## NCERT Solutions Class 9 Maths Chapter 2 – Polynomials Free PDF Download

**NCERT Solutions Class 9 Maths Chapter 2 Polynomials** are provided here. BYJUâ€™S expert faculty create these NCERT Solutions to help students in preparation for their exams. BYJU’S provides NCERT Solutions for Class 9Â Maths which will help students to solve problems easily. They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT textbook for Class 9.

### Download Exclusively Curated Chapter Notes for Class 9 Maths Chapter – 2 Polynomials

### Download Most Important Questions for Class 9 Maths Chapter – 2 Polynomials

In **NCERT Solutions for Class 9,** students are introduced to many important topics that will be helpful for those who wish to pursue Mathematics as a subject in higher studies. NCERT Solutions help students to prepare for their upcoming exams by covering the updated CBSE syllabus for 2023-24 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 – Algebra and has a weightage of 12 marks in the Class 9 Maths CBSE examination. This chapter talks about:

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

Next, it discusses the following topics:

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

**Key Advantages of NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials**

- These
**NCERT Solutions for Class 9 Maths**help you solve and revise the updated CBSE syllabus of Class 9 for 2023-24. - 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 Maths exams.

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

## NCERT Solutions for Class 9 Maths Chapter 2 â€“ Polynomials

## 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

## Access Answers of NCERT Class 9 Maths Chapter 2 – Polynomials

## 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}â€“3x+7**

Solution:

The equation 4x^{2}â€“3x+7 can be written as 4x^{2}â€“3x^{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}â€“3x+7 is a polynomial in one variable.

**(ii) y ^{2}+âˆš2**

Solution:

The equation y^{2}+**âˆš2** can be written as y^{2}+**âˆš**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}+**âˆš**2 is a polynomial in one variable.

**(iii) 3âˆšt+tâˆš2 **

Solution:

The equation 3âˆšt+tâˆš2 can be written as 3t^{1/2}+âˆš2t

Though *t* is the only variable in the given equation, the power 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 can be written as y+2y^{-1}

Though *y *is the only variable in the given equation, the power 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 the coefficient is the number which multiplies the variable.

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

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

**(ii) 2â€“x ^{2}+x^{3}**

Solution:

The equation 2â€“x^{2}+x^{3 }can be written as 2+(â€“1)x^{2}+x^{3}

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

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

Hence, the coefficient of x^{2 }in 2â€“x^{2}+x^{3 }is -1.

**(iii) ( Ï€/2)x^{2}+x**

Solution:

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

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

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

Hence, the coefficient of x^{2 }in (Ï€/2)x^{2 }+x is Ï€/2.

**(iii)âˆš2x-1**

Solution:

The equation âˆš2x-1 can be written as 0x^{2}+âˆš2x-1 [Since 0x^{2} is 0]

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

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

Hence, the coefficient of x^{2 }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.

For example, Â 3x^{35}+5

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

For example, Â 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) 4â€“y ^{2}**

Solution:

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

Here, in 4â€“y^{2},

The power of the variable y is 2

The degree of 4â€“y^{2} 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 5t**â€“âˆš7Â **

The power of the variable t is: 1

The degree of 5t**â€“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— x^{0}

The power of the variable here is: 0

Hence, 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) xâ€“x ^{3}**

Solution:

The highest power of xâ€“x^{3 }is 3

The degree is 3

Hence, xâ€“x^{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}+4 is 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)=5xâˆ’4x ^{2}+3.Â **

**(i) x = 0**

**(ii) xÂ = â€“ 1**

**(iii) xÂ = 2**

Solution:

LetÂ f(x) = 5xâˆ’4x^{2}+3

(i) When x = 0

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

= 3

(ii)Â When x = -1

f(x) = 5xâˆ’4x^{2}+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’4x^{2}+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)=y ^{2}âˆ’y+1**

Solution:

p(y) = y^{2}â€“y+1

âˆ´ p(0) = (0)^{2}âˆ’(0)+1 = 1

p(1) = (1)^{2}â€“(1)+1 = 1

p(2) = (2)^{2}â€“(2)+1 = 3

**(ii) p(t)=2+t+2t ^{2}âˆ’t^{3}**

Solution:

p(t) = 2+t+2t^{2}âˆ’t^{3}

âˆ´ p(0) = 2+0+2(0)^{2}â€“(0)^{3 }= 2

p(1) = 2+1+2(1)^{2}â€“(1)^{3}=2+1+2â€“1 = 4

p(2) = 2+2+2(2)^{2}â€“(2)^{3}=2+2+8â€“8 = 4

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

Solution:

p(x) = x^{3}

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) = x ^{2}âˆ’1, x = 1, âˆ’1**

Solution:

For,Â x = 1, âˆ’1;

p(x) = x^{2}âˆ’1

âˆ´ p(1)=1^{2}âˆ’1=1âˆ’1 = 0

p(âˆ’1)=(-1)^{2}âˆ’1 = 1âˆ’1 = 0

âˆ´ 1, âˆ’1 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, 2 are zeros of p(x).

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

Solution:

For,Â x = 0Â p(x) = x^{2}

p(0) = 0^{2 }= 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) = *l*x+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) = 3x ^{2}âˆ’1, x = -1/âˆš3 , 2/âˆš3 **

Solution:

For,Â x = -1/âˆš3 , 2/âˆš3 ; p(x) = 3x^{2}âˆ’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/âˆš3 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 a 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 x ^{3}+3x^{2}+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€^{2}âˆ’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 x ^{3}âˆ’ax^{2}+6xâˆ’aÂ is divided by x-a.**

Solution:

LetÂ p(x) = x^{3}âˆ’ax^{2}+6xâˆ’a

xâˆ’a = 0

âˆ´ x = a

Remainder:

p(a) = (a)^{3}âˆ’a(a^{2})+6(a)âˆ’a

= a^{3}âˆ’a^{3}+6aâˆ’a = 5a

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

## Exercise 2.4 Page: 43

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

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

Solution:

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

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

Solution:

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

**(iii) x ^{4}+3x^{3}+3x^{2}+x+1Â **

Solution:

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

**(iv) x ^{3 }â€“ x^{2}â€“ (2+âˆš2)x +âˆš2 **

Solution:

Let p(x) =Â x^{3}â€“x^{2}â€“(2+âˆš2)x +âˆš2

The zero of x+1 is -1.

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

= 2âˆš2 â‰ 0

âˆ´ By factor theorem, x+1 is not a factor of x^{3}â€“x^{2}â€“(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) = 2x ^{3}+x^{2}â€“2xâ€“1,Â g(x) =Â x+1**

Solution:

p(x) = 2x^{3}+x^{2}â€“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)^{2}â€“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)=x ^{3}+3x^{2}+3x+1,Â g(x) =Â x+2**

Solution:

p(x) = x^{3}+3x^{2}+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)=x ^{3}â€“4x^{2}+x+6,Â g(x) =Â xâ€“3**

Solution:

p(x) = x^{3}â€“4x^{2}+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)^{3}âˆ’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) = x ^{2}+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) = 2x ^{2}+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) = kx ^{2}–**âˆš

**2x+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)=kx ^{2}â€“3x+k**

Solution:

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

By Factor Theorem

â‡’ k(1)^{2}â€“3(1)+k = 0

â‡’ kâˆ’3+k = 0

â‡’ 2kâˆ’3 = 0

â‡’ k= 3/2

**4. Factorise:**

**(i) 12x ^{2}â€“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]

12x^{2}â€“7x+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 = 2Ã—3 = 6

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

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

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

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

**(iii) 6x ^{2}+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]

6x^{2}+5x-6 = 6x^{2}+9xâ€“4xâ€“6

= 3x(2x+3)â€“2(2x+3)

= (2x+3)(3xâ€“2)

**(iv) 3x ^{2}â€“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]

3x^{2}â€“xâ€“4Â = 3x^{2}â€“4x+3xâ€“4

= x(3xâ€“4)+1(3xâ€“4)

= (3xâ€“4)(x+1)

**5. Factorise:**

**(i) x ^{3}â€“2x^{2}â€“x+2**

Solution:

LetÂ p(x) = x^{3}â€“2x^{2}â€“x+2

Factors of 2 are Â±1 and Â± 2

Now,

p(x) = x^{3}â€“2x^{2}â€“x+2

p(âˆ’1) = (âˆ’1)^{3}â€“2(âˆ’1)^{2}â€“(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)(x^{2}â€“3x+2) = (x+1)(x^{2}â€“xâ€“2x+2)

= (x+1)(x(xâˆ’1)âˆ’2(xâˆ’1))

= (x+1)(xâˆ’1)(x-2)

**(ii) x ^{3}â€“3x^{2}â€“9xâ€“5**

Solution:

LetÂ p(x) =Â x^{3}â€“3x^{2}â€“9xâ€“5

Factors of 5 are Â±1 and Â±5

By the trial method, we find that

p(5)Â = 0

So,Â (x-5)Â is factor ofÂ p(x)

Now,

p(x) =Â x^{3}â€“3x^{2}â€“9xâ€“5

p(5) =Â (5)^{3}â€“3(5)^{2}â€“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)(x^{2}+2x+1) = (xâˆ’5)(x^{2}+x+x+1)

= (xâˆ’5)(x(x+1)+1(x+1))

= (xâˆ’5)(x+1)(x+1)

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

Solution:

LetÂ p(x) =Â x^{3}+13x^{2}+32x+20

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

By the trial method, we find that

p(-1)Â = 0

So,Â (x+1)Â is factor ofÂ p(x)

Now,

p(x)=Â x^{3}+13x^{2}+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)(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)

**(iv) 2y ^{3}+y^{2}â€“2yâ€“1**

Solution:

LetÂ p(y) =Â 2y^{3}+y^{2}â€“2yâ€“1

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

By the trial method, we find that

p(1)Â = 0

So,Â (y-1)Â is factor ofÂ p(y)

Now,

p(y) =Â 2y^{3}+y^{2}â€“2yâ€“1

p(1) =Â 2(1)^{3}+(1)^{2}â€“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)(2y^{2}+3y+1) = (yâˆ’1)(2y^{2}+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

We get,

(x+4)(x+10) = x^{2}+(4+10)x+(4Ã—10)

= x^{2}+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

We get,

(x+8)(xâˆ’10) = x^{2}+(8+(âˆ’10))x+(8Ã—(âˆ’10))

= x^{2}+(8âˆ’10)xâ€“80

= x^{2}âˆ’2xâˆ’80

**(iii) (3x+4)(3xâ€“5)**

Solution:

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

We get,

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

= 9x^{2}+3x(4â€“5)â€“20

= 9x^{2}â€“3xâ€“20

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

Solution:

Using the identity, (x+y)(xâ€“y) = x^{2}â€“y^{ 2}

^{2}and y = 3/2]

We get,

(y^{2}+3/2)(y^{2}â€“3/2) = (y^{2})^{2}â€“(3/2)^{2}

= y^{4}â€“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) = x^{2}+(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) = x^{2}-(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)= a^{2}-b^{2}]

Here, a = 100

b = 4

We get, 104Ã—96 = (100+4)Ã—(100â€“4)

= (100)^{2}â€“(4)^{2}

= 10000â€“16

= 9984

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

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

Solution:

9x^{2}+6xy+y^{2 }= (3x)^{2}+(2Ã—3xÃ—y)+y^{2}

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

Here, x = 3x

y = y

9x^{2}+6xy+y^{2 }= (3x)^{2}+(2Ã—3xÃ—y)+y^{2}

= (3x+y)^{2}

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

**(ii) 4y ^{2}âˆ’4y+1**

Solution:

4y^{2}âˆ’4y+1 = (2y)^{2}â€“(2Ã—2yÃ—1)+1

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

Here, x = 2y

y = 1

4y^{2}âˆ’4y+1 = (2y)^{2}â€“(2Ã—2yÃ—1)+1^{2}

= (2yâ€“1)^{2}

= (2yâ€“1)(2yâ€“1)

**(iii) Â x ^{2}â€“y^{2}/100**

Solution:

x^{2}â€“y^{2}/100 = x^{2}â€“(y/10)^{2}

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

Here, x = x

y = y/10

x^{2}â€“y^{2}/100 = x^{2}â€“(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} = 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}+(2Ã—xÃ—2y)+(2Ã—2yÃ—4z)+(2Ã—4zÃ—x)

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

**(ii) (2xâˆ’y+z) ^{2}Â **

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

Here, x = 2x

y = âˆ’y

z = z

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

= 4x^{2}+y^{2}+z^{2}â€“4xyâ€“2yz+4xz

**(iii) (âˆ’2x+3y+2z) ^{2}**

Solution:

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

= 4x^{2}+9y^{2}+4z^{2}â€“12xy+12yzâ€“8xz

**(iv) (3a â€“7bâ€“c) ^{2}**

Solution:

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

= 9a^{2} + 49b^{2 }+ c^{2}â€“ 42ab+14bcâ€“6ca

**(v) (â€“2x+5yâ€“3z) ^{2}**

Solution:

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

= 4x^{2}+25y^{2 }+9z^{2}â€“ 20xyâ€“30yz+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. Factorise:**

**(i) 4x ^{2}+9y^{2}+16z^{2}+12xyâ€“24yzâ€“16xz**

**(ii) 2x ^{2}+y^{2}+8z^{2}â€“2âˆš2xy+4âˆš2yzâ€“8xz**

Solution:

**(i) 4x ^{2}+9y^{2}+16z^{2}+12xyâ€“24yzâ€“16xz**

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}+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) 2x ^{2}+y^{2}+8z^{2}â€“2âˆš2xy+4âˆš2yzâ€“8xz**

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}â€“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} = x^{3}+y^{3}+3xy(x+y)

(2x+1)^{3}= (2x)^{3}+1^{3}+(3Ã—2xÃ—1)(2x+1)

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

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

**(ii) (2aâˆ’3b) ^{3}**

Using identity,(xâ€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y)

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

= 8a^{3}â€“27b^{3}â€“18ab(2aâ€“3b)

= 8a^{3}â€“27b^{3}â€“36a^{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}+(3Ã—(3/2)xÃ—1)((3/2)x +1)

**(iv) Â (xâˆ’(2/3)y) ^{3}**

Using identity, (x â€“y)^{3} = x^{3}â€“y^{3}â€“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} = x^{3}â€“y^{3}â€“3xy(xâ€“y)

(99)^{3 }= (100â€“1)^{3}

= (100)^{3}â€“1^{3}â€“(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} = x^{3}+y^{3}+3xy(x+y)

(100+2)^{3 }=(100)^{3}+2^{3}+(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} = x^{3}â€“y^{3}â€“3xy(xâ€“y)

(998)^{3 }=(1000â€“2)^{3}

=(1000)^{3}â€“2^{3}â€“(3Ã—1000Ã—2)(1000â€“2)

= 1000000000â€“8â€“6000(1000â€“ 2)

= 1000000000â€“8- 6000000+12000

= 994011992

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

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

**(ii) 8a ^{3}â€“b^{3}â€“12a^{2}b+6ab^{2}**

**(iii) 27â€“125a ^{3}â€“135a +225a^{2}Â Â **

**(iv) 64a ^{3}â€“27b^{3}â€“144a^{2}b+108ab^{2}**

**(v) 27p ^{3}â€“(1/216)âˆ’(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}â€“b^{3}â€“12a^{2}b+6ab^{2}**

Solution:

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

8a^{3}â€“b^{3}âˆ’12a^{2}b+6ab^{2 }= (2a)^{3}â€“b^{3}â€“3(2a)^{2}b+3(2a)(b)^{2}

= (2aâ€“b)^{3}

= (2aâ€“b)(2aâ€“b)(2aâ€“b)

Here, the identity,(xâ€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y) is used.

**(iii) 27â€“125a ^{3}â€“135a+225a^{2}Â **

Solution:

The expression, 27â€“125a^{3}â€“135a +225a^{2} can be written as 3^{3}â€“(5a)^{3}â€“3(3)^{2}(5a)+3(3)(5a)^{2}

27â€“125a^{3}â€“135a+225a^{2 }=

3^{3}â€“(5a)^{3}â€“3(3)^{2}(5a)+3(3)(5a)^{2}

= (3â€“5a)^{3}

= (3â€“5a)(3â€“5a)(3â€“5a)

Here, the identity, (xâ€“y)^{3} = x^{3}â€“y^{3}-3xy(xâ€“y) is used.

**(iv) 64a ^{3}â€“27b^{3}â€“144a^{2}b+108ab^{2}**

Solution:

The expression, 64a^{3}â€“27b^{3}â€“144a^{2}b+108ab^{2}can be written as (4a)^{3}â€“(3b)^{3}â€“3(4a)^{2}(3b)+3(4a)(3b)^{2}

64a^{3}â€“27b^{3}â€“144a^{2}b+108ab^{2}=

(4a)^{3}â€“(3b)^{3}â€“3(4a)^{2}(3b)+3(4a)(3b)^{2}

=(4aâ€“3b)^{3}

=(4aâ€“3b)(4aâ€“3b)(4aâ€“3b)

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

**(v) 27p ^{3}â€“ (1/216)âˆ’(9/2) p^{2}+(1/4)p**

Solution:

The expression, 27p^{3}â€“(1/216)âˆ’(9/2) p^{2}+(1/4)p can be written as

(3p)^{3}â€“(1/6)^{3}âˆ’(9/2) p^{2}+(1/4)p =Â (3p)^{3}â€“(1/6)^{3}âˆ’3(3p)(1/6)(3p – 1/6)

Using (x – y)^{3} = x^{3} – y^{3} – 3xy (x – y)

27p^{3}â€“(1/216)âˆ’(9/2) p^{2}+(1/4)p =Â (3p)^{3}â€“(1/6)^{3}âˆ’3(3p)(1/6)(3p – 1/6)

Taking x = 3p and y = 1/6

= (3pâ€“1/6)^{3}

= (3pâ€“1/6)(3pâ€“1/6)(3pâ€“1/6)

**9. Verify:**

**(i) x ^{3}+y^{3 }= (x+y)(x^{2}â€“xy+y^{2})**

**(ii) x ^{3}â€“y^{3 }= (xâ€“y)(x^{2}+xy+y^{2})**

Solutions:

**(i) x ^{3}+y^{3 }= (x+y)(x^{2}â€“xy+y^{2})**

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

â‡’ x^{3}+y^{3 }= (x+y)^{3}â€“3xy(x+y)

â‡’ x^{3}+y^{3 }= (x+y)[(x+y)^{2}â€“3xy]

Taking (x+y) common â‡’ x^{3}+y^{3 }= (x+y)[(x^{2}+y^{2}+2xy)â€“3xy]

â‡’ x^{3}+y^{3 }= (x+y)(x^{2}+y^{2}â€“xy)

**(ii) x ^{3}â€“y^{3 }= (xâ€“y)(x^{2}+xy+y^{2})Â **

We know that, (xâ€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y)

â‡’ x^{3}âˆ’y^{3 }= (xâ€“y)^{3}+3xy(xâ€“y)

â‡’ x^{3}âˆ’y^{3 }= (xâ€“y)[(xâ€“y)^{2}+3xy]

Taking (x+y) common â‡’ x^{3}âˆ’y^{3 }= (xâ€“y)[(x^{2}+y^{2}â€“2xy)+3xy]

â‡’ x^{3}+y^{3 }= (xâ€“y)(x^{2}+y^{2}+xy)

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

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

**(ii) 64m ^{3}â€“343n^{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}â€“xy+y^{2})

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

= (3y+5z)[(3y)^{2}â€“(3y)(5z)+(5z)^{2}]

= (3y+5z)(9y^{2}â€“15yz+25z^{2})

**(ii) 64m ^{3}â€“343n^{3}**

The expression, 64m^{3}â€“343n^{3}can be written as (4m)^{3}â€“(7n)^{3}

64m^{3}â€“343n^{3 }=

(4m)^{3}â€“(7n)^{3}

We know that, x^{3}â€“y^{3 }= (xâ€“y)(x^{2}+xy+y^{2})

64m^{3}â€“343n^{3 }= (4m)^{3}â€“(7n)^{3}

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

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

**11. Factorise:Â 27x ^{3}+y^{3}+z^{3}â€“9xyz.Â **

Solution:

The expression 27x^{3}+y^{3}+z^{3}â€“9xyzÂ can be written as (3x)^{3}+y^{3}+z^{3}â€“3(3x)(y)(z)

27x^{3}+y^{3}+z^{3}â€“9xyzÂ = (3x)^{3}+y^{3}+z^{3}â€“3(3x)(y)(z)

We know that, x^{3}+y^{3}+z^{3}â€“3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}â€“xy â€“yzâ€“zx)

27x^{3}+y^{3}+z^{3}â€“9xyzÂ = (3x)^{3}+y^{3}+z^{3}â€“3(3x)(y)(z)

= (3x+y+z)[(3x)^{2}+y^{2}+z^{2}â€“3xyâ€“yzâ€“3xz]

= (3x+y+z)(9x^{2}+y^{2}+z^{2}â€“3xyâ€“yzâ€“3xz)

**12. Verify that:**

**x ^{3}+y^{3}+z^{3}â€“3xyz = (1/2) (x+y+z)[(xâ€“y)^{2}+(yâ€“z)^{2}+(zâ€“x)^{2}]**

Solution:

We know that,

x^{3}+y^{3}+z^{3}âˆ’3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}â€“xyâ€“yzâ€“xz)

â‡’ x^{3}+y^{3}+z^{3}â€“3xyz = (1/2)(x+y+z)[2(x^{2}+y^{2}+z^{2}â€“xyâ€“yzâ€“xz)]

= (1/2)(x+y+z)(2x^{2}+2y^{2}+2z^{2}â€“2xyâ€“2yzâ€“2xz)

= (1/2)(x+y+z)[(x^{2}+y^{2}âˆ’2xy)+(y^{2}+z^{2}â€“2yz)+(x^{2}+z^{2}â€“2xz)]

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

**13. IfÂ x+y+z =Â 0, show thatÂ 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}â€“xyâ€“yzâ€“xz)

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

Then, x^{3}+y^{3}+z^{3 }-3xyz = (0)(x^{2}+y^{2}+z^{2}â€“xyâ€“yzâ€“xz)

â‡’ x^{3}+y^{3}+z^{3}â€“3xyz = 0

â‡’ x^{3}+y^{3}+z^{3 }= 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 x^{3}+y^{3}+z^{3}=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 x^{3}+y^{3}+z^{3 }= 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: 25a ^{2}â€“35a+12**

**(ii) Area: 35y ^{2}+13yâ€“12**

Solution:

(i) Area: 25a^{2}â€“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]

25a^{2}â€“35a+12 = 25a^{2}â€“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: 35y^{2}+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]

35y^{2}+13yâ€“12 = 35y^{2}â€“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: 3x ^{2}â€“12x**

**(ii) Volume: 12ky ^{2}+8kyâ€“20k**

Solution:

(i) Volume: 3x^{2}â€“12x

3x^{2}â€“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:

12ky^{2}+8kyâ€“20k

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

12ky^{2}+8kyâ€“20k = 4k(3y^{2}+2yâ€“5)

^{2}+2yâ€“5 can be written as 3y

^{2}+5yâ€“3yâ€“5 using splitting the middle term method.]

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

**Disclaimer:**

**Dropped Topics –**Â 2.4 Remainder theorem.

## 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 polynomials, 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.

Thanks a lot byjus

It makes my homework easyðŸ˜„

Thanks a lot byjus.

It helped me for studying easily. ðŸ˜€