# NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

## NCERT Solutions Class 9 Maths Chapter 2 â€“ CBSE Term II Free PDF Download

NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. BYJUâ€™S expert faculties create these NCERT Solutions to help students in preparation for their second term exams. BYJUâ€™S provides NCERT Solutions for Class 9Â MathsÂ which will help students to solve 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,Â students are introduced to many important topics that will be helpful for those who wish to pursue Mathematics as a subject in further classes.Â NCERT SolutionsÂ help students to prepare for their upcoming Term II exams by covering the updated term wise CBSE syllabus for 2021-22 and its guidelines.

## NCERT Class 9 Maths Chapter 2 Polynomials Topics

As this is one of the important Chapters in Class 9 Maths, it comes under the unit â€“ Algebra and has a weightage of 12 marks in Class 9 Maths CBSE Term II examination. 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 9Â while solving exercise problems and preparing for their Class 9 Maths exams.

## NCERT Class 9 Maths Chapter 2 â€“ Polynomials Summary

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 â€“ Discussion of Linear, Quadratic and Cubic Polynomial.
• Zeroes of a Polynomial â€“ A zero of a polynomial need not be zero and can have more than one zero.
• Real Numbers and their Decimal Expansions â€“ 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 â€“ In this, the solutions for 2 problems in Exercise 2.4.
• Operations on Real Numbers â€“ Here, you explore some of the operations like addition, subtraction, multiplication, and division on irrational numbers.
• Laws of Exponents for Real Numbers â€“ Use these laws of exponents to solve the questions.

Key Advantages of NCERT Solutions for Class 9 Maths Chapter 2 â€“ Polynomials

• These NCERT Solutions for Class 9 Maths helps you solve and revise the updated term wise CBSE syllabus of Class 9 for 2021-22.
• 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 Class 10 CBSE Term II Maths exams.

To learn the NCERT solutions for Class 9 Maths Chapter 2 Polynomials offline, click on the below link:

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## List of Exercises in Class 9 Maths Chapter 2 Polynomials

Class 9 Maths Chapter 2 Polynomials contains 5 exercises. Based on the concept of polynomials, each exercise provides a number of questions. Click on the below links to access the exercise-wise NCERT solutions for class 9 Maths chapter 2 polynomials.

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

## 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) 4x2â€“3x+7

Solution:

The equation 4x2â€“3x+7 can be written as 4x2â€“3x1+7x0

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 4x2â€“3x+7 is a polynomial in one variable.

(ii) y2+âˆš2

Solution:

The equation y2+âˆš2 can be written as y2+âˆš2y0

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 y2+âˆš2 is a polynomial in one variable.

(iii) 3âˆšt+tâˆš2

Solution:

The equation 3âˆšt+tâˆš2 can be written as 3t1/2+âˆš2t

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 3âˆšt+tâˆš2 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) x10+y3+t50

Solution:

Here, in the equation x10+y3+t50

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

x10+y3+t50. Hence, it is not a polynomial in one variable.

2. Write the coefficients of x2 in each of the following:

(i) 2+x2+x

Solution:

The equation 2+x2+x can be written as 2+(1)x2+x

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

Here, the number that multiplies the variable x2 is 1

, the coefficients of x2 in 2+x2+x is 1.

(ii) 2â€“x2+x3

Solution:

The equation 2â€“x2+x3 can be written as 2+(â€“1)x2+x3

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

Here, the number that multiplies the variable x2 is -1

the coefficients of x2 in 2â€“x2+x3 is -1.

(iii) (Ï€/2)x2+x

Solution:

The equation (Ï€/2)x2 +x can be written as (Ï€/2)x2 + x

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

Here, the number that multiplies the variable x2 is Ï€/2.

the coefficients of x2 in (Ï€/2)x2 +x is Ï€/2.

(iii)âˆš2x-1

Solution:

The equation âˆš2x-1 can be written as 0x2+âˆš2x-1 [Since 0x2 is 0]

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

Here, the number that multiplies the variable x2is 0

, the coefficients of x2 in âˆš2x-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., Â 3x35+5

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

Eg., Â 4x100

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

(i) 5x3+4x2+7x

Solution:

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

Here, 5x3+4x2+7x = 5x3+4x2+7x1

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

the degree of 5x3+4x2+7x is 3 as 3 is the highest power of x in the equation.

(ii) 4â€“y2

Solution:

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

Here, in 4â€“y2,

The power of the variable y is 2

the degree of 4â€“y2 is 2 as 2 is the highest power of y in the equation.

(iii) 5tâ€“âˆš7

Solution:

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

Here, in 5tâ€“âˆš7 ,

The power of the variable t is: 1

the degree of 5tâ€“âˆš7 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 = 3Ã—1 = 3Ã— x0

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) x2+x

Solution:

The highest power of x2+x is 2

the degree is 2

Hence, x2+x is a quadratic polynomial

(ii) xâ€“x3

Solution:

The highest power of xâ€“x3 is 3

the degree is 3

Hence, xâ€“x3 is a cubic polynomial

(iii) y+y2+4

Solution:

The highest power of y+y2+4 is 2

the degree is 2

(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) r2

Solution:

The highest power of r2 is 2

the degree is 2

(vii) 7x3

Solution:

The highest power of 7x3 is 3

the degree is 3

Hence, 7x3 is a cubic polynomial.

## Exercise 2.2 Page: 34

1. Find the value of the polynomial (x)=5xâˆ’4x2+3Â

(i) x = 0

(ii) xÂ = â€“ 1

(iii) xÂ = 2

Solution:

LetÂ f(x) = 5xâˆ’4x2+3

(i) When x = 0

f(0) = 5(0)-4(0)2+3

= 3

(ii)Â When x = -1

f(x) = 5xâˆ’4x2+3

f(âˆ’1) = 5(âˆ’1)âˆ’4(âˆ’1)2+3

= âˆ’5â€“4+3

= âˆ’6

(iii)Â When x = 2

f(x) = 5xâˆ’4x2+3

f(2) = 5(2)âˆ’4(2)2+3

= 10â€“16+3

= âˆ’3

2. FindÂ p(0),Â p(1) andÂ p(2) for each of the following polynomials:

(i) p(y)=y2âˆ’y+1

Solution:

p(y) = y2â€“y+1

âˆ´p(0) = (0)2âˆ’(0)+1=1

p(1) = (1)2â€“(1)+1=1

p(2) = (2)2â€“(2)+1=3

(ii) p(t)=2+t+2t2âˆ’t3

Solution:

p(t) = 2+t+2t2âˆ’t3

âˆ´p(0) = 2+0+2(0)2â€“(0)3=2

p(1) = 2+1+2(1)2â€“(1)3=2+1+2â€“1=4

p(2) = 2+2+2(2)2â€“(2)3=2+2+8â€“8=4

(iii) p(x)=x3

Solution:

p(x) = x3

âˆ´p(0) = (0)3 = 0

p(1) = (1)3 = 1

p(2) = (2)3 = 8

(iv) P(x) = (xâˆ’1)(x+1)

Solution:

p(x) = (xâ€“1)(x+1)

âˆ´p(0) = (0â€“1)(0+1) = (âˆ’1)(1) = â€“1

p(1) = (1â€“1)(1+1) = 0(2) = 0

p(2) = (2â€“1)(2+1) = 1(3) = 3

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

(i) p(x)=3x+1, x=âˆ’1/3

Solution:

For,Â x = -1/3, p(x) = 3x+1

âˆ´p(âˆ’1/3) = 3(-1/3)+1 = âˆ’1+1 = 0

âˆ´ -1/3 is a zero of p(x).

(ii) p(x)=5xâ€“Ï€, x = 4/5

Solution:

For,Â x = 4/5, p(x) = 5xâ€“Ï€

âˆ´ p(4/5) = 5(4/5)- Ï€= 4-Ï€

âˆ´ 4/5 is not a zero of p(x).

(iii) p(x)=x2âˆ’1, x=1, âˆ’1

Solution:

For,Â x = 1, âˆ’1;

p(x) = x2âˆ’1

âˆ´p(1)=12âˆ’1=1âˆ’1 = 0

p(âˆ’1)=(-1)2âˆ’1 = 1âˆ’1 = 0

âˆ´1, âˆ’1Â are zeros of p(x).

(iv) p(x) = (x+1)(xâ€“2), x =âˆ’1, 2

Solution:

For,Â x = âˆ’1,2;

p(x) = (x+1)(xâ€“2)

âˆ´p(âˆ’1) = (âˆ’1+1)(âˆ’1â€“2)

= (0)(âˆ’3) = 0

p(2) = (2+1)(2â€“2) = (3)(0) = 0

âˆ´âˆ’1,2Â are zeros of p(x).

(v) p(x) = x2, x = 0

Solution:

For,Â x = 0Â p(x) = x2

p(0) = 02 = 0

âˆ´ 0Â is aÂ zero of p(x).

(vi) p(x) = lx+m, x = âˆ’m/l

Solution:

For,Â x = -m/l ;Â p(x) = lx+m

âˆ´ p(-m/l)= l(-m/l)+m = âˆ’m+m = 0

âˆ´-m/l is a zero of p(x).

(vii) p(x) = 3x2âˆ’1, x = -1/âˆš3 , 2/âˆš3

Solution:

For,Â x = -1/âˆš3 , 2/âˆš3 ; p(x) = 3x2âˆ’1

âˆ´p(-1/âˆš3) = 3(-1/âˆš3)2-1 = 3(1/3)-1 = 1-1 = 0

âˆ´p(2/âˆš3 ) = 3(2/âˆš3)2-1 = 3(4/3)-1 = 4âˆ’1=3 â‰  0

âˆ´-1/âˆš3Â is a zero of p(x) but 2/âˆš3Â Â is not a zero of p(x).

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

Solution:

For,Â x = 1/2 p(x) = 2x+1

âˆ´ p(1/2)=2(1/2)+1 = 1+1 = 2â‰ 0

âˆ´1/2 is not aÂ zero of p(x).

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

(i) p(x) =Â x+5Â

Solution:

p(x) = x+5

â‡’ x+5 = 0

â‡’ x = âˆ’5

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

(ii) p(x) =Â xâ€“5

Solution:

p(x) = xâˆ’5

â‡’ xâˆ’5 = 0

â‡’ x = 5

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

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

Solution:

p(x) = 2x+5

â‡’ 2x+5 = 0

â‡’ 2x = âˆ’5

â‡’ x = -5/2

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

(iv) p(x) = 3xâ€“2Â

Solution:

p(x) = 3xâ€“2

â‡’ 3xâˆ’2 = 0

â‡’ 3x = 2

â‡’x = 2/3

âˆ´x = 2/3 Â is a zero polynomial of the polynomial p(x).

(v) p(x) = 3xÂ

Solution:

p(x) = 3x

â‡’ 3x = 0

â‡’ x = 0

âˆ´0 is a zero polynomial of the polynomial p(x).

(vi) p(x) = ax, aâ‰ 0

Solution:

p(x) = ax

â‡’ ax = 0

â‡’ x = 0

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

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

Solution:

p(x) = cx + d

â‡’ cx+d =0

â‡’ x = -d/c

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

## Exercise 2.3 Page: 40

1. Find the remainder whenÂ x3+3x2+3x+1Â is divided by

(i) x+1

Solution:

x+1= 0

â‡’x = âˆ’1

âˆ´Remainder:

p(âˆ’1) = (âˆ’1)3+3(âˆ’1)2+3(âˆ’1)+1

= âˆ’1+3âˆ’3+1

= 0

(ii) xâˆ’1/2

Solution:

x-1/2 = 0

â‡’ x = 1/2

âˆ´Remainder:

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

âˆ´Remainder:

p(0) = (0)3+3(0)2+3(0)+1

= 1

(iv) x+Ï€

Solution:

x+Ï€ = 0

â‡’ x = âˆ’Ï€

âˆ´Remainder:

p(0) = (âˆ’Ï€)3 +3(âˆ’Ï€)2+3(âˆ’Ï€)+1

= âˆ’Ï€3+3Ï€2âˆ’3Ï€+1

(v) 5+2x

Solution:

5+2x=0

â‡’ 2x = âˆ’5

â‡’ x = -5/2

âˆ´Remainder:

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

= -27/8

2. Find the remainder whenÂ x3âˆ’ax2+6xâˆ’aÂ is divided by x-a.

Solution:

LetÂ p(x) = x3âˆ’ax2+6xâˆ’a

xâˆ’a = 0

âˆ´x = a

Remainder:

p(a) = (a)3âˆ’a(a2)+6(a)âˆ’a

= a3âˆ’a3+6aâˆ’a = 5a

3. Check whether 7+3x is a factor ofÂ 3x3+7x.

Solution:

7+3x = 0

â‡’ 3x = âˆ’7

â‡’ x = -7/3

âˆ´Remainder:

3(-7/3)3+7(-7/3) = -(343/9)+(-49/3)

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

= (-343-147)/9

= -490/9 â‰  0

âˆ´7+3x is not a factor ofÂ 3x3+7x

## Exercise 2.4 Page: 43

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

(i) x3+x2+x+1

Solution:

Let p(x) =Â x3+x2+x+1

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

p(âˆ’1) = (âˆ’1)3+(âˆ’1)2+(âˆ’1)+1

= âˆ’1+1âˆ’1+1

= 0

âˆ´By factor theorem, x+1 is a factor ofÂ x3+x2+x+1

(ii) x4+x3+x2+x+1

Solution:

Let p(x)=Â x4+x3+x2+x+1

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

p(âˆ’1) = (âˆ’1)4+(âˆ’1)3+(âˆ’1)2+(âˆ’1)+1

= 1âˆ’1+1âˆ’1+1

= 1 â‰  0

âˆ´By factor theorem, x+1 is not a factor ofÂ x4Â +Â x3Â +Â x2Â +Â xÂ + 1

(iii) x4+3x3+3x2+x+1Â

Solution:

Let p(x)=Â x4+3x3+3x2+x+1

The zero of x+1 is -1.

p(âˆ’1)=(âˆ’1)4+3(âˆ’1)3+3(âˆ’1)2+(âˆ’1)+1

=1âˆ’3+3âˆ’1+1

=1 â‰  0

âˆ´By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1

(iv) x3 â€“ x2â€“ (2+âˆš2)x +âˆš2

Solution:

Let p(x) =Â x3â€“x2â€“(2+âˆš2)x +âˆš2

The zero of x+1 is -1.

p(âˆ’1) = (-1)3â€“(-1)2â€“(2+âˆš2)(-1) + âˆš2 = âˆ’1âˆ’1+2+âˆš2+âˆš2

= 2âˆš2 â‰  0

âˆ´By factor theorem, x+1 is not a factor ofÂ x3â€“x2â€“(2+âˆš2)x +âˆš2

2. Use the Factor Theorem to determine whetherÂ g(x) is a factor ofÂ p(x) in each of the following cases:

(i) p(x) = 2x3+x2â€“2xâ€“1,Â g(x) =Â x+1

Solution:

p(x) = 2x3+x2â€“2xâ€“1,Â g(x) =Â x+1

g(x) = 0

â‡’ x+1 = 0

â‡’ x = âˆ’1

âˆ´Zero of g(x) is -1.

Now,

p(âˆ’1) = 2(âˆ’1)3+(âˆ’1)2â€“2(âˆ’1)â€“1

= âˆ’2+1+2âˆ’1

= 0

âˆ´By factor theorem, g(x) is a factor of p(x).

(ii) p(x)=x3+3x2+3x+1,Â g(x) =Â x+2

Solution:

p(x) = x3+3x2+3x+1,Â g(x) =Â x+2

g(x) = 0

â‡’ x+2 = 0

â‡’ x = âˆ’2

âˆ´ Zero of g(x) is -2.

Now,

p(âˆ’2) = (âˆ’2)3+3(âˆ’2)2+3(âˆ’2)+1

= âˆ’8+12âˆ’6+1

= âˆ’1 â‰  0

âˆ´By factor theorem, g(x) is not a factor of p(x).

(iii) p(x)=x3â€“4x2+x+6,Â g(x) =Â xâ€“3

Solution:

p(x) = x3â€“4x2+x+6,Â g(x) =Â xÂ -3

g(x) = 0

â‡’ xâˆ’3 = 0

â‡’ x = 3

âˆ´ Zero of g(x) is 3.

Now,

p(3) = (3)3âˆ’4(3)2+(3)+6

= 27âˆ’36+3+6

= 0

âˆ´By factor theorem, g(x) is a factor of p(x).

3. Find the value ofÂ k, ifÂ xâ€“1 is a factor ofÂ p(x) in each of the following cases:

(i) p(x) = x2+x+k

Solution:

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

By Factor Theorem

â‡’ (1)2+(1)+k = 0

â‡’ 1+1+k = 0

â‡’ 2+k = 0

â‡’ k = âˆ’2

(ii) p(x) = 2x2+kx+âˆš2

Solution:

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

â‡’ 2(1)2+k(1)+âˆš2 = 0

â‡’ 2+k+âˆš2 = 0

â‡’ k = âˆ’(2+âˆš2)

(iii) p(x) = kx2â€“âˆš2x+1

Solution:

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

By Factor Theorem

â‡’ k(1)2-âˆš2(1)+1=0

â‡’ k = âˆš2-1

(iv) p(x)=kx2â€“3x+k

Solution:

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

By Factor Theorem

â‡’ k(1)2â€“3(1)+k = 0

â‡’ kâˆ’3+k = 0

â‡’ 2kâˆ’3 = 0

â‡’ k= 3/2

4. Factorize:

(i) 12x2â€“7x+1

Solution:

Using the splitting the middle term method,

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

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

12x2â€“7x+1= 12x2-4x-3x+1

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

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

(ii) 2x2+7x+3

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 7 and product = 2Ã—3 = 6

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

2x2+7x+3 = 2x2+6x+1x+3

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

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

(iii) 6x2+5x-6Â

Solution:

Using the splitting the middle term method,

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

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

6x2+5x-6 = 6x2+9xâ€“4xâ€“6

= 3x(2x+3)â€“2(2x+3)

= (2x+3)(3xâ€“2)

(iv) 3x2â€“xâ€“4Â

Solution:

Using the splitting the middle term method,

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

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

3x2â€“xâ€“4Â = 3x2â€“4x+3xâ€“4

= x(3xâ€“4)+1(3xâ€“4)

= (3xâ€“4)(x+1)

5. Factorize:

(i) x3â€“2x2â€“x+2

Solution:

LetÂ p(x) = x3â€“2x2â€“x+2

Factors of 2 are Â±1 and Â± 2

Now,

p(x) = x3â€“2x2â€“x+2

p(âˆ’1) = (âˆ’1)3â€“2(âˆ’1)2â€“(âˆ’1)+2

= âˆ’1âˆ’2+1+2

= 0

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

Now, Dividend = Divisor Ã— Quotient + Remainder

(x+1)(x2â€“3x+2) = (x+1)(x2â€“xâ€“2x+2)

= (x+1)(x(xâˆ’1)âˆ’2(xâˆ’1))

= (x+1)(xâˆ’1)(x-2)

(ii) x3â€“3x2â€“9xâ€“5

Solution:

LetÂ p(x) =Â x3â€“3x2â€“9xâ€“5

Factors of 5 are Â±1 and Â±5

By trial method, we find that

p(5)Â = 0

So,Â (x-5)Â is factor ofÂ p(x)

Now,

p(x) =Â x3â€“3x2â€“9xâ€“5

p(5) =Â (5)3â€“3(5)2â€“9(5)â€“5

= 125âˆ’75âˆ’45âˆ’5

= 0

Therefore, (x-5) is the factor ofÂ Â p(x)

Now, Dividend = Divisor Ã— Quotient + Remainder

(xâˆ’5)(x2+2x+1) = (xâˆ’5)(x2+x+x+1)

= (xâˆ’5)(x(x+1)+1(x+1))

= (xâˆ’5)(x+1)(x+1)

(iii) x3+13x2+32x+20

Solution:

LetÂ p(x) =Â x3+13x2+32x+20

Factors of 20 are Â±1, Â±2, Â±4, Â±5, Â±10 and Â±20

By trial method, we find that

p(-1)Â = 0

So,Â (x+1)Â is factor ofÂ p(x)

Now,

p(x)=Â x3+13x2+32x+20

p(-1) =Â (âˆ’1)3+13(âˆ’1)2+32(âˆ’1)+20

= âˆ’1+13âˆ’32+20

= 0

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

Now, Dividend = Divisor Ã— QuotientÂ +Remainder

(x+1)(x2+12x+20) = (x+1)(x2+2x+10x+20)

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

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

(iv) 2y3+y2â€“2yâ€“1

Solution:

LetÂ p(y) =Â 2y3+y2â€“2yâ€“1

Factors =Â 2Ã—(âˆ’1)= -2 are Â±1 and Â±2

By trial method, we find that

p(1)Â = 0

So,Â (y-1)Â is factor ofÂ p(y)

Now,

p(y) =Â 2y3+y2â€“2yâ€“1

p(1) =Â 2(1)3+(1)2â€“2(1)â€“1

= 2+1âˆ’2

= 0

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

Now, Dividend = Divisor Ã— QuotientÂ + Remainder

(yâˆ’1)(2y2+3y+1) = (yâˆ’1)(2y2+2y+y+1)

= (yâˆ’1)(2y(y+1)+1(y+1))

= (yâˆ’1)(2y+1)(y+1)

## Exercise 2.5 Page: 48

1. Use suitable identities to find the following products:

(i) (x+4)(xÂ +10)Â

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) = x2+(4+10)x+(4Ã—10)

= x2+14x+40

(ii) (x+8)(xÂ â€“10)Â Â Â Â Â

Solution:

Using the identity, (x+a)(x+b) = x 2+(a+b)x+ab

[Here, a = 8 and b = âˆ’10]

We get,

(x+8)(xâˆ’10) = x2+(8+(âˆ’10))x+(8Ã—(âˆ’10))

= x2+(8âˆ’10)xâ€“80

= x2âˆ’2xâˆ’80

(iii) (3x+4)(3xâ€“5)

Solution:

Using the identity, (x+a)(x+b) = x 2+(a+b)x+ab

[Here, x = 3x, a = 4 and b = âˆ’5]

We get,

(3x+4)(3xâˆ’5) = (3x)2+[4+(âˆ’5)]3x+4Ã—(âˆ’5)

= 9x2+3x(4â€“5)â€“20

= 9x2â€“3xâ€“20

(iv) (y2+3/2)(y2-3/2)

Solution:

Using the identity, (x+y)(xâ€“y) = x2â€“y 2

[Here, x = y2and y = 3/2]

We get,

(y2+3/2)(y2â€“3/2) = (y2)2â€“(3/2)2

= y4â€“9/4

2. Evaluate the following products without multiplying directly:

(i) 103Ã—107

Solution:

103Ã—107= (100+3)Ã—(100+7)

Using identity, [(x+a)(x+b) = x2+(a+b)x+ab

Here, x = 100

a = 3

b = 7

We get, 103Ã—107 = (100+3)Ã—(100+7)

= (100)2+(3+7)100+(3Ã—7)

= 10000+1000+21

= 11021

(ii) 95Ã—96 Â

Solution:

95Ã—96 = (100-5)Ã—(100-4)

Using identity, [(x-a)(x-b) = x2-(a+b)x+ab

Here, x = 100

a = -5

b = -4

We get, 95Ã—96 = (100-5)Ã—(100-4)

= (100)2+100(-5+(-4))+(-5Ã—-4)

= 10000-900+20

= 9120

(iii) 104Ã—96

Solution:

104Ã—96 = (100+4)Ã—(100â€“4)

Using identity, [(a+b)(a-b)= a2-b2]

Here, a = 100

b = 4

We get, 104Ã—96 = (100+4)Ã—(100â€“4)

= (100)2â€“(4)2

= 10000â€“16

= 9984

3. Factorize the following using appropriate identities:

(i) 9x2+6xy+y2

Solution:

9x2+6xy+y2 = (3x)2+(2Ã—3xÃ—y)+y2

Using identity, x2+2xy+y2 = (x+y)2

Here, x = 3x

y = y

9x2+6xy+y2 = (3x)2+(2Ã—3xÃ—y)+y2

= (3x+y)2

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

(ii) 4y2âˆ’4y+1

Solution:

4y2âˆ’4y+1 = (2y)2â€“(2Ã—2yÃ—1)+1

Using identity, x2 â€“ 2xy + y2 = (x â€“ y)2

Here, x = 2y

y = 1

4y2âˆ’4y+1 = (2y)2â€“(2Ã—2yÃ—1)+12

= (2yâ€“1)2

= (2yâ€“1)(2yâ€“1)

(iii) Â x2â€“y2/100

Solution:

x2â€“y2/100 = x2â€“(y/10)2

Using identity, x2-y2 = (x-y)(x+y)

Here, x = x

y = y/10

x2â€“y2/100 = x2â€“(y/10)2

= (xâ€“y/10)(x+y/10)

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

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

(ii) (2xâˆ’y+z)2

(iii) (âˆ’2x+3y+2z)2

(iv) (3a â€“7bâ€“c)2

(v) (â€“2x+5yâ€“3z)2

(vi)((1/4)a-(1/2)b +1)2

Solution:

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

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = x

y = 2y

z = 4z

(x+2y+4z)2 = x2+(2y)2+(4z)2+(2Ã—xÃ—2y)+(2Ã—2yÃ—4z)+(2Ã—4zÃ—x)

= x2+4y2+16z2+4xy+16yz+8xz

(ii) (2xâˆ’y+z)2Â

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = 2x

y = âˆ’y

z = z

(2xâˆ’y+z)2 = (2x)2+(âˆ’y)2+z2+(2Ã—2xÃ—âˆ’y)+(2Ã—âˆ’yÃ—z)+(2Ã—zÃ—2x)

= 4x2+y2+z2â€“4xyâ€“2yz+4xz

(iii) (âˆ’2x+3y+2z)2

Solution:

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = âˆ’2x

y = 3y

z = 2z

(âˆ’2x+3y+2z)2 = (âˆ’2x)2+(3y)2+(2z)2+(2Ã—âˆ’2xÃ—3y)+(2Ã—3yÃ—2z)+(2Ã—2zÃ—âˆ’2x)

= 4x2+9y2+4z2â€“12xy+12yzâ€“8xz

(iv) (3a â€“7bâ€“c)2

Solution:

Using identity (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = 3a

y = â€“ 7b

z = â€“ c

(3a â€“7bâ€“ c)2 = (3a)2+(â€“ 7b)2+(â€“ c)2+(2Ã—3a Ã—â€“ 7b)+(2Ã—â€“ 7b Ã—â€“ c)+(2Ã—â€“ c Ã—3a)

= 9a2 + 49b2 + c2â€“ 42ab+14bcâ€“6ca

(v) (â€“2x+5yâ€“3z)2

Solution:

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = â€“2x

y = 5y

z = â€“ 3z

(â€“2x+5yâ€“3z)2 = (â€“2x)2+(5y)2+(â€“3z)2+(2Ã—â€“2x Ã— 5y)+(2Ã— 5yÃ—â€“ 3z)+(2Ã—â€“3z Ã—â€“2x)

= 4x2+25y2 +9z2â€“ 20xyâ€“30yz+12zx

(vi) ((1/4)a-(1/2)b+1)2

Solution:

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = (1/4)a

y = (-1/2)b

z = 1

5. Factorize:

(i) 4x2+9y2+16z2+12xyâ€“24yzâ€“16xz

(ii ) 2x2+y2+8z2â€“2âˆš2xy+4âˆš2yzâ€“8xz

Solution:

(i) 4x2+9y2+16z2+12xyâ€“24yzâ€“16xz

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2

4x2+9y2+16z2+12xyâ€“24yzâ€“16xzÂ = (2x)2+(3y)2+(âˆ’4z)2+(2Ã—2xÃ—3y)+(2Ã—3yÃ—âˆ’4z)+(2Ã—âˆ’4zÃ—2x)

= (2x+3yâ€“4z)2

= (2x+3yâ€“4z)(2x+3yâ€“4z)

(ii) 2x2+y2+8z2â€“2âˆš2xy+4âˆš2yzâ€“8xz

Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx

We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2

2x2+y2+8z2â€“2âˆš2xy+4âˆš2yzâ€“8xz

= (-âˆš2x)2+(y)2+(2âˆš2z)2+(2Ã—-âˆš2xÃ—y)+(2Ã—yÃ—2âˆš2z)+(2Ã—2âˆš2Ã—âˆ’âˆš2x)

= (âˆ’âˆš2x+y+2âˆš2z)2

= (âˆ’âˆš2x+y+2âˆš2z)(âˆ’âˆš2x+y+2âˆš2z)

6. Write the following cubes in expanded form:

(i) (2x+1)3

(ii) (2aâˆ’3b)3

(iii) ((3/2)x+1)3

(iv) (xâˆ’(2/3)y)3

Solution:

(i) (2x+1)3

Using identity,(x+y)3 = x3+y3+3xy(x+y)

(2x+1)3= (2x)3+13+(3Ã—2xÃ—1)(2x+1)

= 8x3+1+6x(2x+1)

= 8x3+12x2+6x+1

(ii) (2aâˆ’3b)3

Using identity,(xâ€“y)3 = x3â€“y3â€“3xy(xâ€“y)

(2aâˆ’3b)3 = (2a)3âˆ’(3b)3â€“(3Ã—2aÃ—3b)(2aâ€“3b)

= 8a3â€“27b3â€“18ab(2aâ€“3b)

= 8a3â€“27b3â€“36a2b+54ab2

(iii) ((3/2)x+1)3

Using identity,(x+y)3 = x3+y3+3xy(x+y)

((3/2)x+1)3=((3/2)x)3+13+(3Ã—(3/2)xÃ—1)((3/2)x +1)

(iv) Â (xâˆ’(2/3)y)3

Using identity, (x â€“y)3 = x3â€“y3â€“3xy(xâ€“y)

7. Evaluate the following using suitable identities:Â

(i) (99)3

(ii) (102)3

(iii) (998)3

Solutions:

(i) (99)3

Solution:

We can write 99 as 100â€“1

Using identity, (x â€“y)3 = x3â€“y3â€“3xy(xâ€“y)

(99)3 = (100â€“1)3

= (100)3â€“13â€“(3Ã—100Ã—1)(100â€“1)

= 1000000 â€“1â€“300(100 â€“ 1)

= 1000000â€“1â€“30000+300

= 970299

(ii) (102)3

Solution:

We can write 102 as 100+2

Using identity,(x+y)3 = x3+y3+3xy(x+y)

(100+2)3 =(100)3+23+(3Ã—100Ã—2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3

Solution:

We can write 99 as 1000â€“2

Using identity,(xâ€“y)3 = x3â€“y3â€“3xy(xâ€“y)

(998)3 =(1000â€“2)3

=(1000)3â€“23â€“(3Ã—1000Ã—2)(1000â€“2)

= 1000000000â€“8â€“6000(1000â€“ 2)

= 1000000000â€“8- 6000000+12000

= 994011992

8. Factorise each of the following:

(i) 8a3+b3+12a2b+6ab2

(ii) 8a3â€“b3â€“12a2b+6ab2

(iii) 27â€“125a3â€“135a +225a2Â  Â

(iv) 64a3â€“27b3â€“144a2b+108ab2

(v) 27p3â€“(1/216)âˆ’(9/2) p2+(1/4)p

Solutions:

(i) 8a3+b3+12a2b+6ab2

Solution:

The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2

8a3+b3+12a2b+6ab2 = (2a)3+b3+3(2a)2b+3(2a)(b)2

= (2a+b)3

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

Here, the identity, (x +y)3 = x3+y3+3xy(x+y) is used.

Â

(ii) 8a3â€“b3â€“12a2b+6ab2

Solution:

The expression, 8a3â€“b3âˆ’12a2b+6ab2 can be written as (2a)3â€“b3â€“3(2a)2b+3(2a)(b)2

8a3â€“b3âˆ’12a2b+6ab2 = (2a)3â€“b3â€“3(2a)2b+3(2a)(b)2

= (2aâ€“b)3

= (2aâ€“b)(2aâ€“b)(2aâ€“b)

Here, the identity,(xâ€“y)3 = x3â€“y3â€“3xy(xâ€“y) is used.

Â

(iii) 27â€“125a3â€“135a+225a2Â

Solution:

The expression, 27â€“125a3â€“135a +225a2 can be written as 33â€“(5a)3â€“3(3)2(5a)+3(3)(5a)2

27â€“125a3â€“135a+225a2 =
33â€“(5a)3â€“3(3)2(5a)+3(3)(5a)2

= (3â€“5a)3

= (3â€“5a)(3â€“5a)(3â€“5a)

Here, the identity, (xâ€“y)3 = x3â€“y3-3xy(xâ€“y) is used.

(iv) 64a3â€“27b3â€“144a2b+108ab2

Solution:

The expression, 64a3â€“27b3â€“144a2b+108ab2can be written as (4a)3â€“(3b)3â€“3(4a)2(3b)+3(4a)(3b)2

64a3â€“27b3â€“144a2b+108ab2=
(4a)3â€“(3b)3â€“3(4a)2(3b)+3(4a)(3b)2

=(4aâ€“3b)3

=(4aâ€“3b)(4aâ€“3b)(4aâ€“3b)

Here, the identity, (x â€“ y)3 = x3 â€“ y3 â€“ 3xy(x â€“ y) is used.

(v) 7p3â€“ (1/216)âˆ’(9/2) p2+(1/4)p

Solution:

The expression, 27p3â€“(1/216)âˆ’(9/2) p2+(1/4)p

can be written as (3p)3â€“(1/6)3â€“3(3p)2(1/6)+3(3p)(1/6)2

27p3â€“(1/216)âˆ’(9/2) p2+(1/4)p =
(3p)3â€“(1/6)3â€“3(3p)2(1/6)+3(3p)(1/6)2

= (3pâ€“16)3

= (3pâ€“16)(3pâ€“16)(3pâ€“16)

9. Verify:

(i) x3+y3 = (x+y)(x2â€“xy+y2)

(ii) x3â€“y3 = (xâ€“y)(x2+xy+y2)

Solutions:

(i) x3+y3 = (x+y)(x2â€“xy+y2)

We know that, (x+y)3 = x3+y3+3xy(x+y)

â‡’ x3+y3 = (x+y)3â€“3xy(x+y)

â‡’ x3+y3 = (x+y)[(x+y)2â€“3xy]

Taking (x+y) common â‡’ x3+y3 = (x+y)[(x2+y2+2xy)â€“3xy]

â‡’ x3+y3 = (x+y)(x2+y2â€“xy)

(ii) x3â€“y3 = (xâ€“y)(x2+xy+y2)Â

We know that,(xâ€“y)3 = x3â€“y3â€“3xy(xâ€“y)

â‡’ x3âˆ’y3 = (xâ€“y)3+3xy(xâ€“y)

â‡’ x3âˆ’y3 = (xâ€“y)[(xâ€“y)2+3xy]

Taking (x+y) common â‡’ x3âˆ’y3 = (xâ€“y)[(x2+y2â€“2xy)+3xy]

â‡’ x3+y3 = (xâ€“y)(x2+y2+xy)

10. Factorize each of the following:

(i) 27y3+125z3

(ii) 64m3â€“343n3

Solutions:

(i) 27y3+125z3

The expression, 27y3+125z3 can be written as (3y)3+(5z)3

27y3+125z3 = (3y)3+(5z)3

We know that, x3+y3 = (x+y)(x2â€“xy+y2)

27y3+125z3 = (3y)3+(5z)3

= (3y+5z)[(3y)2â€“(3y)(5z)+(5z)2]

= (3y+5z)(9y2â€“15yz+25z2)

(ii) 64m3â€“343n3

The expression, 64m3â€“343n3can be written as (4m)3â€“(7n)3

64m3â€“343n3 =
(4m)3â€“(7n)3

We know that, x3â€“y3 = (xâ€“y)(x2+xy+y2)

64m3â€“343n3 = (4m)3â€“(7n)3

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

= (4m-7n)(16m2+28mn+49n2)

11. Factorise:Â 27x3+y3+z3â€“9xyzÂ

Solution:

The expression27x3+y3+z3â€“9xyzÂ can be written as (3x)3+y3+z3â€“3(3x)(y)(z)

27x3+y3+z3â€“9xyzÂ  = (3x)3+y3+z3â€“3(3x)(y)(z)

We know that, x3+y3+z3â€“3xyz = (x+y+z)(x2+y2+z2â€“xy â€“yzâ€“zx)

27x3+y3+z3â€“9xyzÂ  = (3x)3+y3+z3â€“3(3x)(y)(z)

= (3x+y+z)[(3x)2+y2+z2â€“3xyâ€“yzâ€“3xz]

= (3x+y+z)(9x2+y2+z2â€“3xyâ€“yzâ€“3xz)

12. Verify that:

x3+y3+z3â€“3xyz = (1/2) (x+y+z)[(xâ€“y)2+(yâ€“z)2+(zâ€“x)2]

Solution:

We know that,

x3+y3+z3âˆ’3xyz = (x+y+z)(x2+y2+z2â€“xyâ€“yzâ€“xz)

â‡’ x3+y3+z3â€“3xyz = (1/2)(x+y+z)[2(x2+y2+z2â€“xyâ€“yzâ€“xz)]

= (1/2)(x+y+z)(2x2+2y2+2z2â€“2xyâ€“2yzâ€“2xz)

= (1/2)(x+y+z)[(x2+y2âˆ’2xy)+(y2+z2â€“2yz)+(x2+z2â€“2xz)]

= (1/2)(x+y+z)[(xâ€“y)2+(yâ€“z)2+(zâ€“x)2]

13. IfÂ  x+y+z =Â 0, show thatÂ x3+y3+z3 = 3xyz.

Solution:

We know that,

x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2â€“xyâ€“yzâ€“xz)

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

then, x3+y3+z3 -3xyz = (0)(x2+y2+z2â€“xyâ€“yzâ€“xz)

â‡’ x3+y3+z3â€“3xyz = 0

â‡’ x3+y3+z3 = 3xyz

Hence Proved

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

(i) (âˆ’12)3+(7)3+(5)3

(ii) (28)3+(âˆ’15)3+(âˆ’13)3

Solution:

(i) (âˆ’12)3+(7)3+(5)3

Let a = âˆ’12

b = 7

c = 5

We know that ifÂ x+y+z =Â 0, then x3+y3+z3=3xyz.

Here, âˆ’12+7+5=0

(âˆ’12)3+(7)3+(5)3 = 3xyz

= 3Ã—-12Ã—7Ã—5

= -1260

(ii) (28)3+(âˆ’15)3+(âˆ’13)3

Solution:

(28)3+(âˆ’15)3+(âˆ’13)3

Let a = 28

b = âˆ’15

c = âˆ’13

We know that ifÂ x+y+z =Â 0, then x3+y3+z3 = 3xyz.

Here, x+y+zÂ = 28â€“15â€“13 = 0

(28)3+(âˆ’15)3+(âˆ’13)3 = 3xyz

= 0+3(28)(âˆ’15)(âˆ’13)

= 16380

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:Â

(i) Area :Â 25a2â€“35a+12

(ii) Area :Â 35y2+13yâ€“12

Solution:

(i) Area :Â 25a2â€“35a+12

Using the splitting the middle term method,

We have to find a number whose sum = -35 and product =25Ã—12=300

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

25a2â€“35a+12 = 25a2â€“15aâˆ’20a+12

= 5a(5aâ€“3)â€“4(5aâ€“3)

= (5aâ€“4)(5aâ€“3)

Possible expression for lengthÂ  = 5aâ€“4

Possible expression for breadthÂ  = 5a â€“3

(ii) Area :Â 35y2+13yâ€“12

Using the splitting the middle term method,

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

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

35y2+13yâ€“12 = 35y2â€“15y+28yâ€“12

= 5y(7yâ€“3)+4(7yâ€“3)

= (5y+4)(7yâ€“3)

Possible expression for lengthÂ  =Â (5y+4)

Possible expression for breadthÂ  =Â (7yâ€“3)

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

(i) Volume :Â 3x2â€“12x

(ii) Volume :Â 12ky2+8kyâ€“20k

Solution:

(i) Volume :Â 3x2â€“12x

3x2â€“12xÂ can be written as 3x(xâ€“4) by taking 3x out of both the terms.

Possible expression for lengthÂ =Â 3

Possible expression for breadthÂ =Â x

Possible expression for heightÂ =Â (xâ€“4)

(ii) Volume:
12ky2+8kyâ€“20k

12ky2+8kyâ€“20kÂ can be written as 4k(3y2+2yâ€“5) by taking 4k out of both the terms.

12ky2+8kyâ€“20k = 4k(3y2+2yâ€“5)

[Here, 3y2+2yâ€“5 can be written as 3y2+5yâ€“3yâ€“5 using splitting the middle term method.]

= 4k(3y2+5yâ€“3yâ€“5)

= 4k[y(3y+5)â€“1(3y+5)]

= 4k(3y+5)(yâ€“1)

Possible expression for lengthÂ = 4k

Possible expression for breadthÂ =Â (3y +5)

Possible expression for heightÂ =Â (y -1)

 Also Access: NCERT Solutions for Class 9 Maths Chapter 1 Number System NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Class 9 Maths Chapter 2 Polynomials Notes NCERT Exemplar Class 9 Maths Solutions for Chapter 2 â€“ Polynomials Class 9 Maths Chapter 2 Polynomials Important Questions RD Sharma Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials

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## 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 BYJUâ€™S experts to boost confidence among students in understanding the concepts covered in this chapter.

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