16xz**

**(ii ) 2x ^{2}+y^{2}+8z^{2}โ2โ2xy+4โ2yzโ8xz**

Solution:

**(i) 4x ^{2}+9y^{2}+16z^{2}+12xyโ24yzโ16xz**

Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx

We can say that, x^{2}+y^{2}+z^{2}+2xy+2yz+2zx = (x+y+z)^{2}

4x^{2}+9y^{2}+16z^{2}+12xyโ24yzโ16xz = (2x)^{2}+(3y)^{2}+(โ4z)^{2}+(2ร2xร3y)+(2ร3yรโ4z)+(2รโ4zร2x)

= (2x+3yโ4z)^{2}

= (2x+3yโ4z)(2x+3yโ4z)

**(ii) 2x ^{2}+y^{2}+8z^{2}โ2โ2xy+4โ2yzโ8xz**

Using identity, (x +y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx

We can say that, x^{2}+y^{2}+z^{2}+2xy+2yz+2zx = (x+y+z)^{2}

2x^{2}+y^{2}+8z^{2}โ2โ2xy+4โ2yzโ8xz

= (-โ2x)^{2}+(y)^{2}+(2โ2z)^{2}+(2ร-โ2xรy)+(2รyร2โ2z)+(2ร2โ2รโโ2x)

= (โโ2x+y+2โ2z)^{2}

= (โโ2x+y+2โ2z)(โโ2x+y+2โ2z)

**6. Write the following cubes in expanded form:**

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

**(ii) (2aโ3b) ^{3}**

**(iii) ((3/2)x+1) ^{3}**

**(iv) (xโ(2/3)y) ^{3}**

Solution:

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

Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)

(2x+1)^{3}= (2x)^{3}+1^{3}+(3ร2xร1)(2x+1)

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

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

**(ii) (2aโ3b) ^{3}**

Using identity,(xโy)^{3} = x^{3}โy^{3}โ3xy(xโy)

(2aโ3b)^{3 }= (2a)^{3}โ(3b)^{3}โ(3ร2aร3b)(2aโ3b)

= 8a^{3}โ27b^{3}โ18ab(2aโ3b)

= 8a^{3}โ27b^{3}โ36a^{2}b+54ab^{2}

**(iii) ((3/2)x+1) ^{3}**

Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)

((3/2)x+1)^{3}=((3/2)x)^{3}+1^{3}+(3ร(3/2)xร1)((3/2)x +1)

**(iv) (xโ(2/3)y) ^{3}**

Using identity, (x โy)^{3} = x^{3}โy^{3}โ3xy(xโy)

**7. Evaluate the following using suitable identities: **

**(i) (99) ^{3}**

**(ii) (102) ^{3}**

**(iii) (998) ^{3}**

Solutions:

**(i) (99) ^{3}**

Solution:

We can write 99 as 100โ1

Using identity, (x โy)^{3} = x^{3}โy^{3}โ3xy(xโy)

(99)^{3 }= (100โ1)^{3}

= (100)^{3}โ1^{3}โ(3ร100ร1)(100โ1)

= 1000000 โ1โ300(100 โ 1)

= 1000000โ1โ30000+300

= 970299

**(ii) (102) ^{3}**

Solution:

We can write 102 as 100+2

Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)

(100+2)^{3 }=(100)^{3}+2^{3}+(3ร100ร2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

**(iii) (998) ^{3}**

Solution:

We can write 99 as 1000โ2

Using identity,(xโy)^{3} = x^{3}โy^{3}โ3xy(xโy)

(998)^{3 }=(1000โ2)^{3}

=(1000)^{3}โ2^{3}โ(3ร1000ร2)(1000โ2)

= 1000000000โ8โ6000(1000โ 2)

= 1000000000โ8- 6000000+12000

= 994011992

**8. Factorise each of the following:**

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

**(ii) 8a ^{3}โb^{3}โ12a^{2}b+6ab^{2}**

**(iii) 27โ125a ^{3}โ135a +225a^{2} **

**(iv) 64a ^{3}โ27b^{3}โ144a^{2}b+108ab^{2}**

**(v) 27p ^{3}โ(1/216)โ(9/2) p^{2}+(1/4)p**

Solutions:

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

Solution:

The expression, 8a^{3}+b^{3}+12a^{2}b+6ab^{2} can be written as (2a)^{3}+b^{3}+3(2a)^{2}b+3(2a)(b)^{2}

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

= (2a+b)^{3}

= (2a+b)(2a+b)(2a+b)

Here, the identity, (x +y)^{3} = x^{3}+y^{3}+3xy(x+y) is used.

**(ii) 8a ^{3}โb^{3}โ12a^{2}b+6ab^{2}**

Solution:

The expression, 8a^{3}โb^{3}โ12a^{2}b+6ab^{2} can be written as (2a)^{3}โb^{3}โ3(2a)^{2}b+3(2a)(b)^{2}

8a^{3}โb^{3}โ12a^{2}b+6ab^{2 }= (2a)^{3}โb^{3}โ3(2a)^{2}b+3(2a)(b)^{2}

= (2aโb)^{3}

= (2aโb)(2aโb)(2aโb)

Here, the identity,(xโy)^{3} = x^{3}โy^{3}โ3xy(xโy) is used.

**(iii) 27โ125a ^{3}โ135a+225a^{2} **

Solution:

The expression, 27โ125a^{3}โ135a +225a^{2} can be written as 3^{3}โ(5a)^{3}โ3(3)^{2}(5a)+3(3)(5a)^{2}

27โ125a^{3}โ135a+225a^{2 }=

3^{3}โ(5a)^{3}โ3(3)^{2}(5a)+3(3)(5a)^{2}

= (3โ5a)^{3}

= (3โ5a)(3โ5a)(3โ5a)

Here, the identity, (xโy)^{3} = x^{3}โy^{3}-3xy(xโy) is used.

**(iv) 64a3โ27b3โ144a ^{2}b+108ab^{2}**

Solution:

The expression, 64a^{3}โ27b^{3}โ144a^{2}b+108ab^{2}can be written as (4a)^{3}โ(3b)^{3}โ3(4a)^{2}(3b)+3(4a)(3b)^{2}

64a^{3}โ27b^{3}โ144a^{2}b+108ab^{2}=

(4a)^{3}โ(3b)^{3}โ3(4a)^{2}(3b)+3(4a)(3b)^{2}

=(4aโ3b)^{3}

=(4aโ3b)(4aโ3b)(4aโ3b)

Here, the identity, (x โ y)^{3} = x^{3} โ y^{3} โ 3xy(x โ y) is used.

**(v) 7p ^{3}โ (1/216)โ(9/2) p^{2}+(1/4)p**

Solution:

The expression, 27p^{3}โ(1/216)โ(9/2) p^{2}+(1/4)p

can be written as (3p)^{3}โ(1/6)^{3}โ3(3p)^{2}(1/6)+3(3p)(1/6)^{2}

27p^{3}โ(1/216)โ(9/2) p^{2}+(1/4)p =

(3p)^{3}โ(1/6)^{3}โ3(3p)^{2}(1/6)+3(3p)(1/6)^{2}

= (3pโ16)^{3}

= (3pโ16)(3pโ16)(3pโ16)

**9. Verify:**

**(i) x ^{3}+y^{3 }= (x+y)(x^{2}โxy+y^{2})**

**(ii) x ^{3}โy^{3 }= (xโy)(x^{2}+xy+y^{2})**

Solutions:

**(i) x ^{3}+y^{3 }= (x+y)(x^{2}โxy+y^{2})**

We know that, (x+y)^{3} = x^{3}+y^{3}+3xy(x+y)

โ x^{3}+y^{3 }= (x+y)^{3}โ3xy(x+y)

โ x^{3}+y^{3 }= (x+y)[(x+y)^{2}โ3xy]

Taking (x+y) common โ x^{3}+y^{3 }= (x+y)[(x^{2}+y^{2}+2xy)โ3xy]

โ x^{3}+y^{3 }= (x+y)(x^{2}+y^{2}โxy)

**(ii) x ^{3}โy^{3 }= (xโy)(x^{2}+xy+y^{2}) **

We know that,(xโy)^{3} = x^{3}โy^{3}โ3xy(xโy)

โ x^{3}โy^{3 }= (xโy)^{3}+3xy(xโy)

โ x^{3}โy^{3 }= (xโy)[(xโy)^{2}+3xy]

Taking (x+y) common โ x^{3}โy^{3 }= (xโy)[(x^{2}+y^{2}โ2xy)+3xy]

โ x^{3}+y^{3 }= (xโy)(x^{2}+y^{2}+xy)

**10. Factorize each of the following:**

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

**(ii) 64m ^{3}โ343n^{3}**

Solutions:

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

The expression, 27y^{3}+125z^{3 }can be written as (3y)^{3}+(5z)^{3}

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

We know that, x^{3}+y^{3 }= (x+y)(x^{2}โxy+y^{2})

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

= (3y+5z)[(3y)^{2}โ(3y)(5z)+(5z)^{2}]

= (3y+5z)(9y^{2}โ15yz+25z^{2})

**(ii) 64m ^{3}โ343n^{3}**

The expression, 64m^{3}โ343n^{3}can be written as (4m)^{3}โ(7n)^{3}

64m^{3}โ343n^{3 }=

(4m)^{3}โ(7n)^{3}

We know that, x^{3}โy^{3 }= (xโy)(x^{2}+xy+y^{2})

64m^{3}โ343n^{3 }= (4m)^{3}โ(7n)^{3}

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

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

**11. Factorise: 27x ^{3}+y^{3}+z^{3}โ9xyz **

Solution:

The expression27x^{3}+y^{3}+z^{3}โ9xyz can be written as (3x)^{3}+y^{3}+z^{3}โ3(3x)(y)(z)

27x^{3}+y^{3}+z^{3}โ9xyz = (3x)^{3}+y^{3}+z^{3}โ3(3x)(y)(z)

We know that, x^{3}+y^{3}+z^{3}โ3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}โxy โyzโzx)

27x^{3}+y^{3}+z^{3}โ9xyz = (3x)^{3}+y^{3}+z^{3}โ3(3x)(y)(z)

= (3x+y+z)[(3x)^{2}+y^{2}+z^{2}โ3xyโyzโ3xz]

= (3x+y+z)(9x^{2}+y^{2}+z^{2}โ3xyโyzโ3xz)

**12. Verify that:**

**x ^{3}+y^{3}+z^{3}โ3xyz = (1/2) (x+y+z)[(xโy)^{2}+(yโz)^{2}+(zโx)^{2}]**

Solution:

We know that,

x^{3}+y^{3}+z^{3}โ3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}โxyโyzโxz)

โ x^{3}+y^{3}+z^{3}โ3xyz = (1/2)(x+y+z)[2(x^{2}+y^{2}+z^{2}โxyโyzโxz)]

= (1/2)(x+y+z)(2x^{2}+2y^{2}+2z^{2}โ2xyโ2yzโ2xz)

= (1/2)(x+y+z)[(x^{2}+y^{2}โ2xy)+(y^{2}+z^{2}โ2yz)+(x^{2}+z^{2}โ2xz)]

= (1/2)(x+y+z)[(xโy)^{2}+(yโz)^{2}+(zโx)^{2}]

**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}โxyโyzโxz)

Now, according to the question, let (x+y+z) = 0,

then, x^{3}+y^{3}+z^{3 }-3xyz = (0)(x^{2}+y^{2}+z^{2}โxyโyzโxz)

โ x^{3}+y^{3}+z^{3}โ3xyz = 0

โ x^{3}+y^{3}+z^{3 }= 3xyz

Hence Proved

**14. Without actually calculating the cubes, find the value of each of the following:**

**(i) (โ12) ^{3}+(7)^{3}+(5)^{3}**

**(ii) (28) ^{3}+(โ15)^{3}+(โ13)^{3}**

Solution:

**(i) (โ12) ^{3}+(7)^{3}+(5)^{3}**

Let a = โ12

b = 7

c = 5

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

Here, โ12+7+5=0

(โ12)^{3}+(7)^{3}+(5)^{3 }= 3xyz

= 3ร-12ร7ร5

= -1260

**(ii) (28) ^{3}+(โ15)^{3}+(โ13)^{3}**

Solution:

(28)^{3}+(โ15)^{3}+(โ13)^{3}

Let a = 28

b = โ15

c = โ13

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

Here, x+y+z = 28โ15โ13 = 0

(28)^{3}+(โ15)^{3}+(โ13)^{3 }= 3xyz

= 0+3(28)(โ15)(โ13)

= 16380

**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}โ35a+12**

**(ii) Area : 35y ^{2}+13yโ12**

Solution:

(i) Area : 25a^{2}โ35a+12

Using the splitting the middle term method,

We have to find a number whose sum = -35 and product =25ร12=300

We get -15 and -20 as the numbers [-15+-20=-35 and -15ร-20=300]

25a^{2}โ35a+12 = 25a^{2}โ15aโ20a+12

= 5a(5aโ3)โ4(5aโ3)

= (5aโ4)(5aโ3)

Possible expression for length = 5aโ4

Possible expression for breadth = 5a โ3

(ii) Area : 35y^{2}+13yโ12

Using the splitting the middle term method,

We have to find a number whose sum = 13 and product = 35ร-12 = 420

We get -15 and 28 as the numbers [-15+28 = 13 and -15ร28=420]

35y^{2}+13yโ12 = 35y^{2}โ15y+28yโ12

= 5y(7yโ3)+4(7yโ3)

= (5y+4)(7yโ3)

Possible expression for length = (5y+4)

Possible expression for breadth = (7yโ3)

**16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? **

**(i) Volume : 3x ^{2}โ12x**

**(ii) Volume : 12ky ^{2}+8kyโ20k**

Solution:

(i) Volume : 3x^{2}โ12x

3x^{2}โ12x can be written as 3x(xโ4) by taking 3x out of both the terms.

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (xโ4)

(ii) Volume:

12ky^{2}+8kyโ20k

12ky^{2}+8kyโ20k can be written as 4k(3y^{2}+2yโ5) by taking 4k out of both the terms.

12ky^{2}+8kyโ20k = 4k(3y^{2}+2yโ5)[Here, 3y^{2}+2yโ5 can be written as 3y^{2}+5yโ3yโ5 using splitting the middle term method.]

= 4k(3y^{2}+5yโ3yโ5)

= 4k[y(3y+5)โ1(3y+5)]

= 4k(3y+5)(yโ1)

Possible expression for length = 4k

Possible expression for breadth = (3y +5)

Possible expression for height = (y -1)

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Summary

As this is one of the important topics in Maths, it comes under the unit โ Algebra which has a weightage of 20 marks in Class 9 Maths board exams. This chapter talks about:

- Polynomials in One Variable
- Zeroes of a Polynomial
- Remainder Theorem
- Factorization of Polynomials
- Algebraic Identities

Students can refer to the NCERT Solutions for Class 9 Maths while solving exercise problems and preparing for their Class 9 Maths exams.

**List of Exercises in Class 9 Maths Chapter 2:**

Exercise 2.1 Solutions 5 Questions

Exercise 2.2 Solutions 4 Questions

Exercise 2.3 Solutions 3 Questions

Exercise 2.4 Solutions 5 Questions

Exercise 2.5 Solutions 16 Questions

### NCERT Solutions for Class 9 Maths Chapter 2 โ Polynomials

**NCERT Solutions for Class 9 Maths Chapter 2** Polynomials is the second chapter of Class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses Polynomials and their applications. The introduction of the chapter includes whole numbers, integers, and rational numbers.

The chapter starts with the introduction of Polynomials in section 2.1 followed by two very important topics in sections 2.2 and 2.3

- Polynomials in one Variable โ Discussion of Linear, Quadratic and Cubic Polynomial.
- Zeroes of a Polynomial โ A zero of a polynomial need not be zero and can have more than one zero.
**Real Numbers**and their Decimal Expansions โ Here you study the decimal expansions of real numbers and see whether they can help in distinguishing between rationals and irrationals.

Next, it discusses the following topics:

- Representing Real Numbers on the Number Line โ In this the solutions for 2 problems in Exercise 2.4.
- Operations on Real Numbers โ Here you explore some of the operations like addition, subtraction, multiplication, and division on irrational numbers.
- Laws of Exponents for Real Numbers โ Use these laws of exponents to solve the questions.

### Key Advantages of NCERT Solutions for Class 9 Maths Chapter 2 โ Polynomials

- These
**NCERT Solutions for Class 9 Maths**helps you solve and revise the whole syllabus of Class 9. - After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
- It follows NCERT guidelines which help in preparing the students accordingly.
- It contains all the important questions from the examination point of view.
- It helps in scoring well in Maths in board exams.

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 2

### How many exercises are present in NCERT Solutions for Class 9 Maths Chapter 2?

NCERT Solutions for Class 9 Maths Chapter 2 has 5 exercises. The topics discussed in these exercises are polynomials in one Variable, zeros of a polynomial, real numbers and their decimal expansions, representing real numbers on the number line and operations on real numbers laws of exponents for real numbers. Practice is an essential task to learn and score well in Mathematics. Hence the solutions are designed by BYJUโS experts to boost confidence among students in understanding the concepts covered in this chapter.

### Why should I opt for NCERT Solutions for Class 9 Maths Chapter 2?

The concepts present in NCERT Solutions for Class 9 Maths Chapter 2 are explained in simple language, which makes it possible even for a student not proficient in Maths to understand the subject better. Solutions are prepared by a set of experts at BYJUโS with the aim of helping students boost their exam preparation.

### Is NCERT Solutions for Class 9 Maths Chapter 2 difficult to learn?

No, if you practice regularly with NCERT Solutions for Class 9 Maths Chapter 2 you can achieve your goal by scoring high in finals. These solutions are formulated by a set of Maths experts at BYJUโS. Students can score good marks in the exams by solving all the questions and cross-checking the answers with the NCERT Solutions for Class 9 Maths Chapter 2.