NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials provide comprehensive, step-by-step explanations aligned with the latest CBSE Class 9 Maths syllabus. Chapter 2 – Polynomials is a foundational algebra unit that directly impacts students’ performance in Class 9 examinations and builds the conceptual base required for quadratic equations, coordinate geometry, and higher algebra in Class 10 Syllabus and beyond.

Polynomials are not just textbook concepts — they form the core of algebraic problem-solving. In this chapter, students learn to identify polynomials in one variable, determine their degree, calculate zeroes, apply the Remainder and Factor Theorems, and use algebraic identities for expansion and factorisation. These concepts are frequently tested in school exams and form a recurring theme in higher classes.

This guide offers:

• Exercise-wise NCERT solutions with clear justifications
• Concept-based explanations instead of shortcut memorisation
• Accurate application of theorems as prescribed in NCERT
• Structured methods to reduce common calculation errors

Students often look for clarity on:

• How to find the zeroes of a polynomial correctly
• How to apply the Remainder Theorem step-by-step
• How to factorise expressions using identities
• Where to access reliable Class 9 Maths Chapter 2 PDF solutions

This page addresses each of these with academically sound explanations, ensuring conceptual understanding rather than rote learning.

By mastering these Class 9 Maths NCERT solutions, students strengthen their algebraic foundation, improve analytical thinking, and prepare effectively for both school assessments and future mathematical concepts.

Download NCERT Solutions Class 9 Maths Chapter 2 PDF Online

Downloading the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials PDF helps students revise important concepts in a structured and exam-focused manner. The PDF format allows quick access to exercise-wise answers, formulas, and theorem-based solutions anytime, making it especially useful before school tests and final exams.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials: Overview

Chapter 2 – Polynomials is a core algebra chapter in the CBSE Class 9 Maths syllabus. It introduces students to algebraic expressions, polynomial operations, factorisation techniques, and important theorems that are essential for higher mathematics.

The NCERT Solutions Class 9 Maths Chapter 2 Polynomials simplify these concepts through:

• Detailed explanations based on the latest NCERT textbook
• Exercise-wise step-by-step solutions
• Clear demonstration of methods used in factorisation
• Proper application of Remainder and Factor Theorem
• Logical reasoning behind algebraic identities

By following the Class 9 Maths Chapter 2 Polynomials solutions, students can clearly understand how to approach each question systematically instead of relying on memorisation.

Chapter 2 Polynomials Subtopics of NCERT Class 9 Maths

The following key topics are covered in CBSE Class 9 Maths Chapter 2 – Polynomials:

• Polynomials in One Variable
• Zeroes of a Polynomial
• Factorisation of Polynomials
• Remainder Theorem
• Factor Theorem
• Algebraic Identities

Each topic plays an important role in building a strong algebra foundation and is frequently tested in school examinations.

Polynomials in NCERT Solutions for Class 9 Maths, Chapter 2: All Exercises

The NCERT Solutions for Class 9 Maths Chapter 2 Polynomials cover all textbook exercises with complete working steps. Every question is solved using standard NCERT methods to ensure conceptual clarity and exam accuracy.

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

1. 4x² - 3x + 7
2. y² + √2
3. 3√t + t√2
4. y + 2/y
5. x¹⁰ + y³ + t⁵⁰

Answer:

1. 4x² - 3x + 7

There is only one variable x with whole number power. So, this polynomial is in one variable.

There is only one variable y with whole number power. So, this polynomial is in one variable.

There is only one variable t but in 3√t, power of t is 1/2 which is not a whole number. So, this is not a polynomial.

There is only one variable y but in 2/y, power of y is -1/2 which is not a whole number. So, this is not a polynomial.

There are three variables x, y and z and these powers are whole number. So, this is not a polynomial.

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

1. 2 + x² + x
2. 2 - x² + x³
3. (π/2) × x² + x
4. √2x - 1

Answer:

1. 2 + x² + x = 2 + 1 × x² + x

So, the coefficient of x² is 1.

So, the coefficient of x² is -1.

So, the coefficient of x² is π/2.

So, the coefficient of x² is 0.

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

Answer:

A binomial of degree 35 = x³⁵ + 7

A binomial of degree 100 = 2x¹⁰⁰

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

1. 5x³ + 4x² + 7x
2. 4 - y²
3. 5t - √7
4. 3

Answer:

1. The degree of 5x³ + 4x² + 7x is 3.
2. The degree of 4 - y² is 2.
3. The degree of 5t - √7 = 5t¹ - √7 is 1.
4. The degree of 3 = 3x⁰ is 0.

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

1. x² + x
2. x - x³
3. y + y² + 4
4. 1 + x
5. 3t
6. r²
7. 7x³

Answer:

1. x² + x is a quadratic polynomial.
2. x - x³ is a cubic polynomial.
3. y + y² + 4 is a quadratic polynomial.
4. 1 + x is a linear polynomial.
5. 3t is a linear polynomial.
6. r² is a quadratic polynomial.
7. 7x³ is a cubic polynomial.

NCERT Solutions for Class 9 Maths Polynomials Exercise 2.2

Question 1: Find the value of the polynomial 5x - 4x² + 3 at

1. x = 0
2. x = -1
3. x = 2

Answer:

Let p(x) = 5x - 4x² + 3

1. Put x = 0, we get

p(0) = 5 × 0 – 4 × 0 + 3 = 0 – 0 + 3 = 3

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

p(2) = 5 × 2 - 4 × 2² + 3 = 10 - 4 × 4 + 3 = 10 - 16 + 3 = 13 - 16 = -3

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

1. p(y) = y² – y + 1
2. p(t) = 2 + t + 2t² – t³
3. p(x) = x³
4. p(x) = (x – 1)(x + 1)

Answer:

1. p(y) = y² – y + 1

p(0) = (0)² – (0) + 1 = 0 – 0 + 1 = 1

p(1) = (1)² – (1) + 1 = 1 – 1 + 1 = 1

p(2) = (2)² – 2 + 1 = 4 – 2 + 1 = 3

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

p(1) = 2 + (1) + 2(1)² – (1)³ = 2 + 1 + 2 – 1 = 4

p(2) = 2 + 2 + 2(2)² – (2)³ = 2 + 2 + 8 – 8 = 4

p(0) = (0)³ = 0

p(1) = (1)³ = 1

p(2) = (2)³ = 8

p(0) = (0 – 1)(0 + 1) = –1 × 1 = –1

p(1) = (1 – 1)(1 + 1) = 0 × 2 = 0

p(2) = (2 – 1)(2 + 1) = 1 × 3 = 3

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

1. p(x) = 3x + 1, x = -1/3
2. p(x) = 5x – π, x = 4/5
3. p(x) = x² – 1, x = 1, –1
4. p(x) = (x + 1)(x – 2), x = –1, 2
5. p(x) = x², x = 0
6. p(x) = lx + m, x = -m/l
7. p(x) = 3x² - 1, x = -1/√3, 2/√3
8. p(x) = 2x + 1, x = 1/2

Answer:

1. Put x = -1/3, we get

p(-1/3) = 3 × (-1/3) + 1 = -1 + 1 = 0

Hence, x = -1/3 is a zero of the polynomial p(x) = 3x + 1

p(x) = 5 × 4/5 – π = 4 – π

Hence, x = 4/5 is a zero of the polynomial p(x) = 5x – π

p(1) = 1² – 1 = 1 – 1 = 0

Hence, x = 1 is a zero of the polynomial p(x) = x² – 1

Put x = -1, we get

p(1) = (-1)² – 1 = 1 – 1 = 0

Hence, x = -1 is a zero of the polynomial p(x) = x² – 1

p(-1) = (-1 + 1)(-1 – 2) = 0 × (-3) = 0

Hence, x = -1 is a zero of the polynomial p(x) = (x + 1)(x - 2)

Put x = 2, we get

p(2) = (2 + 1)(2 – 2) = 3 × 0 = 0

Hence, x = 2 is a zero of the polynomial p(x) = (x + 1)(x - 2)

Put x = 0, we get

p(0) = 0² = 0

Hence, x = 0 is a zero of the polynomial p(x) = x²

p(-m/l) = l × (-m/l) + m = -m + m = 0

Hence, x = -m/l is a zero of the polynomial p(x) = lx + m

p(-1/√3) = 3 × (-1/√3)² – 1 = 3 × 1/3 – 1 = 1 – 1 = 0

Hence, x = -1/√3 is a zero of the polynomial 3x² – 1.

Put x = 2/√3, we get

p(2/√3) = 3 × (2/√3)² – 1 = 3 × 4/3 – 1 = 4 – 1 = 3.

It is not equal to zero. Hence, x = 2/√3 is not a zero of the polynomial 3x² – 1.

p(1/2) = 2 × (1/2) + 1 = 1 + 1 = 2

It is not equal to zero. Hence, x = 1/2 is not a zero of the polynomial 2x + 1.

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

1. p(x) = x + 5
2. p(x) = x – 5
3. p(x) = 2x + 5
4. p(x) = 3x – 2
5. p(x) = 3x
6. p(x) = ax, a ≠ 0
7. p(x) = cx + d, c ≠ 0, c, d are real numbers.

Answer:

To find the zero of a polynomial p(x), we have to equate it to zero i.e., p(x) = 0

1. p(x) = 0

=> x + 5 = 0

=> x = -5

Hence, x = -5 is zero of the given polynomials.

=> x – 5 = 0

=> x = 5

Hence, x = 5 is zero of the given polynomials.

=> 2x + 5 = 0

=> 2x = -5

=> x = -5/2

Hence, x = -5/2 is zero of the given polynomials.

=> 3x – 2 = 0

=> 3x = 2

=> x = 2/3

Hence, x = 2/3 is zero of the given polynomials.

=> 3x = 0

=> x = 0/3

=> x = 0

Hence, x = 0 is zero of the given polynomials.

=> ax = 0

=> x = 0/a

=> x = a

Hence, x = -5 is zero of the given polynomials where a ≠ 0.

=> cx + d = 0

=> cx = -d

=> c = -d/c

Hence, x = -d/c is zero of the given polynomials where c ≠ 0.

NCERT Solutions for Class 9 Maths Polynomials Exercise 2.3

Question 1: Find the remainder when x³ + 3x² + 3x + 1 is divided by

1. x + 1
2. x – 1/2
3. x
4. x + π
5. 5 + 2x

Answer:

Let p(x) = x³ + 3x² + 3x + 1

1. put x + 1 = 0, we get

x = -1

Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by x + 1, remainder is given by p(-1).

Now, p(-1) = (-1)³ + 3 × (-1)² + 3 × (-1) + 1

= -1 + 3 – 3 + 1

= -4 + 4

= 0

Hence, the remainder is 0.

x = 1/2

Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by x – 1/2, remainder is given by p(1/2).

Now, p(1/2) = (1/2)³ + 3 × (1/2)² + 3 × (1/2) + 1

= 1/8 + 3 × 1/4 + 3/2 + 1

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

= (1 × 1 + 3 × 2 + 3 × 4 + 1 × 8) ÷ 8

[LCM (8, 4, 2, 1) = 8]

= (1 + 6 + 12 + 8) ÷ 8

= 27/8

Hence, the remainder is 27/8.

Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by x, remainder is given by p(0).

Now, p(-1) = (0)³ + 3 × (0)² + 3 × (0) + 1

= 0 + 0 + 0 + 1

= 1

Hence, the remainder is 1.

Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by x + π, remainder is given by p(-π).

Now, p(-π) = (-π)³ + 3 × (-π)² + 3 × (-π) + 1

= -π³ + 3π³ - 3π + 1

Hence, the remainder is -π³ + 3π³ - 3π + 1.

2x = -5

=> x = -5/2

Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by 5 + 2x, remainder is given by p(-5/2).

Now, p(-5/2) = (-5/2)³ + 3 × (-5/2)² + 3 × (-5/2) + 1

= -125/8 + 3 × 25/4 - 15/2 + 1

= -125/8 + 75/4 - 15/2 + 1

= (-1 × 125 + 75 × 2 - 15 × 4 + 1 × 8) ÷ 8

[LCM (8, 4, 2, 1) = 8]

= (-125 + 150 - 60 + 8) ÷ 8

= -27/8

Hence, the remainder is 27/8.

Question 2: Find the remainder when x³ – ax² + 6x – a is divided by x – a.

Answer:

Let p(x) = x³ – ax² + 6x – a

Put x – a = 0, we get

x = a

Using remainder theorem, when p(x) = x³ – ax² + 6x – a is divided by x - a, remainder is given by p(a).

Now, p(a) = a³ – a × a² + 6 × a – a

= a³ – a³ + 6a – a

= 5a

Hence, the remainder is 5a.

Question 3:

Check whether 7 + 3x is a factor of 3x² + 7x.

Answer:

Given, 3x² + 7x = 3 × x × x + 7 × x

= x (3 × x + 7)

= x (3x + 7)

= x (7 + 3x)

Hence, 7 + 3x is a factor of 3x² + 7x.

NCERT Solutions for Class 9 Maths Polynomials Exercise 2.4

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

1. x³ + x² + x + 1
2. x⁴ + x³ + x² + x + 1
3. x⁴ + 3x³ + 3x² + x + 1
4. x³ – x² – (2 + √2) x + √2

Answer:

1. For x + 1 = 0, we have x = –1

Now, p(–1) = (–1)³ + (–1)² + (–1) + 1

= –1 + 1 – 1 + 1

= 0

i.e., when p(x) is divided by (x + 1), then the remainder is zero.

So, (x + 1) is a factor of x³ + x² + x + 1.

Now, p(–1) = (–1)⁴ + (–1)³ + (–1)² + (–1) + 1

= 1 –1 + 1 – 1 + 1

= 1

i.e., when p(x) is divided by (x + 1), then the remainder is zero.

Since p(x) is not divisible by x + 1

Hence, (x + 1) is not a factor of x⁴ + x³ + x² + x + 1.

Now, p(–1) = (–1)⁴ + 3(–1)³ + 3(–1)² + (-1) + 1

= 1 – 3 + 3 – 1 + 1

= 1

i.e., when p(x) is divided by (x + 1), then the remainder is zero.

Since p(x) is not divisible by x + 1.

Hence, (x + 1) is not a factor of x⁴ + 3x³ + 3x² + x + 1.

For x + 1 = 0, we have x = –1.

Now, p(–1) = (-1)³ – (-1)² – (2 + √2)(-1) + √2

= –1 - 1 + (2 + √2) + √2

= 2√2

i.e., when p(x) is divided by (x + 1), then the remainder is zero.

Since p(x) is not divisible by x + 1

So, (x + 1) is not a factor of x³ – x² – (2 + √2) x + √2.

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

1. p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1
2. p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
3. p(x) = x³ - 4x² + x + 6, g(x) = x - 3

Answer:

1. Given, p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1

Put x + 1 = 0, we get x = -1

Using remainder theorem, when p(x) = 2x³ + x² – 2x – 1 is divided by g(x) = x + 1, remainder is given by p(-1).

Now, p(-1) = 2 × (-1)³ + (-1)² – 2 × (-1) – 1

= -2 + 1 + 2 – 1

= -3 + 3

= 0

Since remainder p(-1) = 0, hence g(x) is a factor of p(x).

Put x + 2 = 0, we get x = -2

Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by g(x) = x + 2, remainder is given by p(-2).

Now, p(-2) = (-2)³ + 3 × (-2)² + 3 × (-2) + 1

= -8 + 3 × 4 – 3 × 2 + 1

= -8 + 12 – 6 + 1

= -14 + 13

= -1

Since remainder p(-2) ≠ 0, hence g(x) is not a factor of p(x).

Put x - 3 = 0, we get x = 3

Using remainder theorem, when p(x) = x³ - 4x² + x + 6 is divided by g(x) = x - 3, remainder is given by p(3).

Now, p(3) = 3³ - 4 × 3² + 3 + 6

= 27 - 4 × 9 + 3 + 6

= 27 - 36 + 3 + 6

= 36 - 36

= 0

Since remainder p(3) = 0, hence g(x) is a factor of p(x).

Question 3: Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

1. p(x) = x² + x + k
2. p(x) = 2x² + kx + √2
3. p(x) = kx² – √2x + 1
4. p(x) = kx² – 3x + k

Answer:

1. If (x - 1) is a factor of p(x) = x² + x + k, then p(1) = 0

=> 1² + 1 + k = 0

=> 2 + k = 0

=> k = -2

So, the value of k is -2.

=> 2 × 1² + k × 1 + √2 = 0

=> 2 + k + √2 = 0

=> k = -2 - √2

=> k = -(2 + √2)

So, the value of k is –(2 + √2)

=> k × 1² – √2 × 1 + 1 = 0

=> k - √2 + 1 = 0

=> k = √2 – 1

So, the value of k is √2 - 1

=> k × 1² – 3 × 1 + k = 0

=> k – 3 + k = 0

=> 2k – 3 = 0

=> k = 3/2

So, the value of k is 3/2.

Question 4: Factorise:

1. 12x² – 7x + 1
2. 2x² + 7x + 3
3. 6x² + 5x – 6
4. 3x² – x – 4

Answer:

1. 12x² – 7x + 1 = 12x² – (4 + 3) x + 1

= 12x² – 4x – 3x + 1

= 4x (3x - 1) – 1(3x - 1)

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

= 2x² + 6x + x + 3

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

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

= 6x² + 9x - 4x – 6

= 3x (2x + 3) – 2(2x + 3)

= (2x + 3)(3x - 2)

= 3x² – 4x + 3x – 4

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

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

Question 5: Factorise:

1. x³ – 2x² – x + 2
2. x³ – 3x² – 9x – 5
3. x³ + 13x² + 32x + 20
4. 2y³ + y² – 2y – 1

Answer:

1. Let p(x) = x³ – 2x² – x + 2

Put x = 1, we get

p(1) = 1³ – 2 × 1² – 1 + 2

= 1 – 2 – 1 + 2

= 3 – 3

= 0

Hence, x = 1 i.e. (x – 1) is a factor of p(x).

Now, divide x³ – 2x² – x + 2 by x – 1, we get

Now, x³ – 2x² – x + 2 = (x² – 3x + 2)(x + 1)

= (x² – x – 2x + 2)(x + 1)

= {x(x - 1) – 2(x - 1)}(x + 1)

= (x - 1)(x - 2)(x + 1)

Put x = 1, we get

p(1) = 1³ – 3 × 1² – 9 × 1 – 5

= 1 – 3 – 9 – 5

= 1 – 17

= -16 ≠ 0

Again, put x = -1, we get

p(1) = (-1)³ – 3 × (-1)² – 9 × (-1) – 5

= -1 – 3 + 9 – 5

= -9 + 9

= 0

Hence, x = -1 i.e. (x + 1) is a factor of p(x)

Now, divide x³ – 3x² – 9x – 5 by (x + 1), we get

Now, x³ – 3x² – 9x – 5 = (x + 1)(x² – 4x - 5)

= (x + 1)(x² – 5x + x - 5)

= (x + 1){x(x – 5) + 1(x - 5)}

= (x + 1)(x + 1)(x - 5)

= (x + 1)²(x - 5)

Put x = -1, we get

p(1) = (-1)³ + 13 × (-1)² + 32 × (-1) + 20

= -1 + 13 - 32 + 20

= -33 + 33

= 0

Hence, x = -1 i.e. (x + 1) is a factor of p(x)

Now, divide x³ + 13x² + 32x + 20 by (x + 1), we get

Now, x³ + 13x² + 32x + 20 = (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)

Put y = 1, we get

p(1) = 2 × (1)³ - (1)² - 2 × (-1) - 1

= 2 - 1 + 2 - 1

= -3 + 3

= 0

Hence, y = 1 i.e. (y - 1) is a factor of p(y)

Now, divide (2y³ + y² – 2y – 1) by (y - 1), we get

Again, 2y³ + y² – 2y – 1 = (y - 1)(2y² + 3y + 1)

= (y - 1)(2y² + 2y + y + 1)

= (y - 1){2y(y + 1) + 1(y + 1)}

= (y - 1)(y + 1)(2y + 1)

NCERT Solutions for Class 9 Maths Polynomials Exercise 2.5

Question 1: Use suitable identities to find the following products:

1. (x + 4)(x + 10)
2. (x + 8)(x – 10)
3. (3x + 4)(3x – 5)
4. (y² + 3/2)(y² – 3/2)
5. (3 – 2x)(3 + 2x)

Answer:

1. (x + 4)(x + 10)

= x² + (4 + 10) x + 4 × 10

[Using (x + a)(x + b) = x² + (a + b) x + a × b]

= x² + 14x + 4

= x² + (8 - 10) x + 8 × (-10)

[Using (x + a)(x + b) = x² + (a + b) x + a × b]

= x² - 2x - 80

= (3x)² + (4 - 5)3x + 4 × (-5)

[Using (x + a)(x + b) = x² + (a + b) x + a × b]

= 9x² - 3x - 20

= (y²)² – (3/2)²

[Using (a + b)(a - b) = a² – b²]

= y⁴ – 9/4

= (3)² – (2x)²

[Using (a + b)(a - b) = a² – b²]

= 9 – 4x²

Question 2: Evaluate the following products without multiplying directly:

1. 103 × 107
2. 95 × 96
3. 104 × 96

Answer:

1. 103 × 107 = (100 + 3)(100 + 7)

= 100² + (3 + 7)100 + 3 × 7

[Using (x + a)(x + b) = x² + (a + b) x + a × b]

= 10000 + 1000 + 21

= 11021

= 100² + (-5 - 4)100 + (-5) × (-4)

[Using (x + a)(x + b) = x² + (a + b) x + a × b]

= 10000 - 900 + 20

= 10020 – 900

= 9120

= 100² – 4²

[Using (a + b)(a - b) = a² – b²]

= 10000 - 16

= 9984

Question 3: Factorise the following using appropriate identities:

1. 9x² + 6xy + y²
2. 4y² – 4y + 1
3. x² – y²/100

Answer:

1. 9x² + 6xy + y² = (3x)² + 2 × 3x × y + y²

= (3x + y)²

[Using a² + 2ab + b² = (a + b)²]

= (2y - 1)²

[Using a² - 2ab + b² = (a - b)²]

= x² – (y/10)²

= (x – y/10)(x + y/10)

[a² – b² = (a - b)(a + b)]

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

1. (x + 2y + 4z)²
2. (2x – y + z)²
3. (–2x + 3y + 2z)²
4. (3a – 7b – c)²
5. (–2x + 5y – 3z)²
6. [a/4 – b/2 + 1]²

Answer:

1. (x + 2y + 4z)²

= x² + (2y)² + (4z)² + 2 × x × 2y + 2 × 2y × 4z + 2 × 4z × x

[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

= x² + 4y² + 16z² + 4xy + 16yz + 8zx

= (2x)² + (-y)² + z² + 2 × 2x × (-y) + 2 × (-y) × z + 2 × 2x × z

[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

= 4x² + y² + z² – 4xy – 2yz + 4xz

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

[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

= 4x² + 9y² + 4z² – 12xy + 12yz - 8xz

= (3a)² + (-7b)² + (-c)² + 2 × 3a × (-7b) + 2 × (-7b) × (-c) + 2 × 3a × (-c)

[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

= 9a² + 49b² + c² – 42ab + 14bc – 6ca

= (-2x)² + (5y)² + (-3z)² + 2 × (-2x) × 5y + 2 × 5y × (-3z) + 2 × (-2x) × (-3z)

[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

= 4x² + 25y² + 9z² – 20xy - 30yz + 12xz

= (a/4)² + (-b/2)² + 1² + 2 × (a/4) × (-b/2) + 2 × (-b/2) × 1 + 2 × 1 × (a/4)

[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

= a²/16 + b²/4 + 1 – ab/4 – b + a/2

Question 5: Factorise:

1. 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
2. 2x² + y² + 8z² – 2√2 xy + 4√2 yz – 8xz

Answer:

1. 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

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

[Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

= (2x + 3y – 4z)²

= (√2x)² + (-y)² + (-2√2z)² + 2 × (√2x) × (-y) + 2 × (-2√2z) × (-y) + 2 × (√2x) × (-2√2z)

[Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

= (√2x – y – 2√2z)²

Question 6: Write the following cubes in expanded form:

1. (2x + 1)³
2. (2a – 3b)³
3. [3x/2 + 1]³
4. [x – 2y/3]³

Answer:

1. (2x + 1)³ = (2x)³ + 1³ + 3 × (2x)² × 1 + 3 × 2x × 1²

[Using (a + b) = a³ + b³ + 3a²b + 3ab²]

= 8x³ + 1 + 12x² + 6x

[Using (a + b) = a³ + b³ + 3a²b + 3ab²]

= 8a³ – 27b³ – 36a²b + 54ab²

[Using (a + b) = a³ + b³ + 3a²b + 3ab²]

= 27x³/8 + 1 + 27x²/4 + 9x/2

[Using (a + b) = a³ + b³ + 3a²b + 3ab²]

= x³ – 8y³/27 – 2x²y + 4xy²/3

Question 7: Evaluate the following using suitable identities:

1. (99)³
2. (102)³
3. (998)³

Answer:

1. (99)³ = (100 - 1)³ = {(100 + (-1)}³

= (100)³ + (-1)³ + 3 × (100)²(-1) + 3 × 100 × (-1)²

[Using (a + b)³ = a³ + b³ + 3a²b + 3ab²]

= 1000000 – 1 – 3 × 10000 + 3 × 100

= 1000000 – 1 – 30000 + 300

= 970299

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

[Using (a + b)³ = a³ + b³ + 3a²b + 3ab²]

= 1000000 + 8 + 3 × 10000 × 2 + 3 × 100 × 4

= 1000000 + 8 + 60000 + 1200

= 1061208

= (1000)³ + (-1)³ + 3 × (1000)²(-2) + 3 × 1000 × (-2)²

[Using (a + b)³ = a³ + b³ + 3a²b + 3ab²]

= 1000000000 – 8 – 3 × 1000000 × 2 + 3 × 1000 × 4

= 1000000000 – 1 – 6000000 + 12000

= 994011992

Question 8: Factorise each of the following:

1. 8a³ + b³ + 12a²b + 6ab²
2. 8a³ – b³ – 12a²b + 6ab²
3. 27 – 125a³ – 135a + 225a²
4. 64a³ – 27b³ – 144a²b + 108ab²
5. 27p³ – 1/216 – 9p²/2 + p/4

Answer:

1. 8a³ + b³ + 12a²b + 6ab² = (2a)³ + b³ + 3 × (2a)² × b + 3 × 2a × b²

= (2a + b)³

[Using a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (2a)³ + (-b)³ + 3 × (2a)² × (-b) + 3 × (2a) × (-b)²

= {2a + (-b)}³

[Using a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (2a - b)³

= 3³ + (-5a)³ + 3 × 3² × (-5a) + 3 × 3 × (-5a)²

= {3 + (-5a)}³

[Using a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (3 - 5a)³

= (4a)³ + (-3b)³ + 3 × (4a)² × (-3b) + 3 × (4a) × (-3b)²

= {4a + (-3b)}³

[Using a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (4a – 3b)³

= (3p)³ + (-1/6)³ + 3 × (3p)² × (-1/6) + 3 × 3p × (-1/6)²

= {3p + (-1/6)}³

[Using a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (3p – 1/6)³

Question 9: Verify:

1. x³ + y³ = (x + y)(x² – xy + y²)
2. x³ - y³ = (x - y)(x² + xy + y²)

Answer:

1. RHS:

(x + y)(x² – xy + y²) = x(x² – xy + y²) + y(x² – xy + y²)

= x³ – x²y + xy² + x²y – xy² + y³

= x³ + y³

= LHS

Hence, x³ + y³ = (x + y)(x² – xy + y²)

(x - y)(x² + xy + y²) = x(x² + xy + y²) - y(x² + xy + y²)

= x³ + x²y + xy² - x²y - xy² - y³

= x³ - y³

= LHS

Hence, x³ - y³ = (x - y)(x² + xy + y²)

Question 10: Factorise each of the following:

1. 27y³ + 125z³
2. 64m³ – 343n³

[Hint: See Question 9.]

Answer:

1. 27y³ + 125z³ = (3y)³ + (5z)³

= (3y + 5x){(3y)² – 3y × 5z + (5z)²}

[a³ + b³ = (a + b)(a² – ab + b²)]

= (3y + 5x)(9y² – 15yz + 25z²)

= (4m – 7n){(4m)² – 4m × 7n + (7n)²}

[a³ – b³ = (a - b)(a² + ab + b²)]

= (4m – 7n)(16m² – 28yz + 49z²)

Question 11: Factorise: 27x³ + y³ + z³ – 9xyz.

Answer:

27x³ + y³ + z³ – 9xyz = (3x)³ + y³ + z³ – 3 × 3x × y × z

= (3x + y + z)[(3x)² + y² + z² – 3x × y – y × z - z × 3x]

[Using a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc - ca)]

= (3x + y + z)[9x² + y² + z² – 3xy – yz - 3zx]

So, 27x³ + y³ + z³ – 9xyz = (3x + y + z)[9x² + y² + z² – 3xy – yz - 3zx]

Question 12: Verify that x³ + y³ + z³ – 3xyz = (1/2) × (x + y + z) × [(x - y)² + (y - z)² + (z - x)²]

Answer:

RHS:

(1/2) × (x + y + z) × [(x - y)² + (y - z)² + (z - x)²]

= (1/2) × (x + y + z) × [x² + y² - 2xy + y² + z² - 2yz + z² + x² - 2zx]

= (1/2) × (x + y + z) × [2x² + 2y² + 2z² – 2xy - 2yz - 2zx]

= (1/2) × 2 × (x + y + z) × [x² + y² + z² – xy - yz - zx]

= (x + y + z) × [x² + y² + z² – xy - yz - zx]

= x³ + y³ + z³ – 3xyz

[Using a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc - ca)]

= LHS

So, x³ + y³ + z³ – 3xyz = (1/2) × (x + y + z) × [(x - y)² + (y - z)² + (z - x)²] = x³ + y³ + z³ – 3xyz

Question 13: If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.

Answer:

We know that

x³ + y³ + z³ - 3xyz = (x + y + z) × [x² + y² + z² – xy - yz - zx]

Given, x + y + z = 0

So, x³ + y³ + z³ - 3xyz = 0 × [x² + y² + z² – xy - yz - zx]

=> x³ + y³ + z³ - 3xyz = 0

=> x³ + y³ + z³ = 3xyz

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

1. (–12)³ + (7)³ + (5)³
2. (28)³ + (–15)³ + (–13)³

Answer:

1. Let a = -12, b = 7, c = 5

So, a + b + c = -12 + 7 + 5

= -12 + 12

= 0

We know that if a + b + c = 0, then

a³ + b³ + c³ = 3abc

So, (–12)³ + (7)³ + (5)³ = 3 × (-12) × 7 × 5 = -1260

So, a + b + c = 28 + (-15) + (-13)

= 28 – 15 – 13

= 28 - 28

= 0

We know that if a + b + c = 0, then

a³ + b³ + c³ = 3abc

So, (28)³ + (-15)³ + (-13)³ = 3 × 28 × (-15) × (-15) = 16380

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

(i) Area: 25a² – 35a + 12

(ii) Area: 35y² + 13y – 12

Answer:

(i) Given, area = 25a² – 35a + 12

= 25a² – 20a – 15a + 12

= 5a(5a – 4) – 3(5a - 4)

= (5a - 4)(5a - 3)

Hence, length = 5a – 4, breadth = 5a – 3

(ii) Given, area = 35y² + 13y - 12

= 35y² + 28y – 15y – 12

= 7y(5y + 4) – 3(5y + 4)

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

Hence, length = 5y + 4, breadth = 7y – 3

Question 16: What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume: 3x² – 12x

(ii) Volume: 12ky² + 8ky – 20k

Answer:

(i) Given volume = 3x² – 12x

= 3x(x - 4)

= 3 × x × (x - 4)

Hence, the possible dimensions of cuboids are 3, x and (x – 4).

(ii) Given, volume = 12ky² + 8ky – 20k

= 4k(3y² + 2y - 5)

= 4k(3y² + 5y – 3y - 5)

= 4k{y(3y + 5) – 1(3y + 5)}

= 4k(3y + 5)(y - 1)

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