math 2nd assignment class 9

Download Sample Questions and Assignments for Class 9 Mathematics and Science with important Topic- Wise Questions. These Sample Questions and Assignments have been prepared as per the syllabus issued by CBSE, KVS and topics given in NCERT Book. Solving this Sample questions and Assignments will help in scoring in more marks in your class tests and school examinations.

Click on links below for Assignments with important questions for Class 9 Mathematics and Science chapter wise.

Assignment For Class 9th Mathematics Topic -wise- PDF Download

Chapter 1: number system.

Assignment of Mathematics Chapter 1 Number System Assignment

Chapter 2: Polynomials

Assignment of Mathematics Chapter 2 Polynomials Assignment 1

Assignment of Mathematics Chapter 2 Polynomials Assignment 2

Chapter 3: Linear Equation in Two Variables

Assignment of Mathematics Chapter 3 Linear Equation in Two Variables Assignment

Chapter 4: Coordinate Geometry

Assignment of Mathematics Chapter 4 Coordinate Geometry Assignment

Assignment For Class 9th Biology Topic -wise- PDF Download

Chapter 1: the fundamental unit of life: cell.

Assignment of Biology Chapter 1 The Fundamental Unit of Life: Cell Assignment 1

Assignment of Biology Chapter 1 The Fundamental Unit of Life: Cell  Assignment 2

Assignment of Biology Chapter 1 The Fundamental Unit of Life: Cell  Assignment 3

Assignment of Biology Chapter 1 The Fundamental Unit of life: Cell  Assignment

Chapter 2: Tissue

Assignment of Biology Chapter 2   Tissue  Assignment 1

Assignment of Biology Chapter 2 Tissue  Assignment 2

Assignment of Biology Chapter 2 Tissue Assignment 3

Assignment of Biology Chapter 2 Tissue  Assignment 4

Assignment For Class 9th Chemistry Topic -wise- PDF Download

Chapter 1: matter in our surrounding.

Assignment of Chemistry Chapter 1 Matter In Our Surrounding Assignment 1

Assignment of Chemistry Chapter 1 Matter In Our Surrounding Assignment 2

Assignment of Chemistry Chapter 1 Matter In Our Surrounding Assignment 3

         2. Is Matter Around Us Pure

Assignment of Chemistry Chapter 2  Is Matter Around Us Pure  Assignment 1

Assignment of Chemistry Chapter 2 Is Matter Around Us Pure  Assignment 2

Assignment of Chemistry Chapter 2 Is Matter Around Us Pure  Assignment 3

Assignment For Class 9th Physics Topic -wise- PDF Download

Chapter 1: motion.

Assignment of Physics Chapter 1 Motion Assignment 1

Sample Question for Physics Motion Assignment 2

Assignment of Physics Chapter 1 Motion Assignment 3

Assignment of Physics Chapter 1 Motion Assignment 4

Assignment of Physics Chapter 1 Motion Assignment 5

Chapter 2: Force and Laws Of Motion

Assignment of Physics Chapter 2 Force and Laws Of Motion Assignment 1

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Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers

November 22, 2021 by Prasanna

Here we are providing Polynomials Class 9 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 9 Maths  was designed by subject expert teachers.

Extra Questions for Class 9 Maths Polynomials with Answers Solutions

Extra Questions for Class 9 Maths Chapter 2 Polynomials with Solutions Answers

Polynomials Class 9 Extra Questions Very Short Answer Type

Question 1. Factorise : 125x 3 – 64y 3 Solution: 125x 3 – 6443 = (5x) 3 – (4y) 3 By using a3 – b3 = (a – b) (a 2 + ab + b 2 ), we obtain 125x 3 – 64y 3 = (5x – 4y) (25x 2 + 20xy + 16y 2 )

Question 2. Find the value of (x + y) 2 + (x – y) 2 . Solution: (x + y) 2 + (x – y) 2 = x 2 + y 2 + 2xy + x 2 + y 2 – 2xy = 2x 2 + 2y 2 = 21x 2 + y 2 )

Polynomial Division Calculator is a free online tool that shows the division of two polynomials with Polynomial long division method.

Question 3. If p(x)= x 2 – 2√2x+1, then find the value of p(2√2) Solution: Put x = 2√2 in p(x), we obtain p(2√2) = (2√2) 2 – 2√2(2√2) + 1 = (2√2) 2 – (2√2) 2 + 1 = 1

Question 4. Find the value of m, if x + 4 is a factor of the polynomial x 2 + 3x + m. Solution: Let p(x) = x 2 + 3x + m Since (x + 4) or (x – (-4)} is a factor of p(x). ∴ p(-4) = 0 ⇒ (-4) 2 + 3(-4) + m = 0 ⇒ 16 – 12 + m = 0 ⇒ m = -4

Question 6. Find the common factor in the quadratic polynomials x 2 + 8x + 15 and x 2 + 3x – 10. Solution: x 2 + 8x + 15 = x 2 + 5x + 3x + 15 = (x + 3) (x + 5) x 2 + 3x – 10 = x 2 + 5x – 2x – 10 = (x – 2) (x + 5) Clearly, the common factor is x + 5.

Polynomials Class 9 Extra Questions Short Answer Type 1

Question 1. Expand : (i) (y – √3) 2 (ii) (x – 2y – 3z) 2 Solution: (i) (y – √3) 2 = y 2 -2 × y × √3 + (√3) 2 = y 2 – 2√3 y + 3 (x – 2y – 3z) 2 = x 2 + 1 – 2y) 2 + (-3z) 2 + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x = x 2 + 4y 2 + 9z 2 – 4xy + 12yz – 6zx

Question 3. Show that p – 1 is a factor of p 10 + p 8 + p 6 – p 4 – p 2 – 1. Solution: Let f(p) = p 10 + p 8 + p 6 – p 4 – p 2 – 1 Put p = 1, we obtain f(1) = 1 10 + 1 8 + 1 6 – 1 4 – 1 2 – 1 = 1 + 1 + 1 – 1 – 1 – 1 = 0 Hence, p – 1 is a factor of p 10 + p 8 + p 6 – p 4 – p 2 – 1.

Question 4. If 3x + 2y = 12 and xy = 6, find the value of 27x 3 + 8y 3 Solution: We have 3x + 2y = 12 On cubing both sides, we have ⇒ (3x + 2y) 3 = 12 3 ⇒ (3x) 3 +(2y) 3 + 3 × 3x × 2y(3x + 2y) = √728 ⇒ 27x 3 + 8y 3 + 18xy(3x + 2y) = √728 ⇒ 27x 3 + 8y 3 + 18 × 6 × 12 = √728 ⇒ 27x 3 + 8y 3 + 1296 = √728 ⇒ 27x 3 + 8y 3 = √728 – 1296 ⇒ 27x 3 + 8y 3 = 432

Question 5. Factorise : 4x 2 + 9y 2 + 16z 2 2 + 12xy – 24 yz – 16xz. Solution: 4x 2 + 9y 2 + 16z 2 2 + 12xy – 24yz – 16xz = (2x) 2 + (3y) 2 + (-4z) 2 + 2(2x)(3y) + 2(3y)(= 42) + 2(- 42)(2x) By using a 2 + b 2 + 2ab + 2bc + 2ca = (a + b + c) 2 , we obtain = (2x + 3y – 4z) 2 = (2x + 3y – 4z) (2x + 3y – 4z)

Question 6. Factorise : 1 – 2ab – (a 2 + b 2 ). Solution: 1 – 2ab – (a 2 + b 2 ) = 1 – (a 2 + b 2 + 2ab) = 1 2 – (a + b) 2 = (1 + a + b) (1 – a – b) [∵ x 2 – y 2 = (x + y)(x – y)]

Polynomials Class 9 Extra Questions Short Answer Type 2

Question 2. Factorise 64a 3 – 27b 3 – 144a 2 b + 108ab 2 . Solution: 64a 2 – 27b 2 – 144a 2 b + 108ab 2 = (4a) 3 – (3b) 3 – 36ab(4a – 3b) = (40) 2 – (3b) 3 – 3 × 4a × 3b (4a – 3b) = (4a – 3b) 3 [∵ (x – y) 3 = x 3 – y 3 – 3xy(x – y)] = (40 – 3b) (4a – 3b) (4a – 3b)

Question 3. What are the possible expressions for the dimensions of a cuboid whose volume is given below ? Volume = 12ky 2 + 8ky – 20k. Solution: We have, volume = 12ky 2 + 8ky – 20k = 4k(3y 2 + 2y – 5) = 4k(3y 2 + 5y – 3y – 5) = 4k[y(3y + 5) – 1(3y + 5)] = 4k(3y + 5) (y – 1) ∴Possible expressions for the dimensions of cuboid are 4k units, (3y + 5) units and (y – 1) units.

Question 4. If p(x) = x 3 + 3x 2 – 2x + 4, then find the value of p(2) + p(-2) – P(0). Solution: Here, p(x) = x 3 + 3x 2 – 2x + 4 Now, p(2) = 2 3 + 3(2) 2 – 2(2) + 4 = 8 + 12 – 4 + 4 = 20 p(-2) = (-2) 3 + 3(-2) 2 – 21 – 2) + 4 = 8 + 12 + 4 + 4 = 12 and p(0) = 0 + 0 – 0 + 4 = 4 ∴ p(2) + p(-2) – p(0) = 20 + 12 – 4 = 28.

Question 5. If one zero of the polynomial x 2 – √3x + 40 is 5, which is the other zero ? Solution: Let p(x) = x 2 – √3x + 40 = x 2 – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8) Now, for zeroes of given polynomial, put p(x) = 0 ∴ (x – 5) (x – 8) = 0 ⇒ x = 5 or x = 8 Hence, other zero is 8.

Question 7. If one zero of the polynomial x 2 – √3x + 40 is 5, which is the other zero ? Solution: Let p(x) = x 2 – √3x + 40 = x 2 – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8) Now, for zeroes of given polynomial, put p(x) = 0 ∴ (x – 5) (x – 8) = 0 x = 5 or x = 8 ⇒ Hence, other zero is 8.

Polynomials Class 9 Extra Questions Long Answer Type

Question 1. Prove that (a + b + c) 3 – a 3 – b 3 – c 3 = 3(a + b) (b + c) (c + a). Solution: L.H.S. = (a + b + c) 3 – a 3 – b 3 – c 3 = {(a + b + c) 3 – 3} – {b 3 + c 3 } = (a + b + c – a) {(a + b + c) 2 + a 2 + a(a + b + c)} – (b + c) (b 2 + c 2 – bc) = (b + c) {a 2 + b 2 + 2 + 2ab + 2bc + 2ca + a 2 + a 2 + ab + ac – b 2 – a 2 + bc) = (b + c) (3a 2 + 3ab + 3bc + 3ca} = 3(b + c) {a 2 + ab + bc + ca} = 31b + c) {{a 2 + ca) + (ab + bc)} = 3(b + c) {a(a + c) + b(a + c)} = 3(b + c)(a + c) (a + b) = 3(a + b)(b + c) (c + a) = R.H.S.

Question 2. Factorise : (m + 2n) 2 x 2 – 22x (m + 2n) + 72. Solution: Let m + 2n = a ∴ (m + 2n) 2 x 2 – 22x (m + 2n) + 72 = a 2 x 2 – 22ax + 72 = a 2 x 2 – 18ax – 4ax + 72 = ax(ax – 18) – 4(ax – 18) = (ax – 4) (ax – 18) = {(m + 2n)x – 4)} {(m + 2n)x – 18)} = (mx + 2nx – 4) (mx + 2nx – 18).

Question 3. If x – 3 is a factor of x 2 – 6x + 12, then find the value of k. Also, find the other factor of the – polynomial for this value of k. Solution: Here, x – 3 is a factor of x 2 – kx + 12 ∴ By factor theorem, putting x = 3, we have remainder 0. ⇒ (3) 2 – k(3) + 12 = 0 ⇒ 9 – 3k + 12 = 0 ⇒ 3k = 21 ⇒ k = 7 Now, x 2 – 7x + 12 = x 2 – 3x – 4x + 12 = x(x – 3) – 4(x – 3) = (x – 3) (x – 4) Hence, the value of k is 7 and other factor is x – 4.

Question 4. Find a and b so that the polynomial x 3 – 10x 2 + ax + b is exactly divisible by the polynomials (x – 1) and (x – 2). Solution: Let p(x) = x 3 – 10x 2 + ax + b Since p(x) is exactly divisible by the polynomials (x – 1) and (x – 2). ∴ By putting x = 1, we obtain (1) 3 – 10(1) 2 + a(1) + b = 0 ⇒ a + b = 9 And by putting x = 2, we obtain (2) 3 – 10(2) 2 + a(2) + b = 0 8 – 40 + 2a + b = 0 ⇒ 2a + b = 32 Subtracting (i) from (ii), we have a = 23 From (i), we have 23 + b = 9 = b = -14 Hence, the values of a and b are a = 23 and b = -14

Question 5. Factorise : x 2 – 6x 2 + 11x – 6. Solution: Let p(x) = x 2 – 6x 2 + 11x – 6 Here, constant term of p(x) is -6 and factors of -6 are ± 1, ± 2, ± 3 and ± 6 By putting x = 1, we have p(1) = (1) 3 – 6(1) 2 + 11(1) – 6 = 1 – 6 + 11 -6 = 0 ∴ (x – 1) is a factor of p(x) By putting x = 2, we have p(2) = (2) 3 – 6(2) 2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 0 ∴ (x – 2) is a factor of p(x) By putting x = 3, we have p(3) = (3) 3 – 6(3) 2 + 11(3) – 6 = 27 – 54 + 33 – 6 = 0 ∴ (x – 3) is a factor of p(x) Since p(x) is a polynomial of degree 3, so it cannot have more than three linear factors. ∴ x 3 – 6x 2 + 11x – 6 = k (x – 1) (x – 2) (x – 3) By putting x = 0, we obtain 0 – 0 + 0 – 6 = k (-1) (-2) (3) -6 = -6k k = 1 Hence, x 3 – 6x 2 + 11x – 6 = (x – 1) (x – 2)(x – 3).

Polynomials Class 9 Extra Questions HOTS

Question 2. If x = 2 – √3, y = √3 – √7 and 2 = √7 – √4, find the value of x’ + 43 + 2?. Solution: Here, x + y + z = 2 – √3+ √3 – √7+√7 – 2 = 0 x 3 + √3 + x 3 = 3(x)(y)(z) = 3(2 – √3)(√3 – √7)(√7 – 2) = 3(2√3 – 2√7 – 3 + √21)(√7 – 2) = 3(2√21 – 14 – 3√7 + 7√3 – 4√3 + 4√7 + 6 – 2√21) = 3(3√3 + √7 – 8)

Question 3. If (x – a) is a factor of the polynomials x 2 + px – q and x 2 + rx – t, then prove that a = \(\frac{t-q}{r-p}\) Solution: Let f(x) = x + px -q and g(x) = x 2 + x – t Since x-a is factor of both f(x) and g(x) ⇒ f(a) = g(a) = 0 Now, here f(a) = a 2 + pa – q and g(a) = a 2 + ra- t ⇒ a 2 + pa – q = a + ra – t (considering f(a) = g(a)] ⇒ pa – q = ra – t ⇒ ra – pa = t – q ⇒ a(r – p) = t – q a = \(\frac{t-q}{r-p}\)

Polynomials Class 9 Extra Questions Value Based (VBQs)

Question 1. If a teacher divides a material of volume 27x 3 + 54x 2 + 36x + 8 cubic units among three students. Is it possible to find the quantity of material ? Can you name the shape of the figure teacher obtained ? Which value is depicted by the teacher ? Solution: We know that, √olume = Length × Breadth × Height Now, 27x 3 + 54x 2 + 36x + 8 = (3x) 3 + 3(3x) 2 (2) + 3(3x)(2) 2 + (2) 3 = (3x + 2) 2 = (3x + 2) (3x + 2) (3x + 2) Thus, volume = (3x + 2) (3x + 2) (3x + 2) Yes, it is possible to find the quantity of material. (3x + 2) units. Cube. Apply knowledge and use of example for clarity of subject, student friendly.

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Mathematics Assignment on Polynomials 

Class ix chapter 2.

Ans: P(-2) = 0  ⇒ x + 2 is the factor of P(x)

Hint: If remainder = 0 then given function is the factor of the given polynomial otherwise not.

a) x 2  + 11x + 30      : Ans  (x + 5)(x + 6)

a)  8x 3  - 125y 3  - z 3  - 30xyz                            

a)  (-15) 3 + (-13) 3 + (28) 3      Ans: 16380

  4a -  b  = 0 -9a +  b  = -5 ----------------     -5a   = -5       ⇒  a = 1

5a +  b  = -21 3a   +  b  =  ∓  3 ----------------- 2a       =  -18       a   =  - 9

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Very very nice assigament, thank ku so much.

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Class 9 Maths Chapter 2 Solution for CBSE Board Class 9 Maths Exercise 2.1 in English Class 9 Maths Exercise 2.2 in English Class 9 Maths Exercise 2.3 in English Class 9 Maths Exercise 2.4 in English

Class 9 Maths Chapter 2 Solution for State Boards Class 9 Maths Chapter 2 Exercise 2.1 Class 9 Maths Chapter 2 Exercise 2.2 Class 9 Maths Chapter 2 Exercise 2.3 Class 9 Maths Chapter 2 Exercise 2.4 Class 9 Maths Chapter 2 Exercise 2.5

In addition to solutions, Tiwari Academy may also provide practice questions and additional exercises related to each topic. This extra practice can be valuable for reinforcing learning and improving problem-solving skills. Class 9 Maths Chapter 2 Solution in Hindi Class 9 Maths Exercise 2.1 in Hindi Class 9 Maths Exercise 2.2 in Hindi Class 9 Maths Exercise 2.3 in Hindi Class 9 Maths Exercise 2.4 in Hindi

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials in Hindi and English Medium revised and modified for academic year 2024-25. The solutions offered by Tiwari Academy often aim to simplify complex mathematical concepts, making it easier for students to understand and grasp the subject matter. Questions are based on latest NCERT books following new syllabus for 2024-25 exams.

As per the Uttar Pradesh Board , Prayagraj – the NCERT Books are now implemented for class 9 Maths for 2024-25. So, the students of class 9 (High School) can download UP Board Solutions for Class 9 Maths Chapter 2 from the links given below. 9th Maths Chapter 2 Solutions are available in Hindi and English Medium.

Videos related to each exercise are also given below, which help the students to know the steps of solutions. Download NCERT (https://ncert.nic.in/) Solutions Apps and UP Board Solutions app 2024-25 and Offline Solutions based on latest Curriculum for 2024-25.

Study Materials on 9th Maths Chapter 2

• Study Material 9th Maths Chapter 2 in English Medium
• Study Material 9th Maths Chapter 2 in Hindi Medium
• Class 9 Mathematics Solutions Main Page

Important Questions on 9th Maths Chapter 2

Determine whether polynomials x³ + x² + x + 1 has x + 1 a factor.

Let p(x) = x³ + x² + x + 1 Putting x + 1 = 0, we get, x = – 1 Using remainder theorem, when p(x)= x³ + x² + x + 1 is divided by x + 1, remainder is given by p(-1) = (-1)³ + (-1)² + (-1) + 1 = – 1 + 1 – 1 + 1 = 0 Since, remainder p(-1) = 0, Hence x + 1 is a factor of x³ + x² + x + 1.

Use the Factor Theorem to determine whether g(x) is a factor of p(x): p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1.

p(x) = 2x³ + x² – 2x – 1 and g(x) = x + 1 Putting 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) = (-1)³ + (-1)² + (-1) + 1 = – 1 + 1 – 1 + 1 = 0 Since, remainder p(-1) = 0, hence g(x) is a factor of p(x).

Find the value of k, if x – 1 is a factor of p(x) = x² + x + k.

p(x) = x² + x + k Putting x – 1 = 0, we get, x = 1 Using remainder theorem, When p(x) = x² + x + k is divided by x – 1, remainder is given by p(1) = (1)² + (1) + k = 2 + k Since x – 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ 2 + k = 0 ⇒ k = -2

Find the zero of the polynomial p(x) = x + 5.

p(x) = x + 5 Putting p(x) = 0, we get x + 5 = 0 ⇒ x = – 5 Hence, x = – 5 is a zero of the polynomial p(x).

1. A combination of constants and variables, connected by four fundamental arithmetical operations +, -, x and / is called an algebraic expression. e.g. 6x² – 5y² + 2xy 2. An algebraic expression which have only whole numbers as the exponent of one variable, is called polynomial in one variable. e.g. 3x³ + 2x² – 7x + 5 etc. 3. The part of a polynomial separated from each other by + or – sign is called a term and each term of a polynomial has a coefficient. 4. Highest power of the variable in a polynomial, is known as degree of that polynomial. 5. The value obtained on putting a particular value of the variable in polynomial is called value of the polynomial at the value of variable. 6. Zero of a polynomial p(x) is a number alpha, such that p(alpha) = 0. It is also called root pf polynomial equation p(x) = 0. 7. Let f(x) be any polynomial of degree n,(n ≥ 1) and a be any real number. If f(x) is divided by the linear polynomial (x-a), then the remainder is f(a). 8. Let f(x) be a polynomial of degree n,(n ≥ 1) and a be any real number. Then, i). If f(a) = 0, then (x – a) is a factor of f(x). ii). If (x – a) is a factor of f(x), then f(a) = 0.

Polynomial on the Basis of Number of Terms

Polynomial on the Basis of Number of Terms A polynomial containing one non-zero term, is called a monomial. A polynomial containing two non-zero terms, is called a binomial. A polynomial containing three non-zero terms, is called a trinomial.

Polynomial on the Basis of Degree of Variables

A polynomial of degree 0, is called a constant polynomial. A polynomial of degree 1, is called a linear polynomial . A polynomial of degree 2, is called a quadratic polynomial. A polynomial of degree 3, is called a cubic polynomial. A polynomial of degree 4, is called a biquadratic polynomial.

What is the core motive of the chapter 2 Polynomials of class 9 Maths?

The core motive of chapter 2 Polynomials of class 9 Maths is to make the meaning of the following things clear to students.

• 1. Meaning of polynomials in one variable.
• 2. Terms of the polynomials.
• 3. Meaning of coefficients.
• 4. Meaning of zero polynomial.
• 5. Degree of the polynomial.
• 6. Types of the polynomial (Linear, Quadratic, Cubic).
• 7. Zeroes of a polynomial.
• 8. Factorization of polynomial, Factor theorem.
• 9. Algebraic identities.

Is the chapter 2 Polynomials of class 9th Maths complicated?

No, chapter 2 Polynomials of class 9th Maths is not that complicated. Basically, in exercises 2.3 and 2.4, most of the students face a little problem. Exercises 2.1, and 2.2 are quite easy compared to exercise 2.3 and 2.4.

How many exercises are there in 9th Maths Chapter 2 with least number of difficult questions?

In chapter 2 of class 9th Maths, there are four exercises. The first exercise (Ex 2.1) contains five questions, the second exercise (Ex 2.2) contains four questions, the third exercise contains five questions, and the fourth exercise (Ex 2.4) contains sixteen questions. So the second exercise (Ex 2.2) has the least number of questions.

How many theorems are there in the chapter 2 Polynomials of class 9 Maths?

There are two theorems (Remainder theorem and Factor theorem) in chapter 2 Polynomials of class 9 Maths. Both the theorems are important for the exams. The remainder theorem is used in exercise 2.3, and the Factor theorem is used in exercise 2.4.

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Class 9 Mathematics Polynomials Assignments

We have provided below free printable Class 9 Mathematics Polynomials Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 9 Mathematics Polynomials . These Assignments for Grade 9 Mathematics Polynomials cover all important topics which can come in your standard 9 tests and examinations. Free printable Assignments for CBSE Class 9 Mathematics Polynomials , school and class assignments, and practice test papers have been designed by our highly experienced class 9 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Polynomials Class 9 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Polynomials Class 9. Students can click on the links below and download all Pdf Assignments for Mathematics Polynomials class 9 for free. All latest Kendriya Vidyalaya Class 9 Mathematics Polynomials Assignments with Answers and test papers are given below.

Mathematics Polynomials Class 9 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 9 Mathematics Polynomials . Students and teachers can download and save all free Mathematics Polynomials assignments in Pdf for grade 9th. Our expert faculty have covered Class 9 important questions and answers for Mathematics Polynomials as per the latest syllabus for the current academic year. All test papers and question banks for Class 9 Mathematics Polynomials and CBSE Assignments for Mathematics Polynomials Class 9 will be really helpful for standard 9th students to prepare for the class tests and school examinations. Class 9th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 9 Mathematics Polynomials Download in Pdf

Advantages of Class 9 Mathematics Polynomials Assignments

• As we have the best and largest collection of Mathematics Polynomials assignments for Grade 9, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Mathematics Polynomials textbooks for Class 9 .
• All Mathematics Polynomials assignments for Class 9 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 9 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Mathematics Polynomials chapter wise worksheets and assignments for free in Pdf
• Class 9 Mathematics Polynomials question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 9 Mathematics Polynomials students

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Ncert solutions class 9 maths chapter 2 – polynomials free pdf download.

NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. BYJU’S expert faculty create these NCERT Solutions to help students in preparation for their exams. BYJU’S provides NCERT Solutions for Class 9 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.

• Chapter 1: Number System
• Chapter 2: Polynomials
• Chapter 3: Coordinate Geometry
• Chapter 4: Linear Equations in Two Variables
• Chapter 5: Introduction to Euclid’s Geometry
• Chapter 6: Lines and Angles
• Chapter 7: Triangles
• Chapter 8: Quadrilaterals
• Chapter 9: Areas of Parallelograms and Triangles
• Chapter 10: Circles
• Chapter 11: Constructions
• Chapter 12: Heron’s Formula
• Chapter 13: Surface Areas and Volumes
• Chapter 14: Statistics

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 9  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 – Polynomials

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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

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 –3x+7

The equation 4x 2 –3x+7 can be written as 4x 2 –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 –3x+7 is a polynomial in one variable.

(ii) y 2 +√2

The equation y 2 + √2 can be written as y 2 + √ 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 + √ 2 is a polynomial in one variable.

(iii) 3√t+t√2

The equation 3√t+t√2 can be written as 3t 1/2 +√2t

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 3√t+t√2 is not a polynomial in one variable.

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

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

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) 2–x 2 +x 3

The equation 2–x 2 +x 3 can be written as 2+(–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 2–x 2 +x 3 is -1.

(iii) ( π /2)x 2 +x

The equation (π/2)x 2 +x can be written as (π/2)x 2 + x

Here, the number that multiplies the variable x 2 is π/2.

Hence, the coefficient of x 2 in (π/2)x 2 +x is π/2.

Here, the number that multiplies the variable x 2 is 0

Hence, the coefficient of x 2 in √2x-1 is 0.

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

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

For example,  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,  4x 100

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

(i) 5x 3 +4x 2 +7x

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.

Here, in 4–y 2 ,

The power of the variable y is 2

The degree of 4–y 2 is 2, as 2 is the highest power of y in the equation.

(iii) 5t–√7

Here, in 5t –√7 

The power of the variable t is: 1

The degree of 5t –√7 is 1, as 1 is the highest power of y in the equation.

Here, 3 = 3×1 = 3× 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:

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.

The highest power of x 2 +x is 2

The degree is 2

Hence, x 2 +x is a quadratic polynomial

The highest power of x–x 3 is 3

The degree is 3

Hence, x–x 3 is a cubic polynomial

(iii) y+y 2 +4

The highest power of y+y 2 +4 is 2

Hence, y+y 2 +4 is a quadratic polynomial

The highest power of 1+x is 1

The degree is 1

Hence, 1+x is a linear polynomial.

The highest power of 3t is 1

Hence, 3t is a linear polynomial.

The highest power of r 2 is 2

Hence, r 2 is a quadratic polynomial.

The highest power of 7x 3 is 3

Hence, 7x 3 is a cubic polynomial.

Exercise 2.2 Page: 34

1. Find the value of the polynomial (x)=5x−4x 2 +3. 

(ii) x = – 1

(iii) x = 2

Let f(x) = 5x−4x 2 +3

(i) When x = 0

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

(ii) When x = -1

f(x) = 5x−4x 2 +3

f(−1) = 5(−1)−4(−1) 2 +3

(iii) When x = 2

f(2) = 5(2)−4(2) 2 +3

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

(i) p(y)=y 2 −y+1

p(y) = y 2 –y+1

∴ p(0) = (0) 2 −(0)+1 = 1

p(1) = (1) 2 –(1)+1 = 1

p(2) = (2) 2 –(2)+1 = 3

(ii) p(t)=2+t+2t 2 −t 3

p(t) = 2+t+2t 2 −t 3

∴ p(0) = 2+0+2(0) 2 –(0) 3 = 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

(iii) p(x)=x 3

∴ p(0) = (0) 3 = 0

p(1) = (1) 3 = 1

p(2) = (2) 3 = 8

(iv) P(x) = (x−1)(x+1)

p(x) = (x–1)(x+1)

∴ 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

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

(i) p(x)=3x+1, x = −1/3

For, x = -1/3, p(x) = 3x+1

∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0

∴ -1/3 is a zero of p(x).

(ii) p(x) = 5x–π, x = 4/5

For, x = 4/5, p(x) = 5x–π

∴ p(4/5) = 5(4/5)- π = 4-π

∴ 4/5 is not a zero of p(x).

(iii) p(x) = x 2 −1, x = 1, −1

For, x = 1, −1;

p(x) = x 2 −1

∴ p(1)=1 2 −1=1−1 = 0

p(−1)=(-1) 2 −1 = 1−1 = 0

∴ 1, −1 are zeros of p(x).

(iv) p(x) = (x+1)(x–2), x =−1, 2

For, x = −1,2;

p(x) = (x+1)(x–2)

∴ p(−1) = (−1+1)(−1–2)

= (0)(−3) = 0

p(2) = (2+1)(2–2) = (3)(0) = 0

∴ −1, 2 are zeros of p(x).

(v) p(x) = x 2 , x = 0

For, x = 0 p(x) = x 2

p(0) = 0 2 = 0

∴ 0 is a zero of p(x).

(vi) p(x) = lx +m, x = −m/ l

For, x = -m/ l ; p(x) = l x+m

∴ p(-m/ l) = l (-m/ l )+m = −m+m = 0

∴ -m/ l is a zero of p(x).

(vii) p(x) = 3x 2 −1, x = -1/√3 , 2/√3

For, x = -1/√3 , 2/√3 ; p(x) = 3x 2 −1

∴ p(-1/√3) = 3(-1/√3) 2 -1 = 3(1/3)-1 = 1-1 = 0

∴ p(2/√3 ) = 3(2/√3) 2 -1 = 3(4/3)-1 = 4−1 = 3 ≠ 0

∴ -1/√3 is a zero of p(x), but 2/√3  is not a zero of p(x).

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

For, x = 1/2 p(x) = 2x+1

∴ p(1/2) = 2(1/2)+1 = 1+1 = 2≠0

∴ 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) = x+5 

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

(ii) p(x) = x–5

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

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

p(x) = 2x+5

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

(iv) p(x) = 3x–2 

p(x) = 3x–2

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

(v) p(x) = 3x 

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

(vi) p(x) = ax, a≠0

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

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

p(x) = cx + d

∴ 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

∴ Remainder:

p(−1) = (−1) 3 +3(−1) 2 +3(−1)+1

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

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

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

p(0) = (−π) 3 +3(−π) 2 +3(−π)+1

= −π 3 +3π 2 −3π+1

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

2. Find the remainder when x 3 −ax 2 +6x−a is divided by x-a.

Let p(x) = x 3 −ax 2 +6x−a

p(a) = (a) 3 −a(a 2 )+6(a)−a

= a 3 −a 3 +6a−a = 5a

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

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

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

= (-343-147)/9

= -490/9 ≠ 0

∴ 7+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

Let p(x) = x 3 +x 2 +x+1

p(−1) = (−1) 3 +(−1) 2 +(−1)+1

∴ By factor theorem, x+1 is a factor of x 3 +x 2 +x+1

(ii) x 4 +x 3 +x 2 +x+1

Let p(x)= x 4 +x 3 +x 2 +x+1

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

= 1−1+1−1+1

∴ By 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 

Let p(x)= x 4 +3x 3 +3x 2 +x+1

The zero of x+1 is -1.

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

∴ By factor theorem, x+1 is not a factor of x 4 +3x 3 +3x 2 +x+1

(iv) x 3 – x 2 – (2+√2)x +√2

Let p(x) = x 3 –x 2 –(2+√2)x +√2

p(−1) = (-1) 3 –(-1) 2 –(2+√2)(-1) + √2 = −1−1+2+√2+√2

∴ By factor theorem, x+1 is not a factor of x 3 –x 2 –(2+√2)x +√2

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 –2x–1, g(x) = x+1

p(x) = 2x 3 +x 2 –2x–1, g(x) = x+1

∴ Zero of g(x) is -1.

p(−1) = 2(−1) 3 +(−1) 2 –2(−1)–1

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

(ii) p(x)=x 3 +3x 2 +3x+1, g(x) = x+2

p(x) = x 3 +3x 2 +3x+1, g(x) = x+2

∴ Zero of g(x) is -2.

p(−2) = (−2) 3 +3(−2) 2 +3(−2)+1

= −8+12−6+1

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

(iii) p(x)=x 3 –4x 2 +x+6, g(x) = x–3

p(x) = x 3 –4x 2 +x+6, g(x) = x -3

∴ Zero of g(x) is 3.

p(3) = (3) 3 −4(3) 2 +(3)+6

= 27−36+3+6

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

(i) p(x) = x 2 +x+k

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

By Factor Theorem

⇒ (1) 2 +(1)+k = 0

⇒ 1+1+k = 0

(ii) p(x) = 2x 2 +kx+ √2

⇒ 2(1) 2 +k(1)+√2 = 0

⇒ 2+k+√2 = 0

⇒ k = −(2+√2)

(iii) p(x) = kx 2 – √ 2x+1

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

⇒ k(1) 2 -√2(1)+1=0

(iv) p(x)=kx 2 –3x+k

⇒ k(1) 2 –3(1)+k = 0

⇒ k−3+k = 0

4. Factorise:

(i) 12x 2 –7x+1

Using the splitting the middle term method,

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

12x 2 –7x+1= 12x 2 -4x-3x+1

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

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

(ii) 2x 2 +7x+3

We have to find a number whose sum = 7 and product = 2×3 = 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 

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

6x 2 +5x-6 = 6x 2 +9x–4x–6

= 3x(2x+3)–2(2x+3)

= (2x+3)(3x–2)

(iv) 3x 2 –x–4 

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

3x 2 –x–4 = 3x 2 –4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

5. Factorise:

(i) x 3 –2x 2 –x+2

Let p(x) = x 3 –2x 2 –x+2

Factors of 2 are ±1 and ± 2

p(x) = x 3 –2x 2 –x+2

p(−1) = (−1) 3 –2(−1) 2 –(−1)+2

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

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x 2 –3x+2) = (x+1)(x 2 –x–2x+2)

= (x+1)(x(x−1)−2(x−1))

= (x+1)(x−1)(x-2)

(ii) x 3 –3x 2 –9x–5

Let p(x) = x 3 –3x 2 –9x–5

Factors of 5 are ±1 and ±5

By the trial method, we find that

So, (x-5) is factor of p(x)

p(x) = x 3 –3x 2 –9x–5

p(5) = (5) 3 –3(5) 2 –9(5)–5

= 125−75−45−5

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

(x−5)(x 2 +2x+1) = (x−5)(x 2 +x+x+1)

= (x−5)(x(x+1)+1(x+1))

= (x−5)(x+1)(x+1)

(iii) x 3 +13x 2 +32x+20

Let p(x) = x 3 +13x 2 +32x+20

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

So, (x+1) is factor of p(x)

p(x)= x 3 +13x 2 +32x+20

p(-1) = (−1) 3 +13(−1) 2 +32(−1)+20

= −1+13−32+20

Now, Dividend = Divisor × Quotient +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 –2y–1

Let p(y) = 2y 3 +y 2 –2y–1

Factors = 2×(−1)= -2 are ±1 and ±2

So, (y-1) is factor of p(y)

p(y) = 2y 3 +y 2 –2y–1

p(1) = 2(1) 3 +(1) 2 –2(1)–1

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

Now, Dividend = Divisor × Quotient + Remainder

(y−1)(2y 2 +3y+1) = (y−1)(2y 2 +2y+y+1)

= (y−1)(2y(y+1)+1(y+1))

= (y−1)(2y+1)(y+1)

Exercise 2.5 Page: 48

1. Use suitable identities to find the following products:

(i) (x+4)(x +10) 

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

(x+4)(x+10) = x 2 +(4+10)x+(4×10)

= x 2 +14x+40

(ii) (x+8)(x –10)     

(x+8)(x−10) = x 2 +(8+(−10))x+(8×(−10))

= x 2 +(8−10)x–80

= x 2 −2x−80

(iii) (3x+4)(3x–5)

(3x+4)(3x−5) = (3x) 2 +[4+(−5)]3x+4×(−5)

= 9x 2 +3x(4–5)–20

= 9x 2 –3x–20

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

Using the identity, (x+y)(x–y) = x 2 –y 2

(y 2 +3/2)(y 2 –3/2) = (y 2 ) 2 –(3/2) 2

2. Evaluate the following products without multiplying directly:

(i) 103×107

103×107= (100+3)×(100+7)

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

Here, x = 100

We get, 103×107 = (100+3)×(100+7)

= (100) 2 +(3+7)100+(3×7)

= 10000+1000+21

(ii) 95×96  

95×96 = (100-5)×(100-4)

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

We get, 95×96 = (100-5)×(100-4)

= (100) 2 +100(-5+(-4))+(-5×-4)

= 10000-900+20

(iii) 104×96

104×96 = (100+4)×(100–4)

Here, a = 100

We get, 104×96 = (100+4)×(100–4)

= (100) 2 –(4) 2

3. Factorise the following using appropriate identities:

(i) 9x 2 +6xy+y 2

9x 2 +6xy+y 2 = (3x) 2 +(2×3x×y)+y 2

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

Here, x = 3x

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

(ii) 4y 2 −4y+1

4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1

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

Here, x = 2y

4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1 2

= (2y–1)(2y–1)

(iii)  x 2 –y 2 /100

x 2 –y 2 /100 = x 2 –(y/10) 2

Using identity, x 2 -y 2 = (x-y)(x+y)

Here, x = x

= (x–y/10)(x+y/10)

4. Expand each of the following using suitable identities:

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

(ii) (2x−y+z) 2

(iii) (−2x+3y+2z) 2

(iv) (3a –7b–c) 2

(v) (–2x+5y–3z) 2

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

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

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

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

(ii) (2x−y+z) 2  

Here, x = 2x

(2x−y+z) 2 = (2x) 2 +(−y) 2 +z 2 +(2×2x×−y)+(2×−y×z)+(2×z×2x)

= 4x 2 +y 2 +z 2 –4xy–2yz+4xz

Here, x = −2x

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

= 4x 2 +9y 2 +4z 2 –12xy+12yz–8xz

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

Here, x = 3a

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

= 9a 2 + 49b 2 + c 2 – 42ab+14bc–6ca

Here, x = –2x

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

= 4x 2 +25y 2 +9z 2 – 20xy–30yz+12zx

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

Here, x = (1/4)a

y = (-1/2)b

(i) 4x 2 +9y 2 +16z 2 +12xy–24yz–16xz

(ii) 2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz

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

4x 2 +9y 2 +16z 2 +12xy–24yz–16xz = (2x) 2 +(3y) 2 +(−4z) 2 +(2×2x×3y)+(2×3y×−4z)+(2×−4z×2x)

= (2x+3y–4z) 2

= (2x+3y–4z)(2x+3y–4z)

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

2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz

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

= (−√2x+y+2√2z) 2

= (−√2x+y+2√2z)(−√2x+y+2√2z)

6. Write the following cubes in expanded form:

(i) (2x+1) 3

(ii) (2a−3b) 3

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

(iv) (x−(2/3)y) 3

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

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

= 8x 3 +1+6x(2x+1)

= 8x 3 +12x 2 +6x+1

Using identity,(x–y) 3 = x 3 –y 3 –3xy(x–y)

(2a−3b) 3 = (2a) 3 −(3b) 3 –(3×2a×3b)(2a–3b)

= 8a 3 –27b 3 –18ab(2a–3b)

= 8a 3 –27b 3 –36a 2 b+54ab 2

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

(iv)  (x−(2/3)y) 3

Using identity, (x –y) 3 = x 3 –y 3 –3xy(x–y)

7. Evaluate the following using suitable identities: 

(ii) (102) 3

(iii) (998) 3

We can write 99 as 100–1

(99) 3 = (100–1) 3

= (100) 3 –1 3 –(3×100×1)(100–1)

= 1000000 –1–300(100 – 1)

= 1000000–1–30000+300

We can write 102 as 100+2

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

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

We can write 99 as 1000–2

(998) 3 =(1000–2) 3

=(1000) 3 –2 3 –(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992

8. Factorise each of the following:

(i) 8a 3 +b 3 +12a 2 b+6ab 2

(ii) 8a 3 –b 3 –12a 2 b+6ab 2

(iii) 27–125a 3 –135a +225a 2    

(iv) 64a 3 –27b 3 –144a 2 b+108ab 2

(v) 27p 3 –(1/216)−(9/2) p 2 +(1/4)p

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

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

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)(2a–b)(2a–b)

Here, the identity,(x–y) 3 = x 3 –y 3 –3xy(x–y) is used.

(iii) 27–125a 3 –135a+225a 2  

The expression, 27–125a 3 –135a +225a 2 can be written as 3 3 –(5a) 3 –3(3) 2 (5a)+3(3)(5a) 2

27–125a 3 –135a+225a 2 = 3 3 –(5a) 3 –3(3) 2 (5a)+3(3)(5a) 2

= (3–5a)(3–5a)(3–5a)

Here, the identity, (x–y) 3 = x 3 –y 3 -3xy(x–y) is used.

The expression, 64a 3 –27b 3 –144a 2 b+108ab 2 can be written as (4a) 3 –(3b) 3 –3(4a) 2 (3b)+3(4a)(3b) 2

64a 3 –27b 3 –144a 2 b+108ab 2 = (4a) 3 –(3b) 3 –3(4a) 2 (3b)+3(4a)(3b) 2

=(4a–3b)(4a–3b)(4a–3b)

Here, the identity, (x – y) 3 = x 3 – y 3 – 3xy(x – y) is used.

(v) 27p 3 – (1/216)−(9/2) p 2 +(1/4)p

The expression, 27p 3 –(1/216)−(9/2) p 2 +(1/4)p can be written as

(3p) 3 –(1/6) 3 −(9/2) p 2 +(1/4)p = (3p) 3 –(1/6) 3 −3(3p)(1/6)(3p – 1/6)

Using (x – y) 3 = x 3 – y 3 – 3xy (x – y)

27p 3 –(1/216)−(9/2) p 2 +(1/4)p = (3p) 3 –(1/6) 3 −3(3p)(1/6)(3p – 1/6)

Taking x = 3p and y = 1/6

= (3p–1/6) 3

= (3p–1/6)(3p–1/6)(3p–1/6)

(i) x 3 +y 3 = (x+y)(x 2 –xy+y 2 )

(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 )

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

⇒ x 3 +y 3 = (x+y) 3 –3xy(x+y)

⇒ x 3 +y 3 = (x+y)(x 2 +y 2 –xy)

(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 ) 

We know that, (x–y) 3 = x 3 –y 3 –3xy(x–y)

⇒ x 3 −y 3 = (x–y) 3 +3xy(x–y)

⇒ x 3 +y 3 = (x–y)(x 2 +y 2 +xy)

10. Factorise each of the following:

(i) 27y 3 +125z 3

(ii) 64m 3 –343n 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 –xy+y 2 )

= (3y+5z)(9y 2 –15yz+25z 2 )

The expression, 64m 3 –343n 3 can be written as (4m) 3 –(7n) 3

64m 3 –343n 3 = (4m) 3 –(7n) 3

We know that, x 3 –y 3 = (x–y)(x 2 +xy+y 2 )

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

11. Factorise: 27x 3 +y 3 +z 3 –9xyz. 

The expression 27x 3 +y 3 +z 3 –9xyz can be written as (3x) 3 +y 3 +z 3 –3(3x)(y)(z)

27x 3 +y 3 +z 3 –9xyz  = (3x) 3 +y 3 +z 3 –3(3x)(y)(z)

We know that, x 3 +y 3 +z 3 –3xyz = (x+y+z)(x 2 +y 2 +z 2 –xy –yz–zx)

= (3x+y+z)(9x 2 +y 2 +z 2 –3xy–yz–3xz)

12. Verify that:

x 3 +y 3 +z 3 –3xyz = (1/2) (x+y+z)[(x–y) 2 +(y–z) 2 +(z–x) 2 ]

x 3 +y 3 +z 3 −3xyz = (x+y+z)(x 2 +y 2 +z 2 –xy–yz–xz)

= (1/2)(x+y+z)(2x 2 +2y 2 +2z 2 –2xy–2yz–2xz)

13. If  x+y+z = 0, show that x 3 +y 3 +z 3 = 3xyz.

x 3 +y 3 +z 3 -3xyz = (x +y+z)(x 2 +y 2 +z 2 –xy–yz–xz)

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 –xy–yz–xz)

⇒ x 3 +y 3 +z 3 –3xyz = 0

⇒ x 3 +y 3 +z 3 = 3xyz

Hence Proved

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

(i) (−12) 3 +(7) 3 +(5) 3

(ii) (28) 3 +(−15) 3 +(−13) 3

Let a = −12

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

Here, −12+7+5=0

(−12) 3 +(7) 3 +(5) 3 = 3xyz

= 3×-12×7×5

(28) 3 +(−15) 3 +(−13) 3

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

Here, x+y+z = 28–15–13 = 0

(28) 3 +(−15) 3 +(−13) 3 = 3xyz

= 0+3(28)(−15)(−13)

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 –35a+12

(ii) Area: 35y 2 +13y–12

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

25a 2 –35a+12 = 25a 2 –15a−20a+12

= 5a(5a–3)–4(5a–3)

= (5a–4)(5a–3)

Possible expression for length  = 5a–4

Possible expression for breadth  = 5a –3

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

35y 2 +13y–12 = 35y 2 –15y+28y–12

= 5y(7y–3)+4(7y–3)

= (5y+4)(7y–3)

Possible expression for length  = (5y+4)

Possible expression for breadth  = (7y–3)

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

(i) Volume: 3x 2 –12x

(ii) Volume: 12ky 2 +8ky–20k

3x 2 –12x can be written as 3x(x–4) by taking 3x out of both the terms.

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x–4)

12ky 2 +8ky–20k can be written as 4k(3y 2 +2y–5) by taking 4k out of both the terms.

12ky 2 +8ky–20k = 4k(3y 2 +2y–5)

= 4k(3y 2 +5y–3y–5)

= 4k(3y+5)(y–1)

Possible expression for length = 4k

Possible expression for breadth = (3y +5)

Possible expression for height = (y -1)

Disclaimer:

Dropped Topics –  2.4 Remainder theorem.

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NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials

• NCERT Solutions
• Chapter 2 Polynomials

NCERT Solutions for Maths Chapter 2 Polynomials Class 9 - Free PDF Download

Vedantu specialists have put up NCERT Answers Class 9 Mathematics Chapter 2 Polynomials to satisfy the long-standing need of CBSE students studying for Board and other competitive Examinations. This answer has been rigorously reviewed in compliance with the CBSE's newly modified syllabus. CBSE Class 9 Mathematics Chapter 2 Polynomials solutions include a substantial number of solved questions that span the complete syllabus in the form of graded exercises and step-by-step explanations. Vedantu's goal is to clarify the chapter's key subject and to help students build problem-solving abilities.

Glance of NCERT Solutions for Class 9 Maths Chapter 2 Polynomials | Vedantu

Chapter 2 of Class 9 Maths deals with Polynomials, which are basically algebraic expressions built using variables (like x, y), constants (numbers like 2, 3), and exponents (whole numbers like $x^2, y^3$).

Learn about Degree of a Polynomial and types of Polynomials.

Polynomials are classified based on the highest exponent of the variable:

Linear Polynomial (degree 1) (e.g., 5x + 2)

Quadratic Polynomial (degree 2) (e.g., x^2 + 3x - 4)

Cubic Polynomial (degree 3) (e.g., 2x^3 - x^2 + 5x + 1)

Covered concepts duch as Degree of a Polynomial, Zero Polynomial, Operations on Polynomials and Monomial, Binomial, Trinomial.

This article contains chapter notes, formulas, exercise links and important questions for chapter 2 -  Polynomials. 

There are five exercises (33 fully solved questions) in Class 9th Maths Chapter 2 Polynomials.

Access Exercise Wise NCERT Solutions for Chapter 2 Maths Class 9

Exercises Under NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Exercise 2.1: This exercise covers the definition and basic concepts of polynomials. The questions in this exercise aim to familiarise students with terms like coefficients, exponents, degrees, and standard forms of polynomials. Students are also required to classify polynomials based on their degrees. They will have to solve problems related to the addition, subtraction, and multiplication of polynomials and also learn how to factorise polynomials.

Exercise 2.2: This exercise deals with the factors and zeros of polynomials. The questions in this exercise require students to find the factors and zeros of given polynomials. They will also learn how to use the factor theorem and remainder theorem to factorise polynomials and find their zeros.

Exercise 2.3: This exercise covers the division algorithm for polynomials. The questions in this exercise require students to divide a polynomial by another polynomial using the long division method. They will also learn how to use the remainder theorem to find the remainder when a polynomial is divided by another polynomial.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomial - PDF Download

Variables - the unknown value.

Have you ever wondered why children have different heights? Some children grow taller and some end up being shorter than average. To answer this question Scientists have come closer and researched the parameters in the form of variables that are the cause of height.

The word ‘variable’ is derived from the word ‘vary’ which means changing. Therefore, a variable can be any trait, condition, or factor that can change by only differing amounts or it is the unknown term whose value is not known. Example: A child’s height is dependent on the amount of protein and nutrients he or she consumes. Not only that, the height of kids is also dependent on their DNA which means if their parents are tall then there are more chances of them being tall whereas short parents usually have short kids. The height of the kid is also dependent on the rate of work or activities. It is believed that children with more activities like jumping, running, skipping, etc tend to grow faster. Thus, nutrients, DNA, and activities are the three variables that control the height in our body. These variables keep changing from body to body.

For example, while cooking dal we know that the quantity of water is thrice the number of lentils. That you can add 1 cup of lentils to three cups of water. This process can be expressed as,

Here, the quantity of lentils is variable. That means if the quantity of lentils changes then the quantity of water also changes.

In a World Full of Variables, You Will Always Find Constant.

There is one interesting thing about constants and that is this it never changes. A constant is actually a value that is a fixed number on its own. For example - In the equation 9 - x = 5, 9 and 5 are two constants whose values will not change whereas the value of x is not known. Thus, x is a variable.

Can Constant be a Coefficient To?

Since now we already know about variables, it is easier for us to understand the constant and coefficient. A coefficient is usually the number that is multiplied by the variable or letters. For example in ‘5x + y - 7’, 5 is a coefficient of x in the term 5x because it is a number that is multiplied by the unknown variable x. Also, in the term y, it can be considered as the coefficient of y because y can be written as 1xy.

The coefficient is the number that is always multiplied by the variables but constants are terms without variables. Therefore, coefficients cannot be called constants and vice versa. In the aforementioned example, -7 is constant.

Like terms are the terms having the same variables raised to the same power. In 5x + y - 7, no variable is common therefore no like terms. 

In 5a + 2b - 3a + 4 the terms like 5a and -3a are like terms whereas 4 is constant.

What is a Polynomial?

The word Polynomial is derived from the word poly ("many") and nominal ("term"). It is an expression consisting of many terms such that each term holds at least one variable. The variables can be raised to the power and further multiplied by a coefficient but the simplest polynomials hold one variable. The terms are separated by signs( + or - ). Also, the variables and numbers can be combined using addition, subtraction, multiplication or division but it can never be divided by a variable which means a term can never be like  2/x. A polynomial can also not have infinite terms. It always has a finite sum of terms with all variables having whole-number exponents and no variable as a denominator.

Polynomials are composed of the following:

Constants such as 3, −20, or ½, etc.

Variables such as g, h, x, y, etc.

Exponents such as 2 in y 2 or 5 in x5  etc

Examples of Polynomials: 5x 3 – 2x 2 + x – 13 and  x 2 y 3 + xy.

Degree of a Polynomial

It is simply the highest of the powers or exponents on the terms present in the algebraic expression.

Example: In 7x – 5, the first term is 7x, whereas the second term is -5. The power on the variable of the given first term is one and on the second term is zero. Since the highest exponent is one, the degree of the polynomial is also 1.

Types of Polynomials

Polynomials can be classified on the basis of

Number of Terms.

Degree of a polynomial.

Classification on the Basis of Terms

A polynomial either has one term, two terms, three terms, or more than three terms.

Monomials- ‘Mono’ stands for one and ‘mial’ stands for terms thus an algebraic expression with one term is called a monomial. 

Binomials- ‘Bi’ stands for two and ‘mial’ stands for terms therefore an algebraic expression with two, unlike terms is called binomials. 

Trinomials- ‘Tri’ stands for three and ‘mial’ stands for terms thus an algebraic expression with three unlike terms is called trinomials.

Classification on the Basis of Degrees

The Degree of Polynomial is considered as the highest value of the exponent in the expression because it is the largest exponent. We can also call it an order of the polynomial. While finding the degree of the polynomial, remember that the polynomial powers of the variables must be either in an ascending or descending order.

Linear Polynomial: If the expression holds degree 1 then we can call it a linear polynomial. 

Quadratic Polynomial: If the expression holds degree 2 then it can be called a quadratic polynomial.

Cubic Polynomial: If the expression holds degree 3 then it will be called a cubic polynomial.

Zeros of Polynomials

If the value of every coefficient of a variable is zero then it is called the zeros of a Polynomial. In order to find the relationship between the zeroes and coefficients of a given quadratic polynomial, we can find the zeros of the polynomial by the factorization method that is, by taking the sum and product of these zeros.

Operations on Polynomial

There are four main polynomial operations which are:

Addition of Polynomials

Subtraction of Polynomials

Multiplication of Polynomials

Division of Polynomials

NCERT Solutions Class 9 Maths Chapter 2 Polynomials All Exercises

NCERT Maths Class 9 Solutions Vedantu's polynomials provide a thorough grasp of this significant subject. Students can build a solid foundation in algebra by concentrating on important ideas such as polynomial expressions, degree of polynomials, and polynomial operations.It's important to pay attention to the step-by-step solutions provided in the NCERT Solutions, as they help clarify concepts and reinforce problem-solving techniques. Understanding polynomials is crucial as they form the basis for understanding more complex algebraic concepts. Approximately four to five questions from this chapter have usually been included in previous year's question papers. As a result, practicing a range of issues from NCERT Solutions and past test papers helps improve exam readiness and confidence.

Other Related Links for CBSE Class 9 Maths Chapter 2

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths . Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials

1. What is Polynomial and How is It Classified on the Basis of the Number of Terms and Degrees ? 

A polynomial in a variable x is an algebraic expression  of the form

p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + . . . . .+ a₂x²+ a₁x + a₀ ,where a₀, a₁, a₂, . . . . ., aₙ are constants and aₙ ≠ 0.

In the above expression of  polynomial p(x),  a₀, a₁, a₂, . . . . ., aₙ are respectively the coefficients of variables  x⁰, x, x² , . . . . ., xⁿ, and the degree of polynomial p(x) is the non-negative (i.e. n 0) power n to which the variable x is raised in the expression.

Each of the expressions aₙxⁿ + aₙ₋₁xⁿ⁻¹ + . . . . .+ a₂x²+ a₁x + a₀  is called terms of the polynomial p(x).

Polynomial class 9 are classified on the basis of the number of terms and degrees they have as follows:

Classification of Polynomials Based on the Number of Terms:

Monomial: A polynomial having only one term is known as a monomial.

Example: 2x², – 3,  –3/2 (y) 

Binomial: A polynomial having two terms is known as a binomial.

Example: 2x² + 1, x – 3,  –3/2(y) + x

Trinomial: A polynomial having three terms is known as a trinomial.

Example: x² - 2x + 2

Classification of Polynomials Based on the Degrees of Its Variable:

Linear polynomial: A polynomial having degree one is known as a linear polynomial. 

Example: 2x + y, x – 3

Quadratic polynomial: A polynomial having degree two is known as a quadratic polynomial.

Example: 2x² + 1, x² - 2x + 2

Cubic polynomial: A polynomial of degree three is known as a cubic polynomial.

Example: x³ + 4x² + 7x - 3

2. Give the Difference between Remainder and Factor Theorem?

The remainder theorem states that if a polynomial p(x) is divided by (x - a), then the remainder is obtained by evaluating the expression p(a).

While, the Factor theorem states that (x - a) will be a factor of polynomial p(x) only if the remainder obtained by evaluating the expression p(a) equals zero i.e; p(a) = 0.

For Example: let p(x) = x² - 6x + 9, then find the remainder when it is divided by (x - 1).

So, according to the remainder theorem, remainder is obtained by evaluating the expression p(1)

p(1) = 1² - 6(1) + 9 

       = 1 - 6 + 9 = 4.

Therefore, the remainder is 4 when  p(x) = x² - 6x + 9 is divided by (x - 1).

Now, check whether (x - 3) is the factor of p(x) = x² - 6x + 9.

So, according to factor theorem  (x - 3) is the factor of p(x) = x² - 6x + 9 only if p(3) = 0.

p(3) = 3² - 6(3) + 9

        = 9 - 18 + 9 = 0.

Therefore,  (x - 3) is the factor of p(x) = x² - 6x + 9.

3. What is an Algebraic Identity and What are Its Uses?

An algebraic identity is an algebraic equation that is true for any values of the variables occurring in it.

Algebraic identities are used to factorize the algebraic expressions and to compute the products of some arithmetic expressions without multiplying them directly.

Most commonly used algebraic identities are, (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b², etc.

4. What are Polynomials in One Variable?

Polynomials are algebraic expressions in which variables and constants terms are connected by various arithmetic operators. For ‘Polynomials in one variable’ the terms of the polynomial have the same common variable, with numeric coefficients. In ‘Polynomials in one variable’, the variables are raised to powers and the degree of the equation can be determined with the highest power of the variable. Also, the degree of a polynomial is always a positive integer. Hence, if there is a variable term raised to a negative power in an algebraic expression, then it is not counted as a polynomial.

Examples of Polynomials: 2x³ - 7x² + 3x - 2, in this algebraic expression, there are four terms, and all the variable terms have positive integer exponents. So, this algebraic expression is a polynomial.

5. What do We Learn in the Class 9 Maths Chapter 2 Polynomials?

‘Polynomials’ is one of the most important chapters in the Maths syllabus of class 9. The basic highlights of this chapter are listed below.

Polynomial is an algebraic expression consisting of variables, preceded by coefficients, and connected by arithmetic operators. In this chapter, you will learn about polynomials in one variable, that is, the variable coefficient for all the terms in a polynomial, here, will be the same.

The variables along with their coefficients in the algebraic expression are called terms of the polynomial. Some variable terms in a polynomial are raised to exponents.

The degree of a polynomial refers to the greatest exponent of its variable term. If there is a constant term in the polynomial, then it has to be assumed that the exponential value for its variable coefficient is zero.

6. What are the Important Topics Discussed in Class 9 Maths Chapter 2 Polynomials?

Remainder theorem, degree of polynomials, factorization, zeros of a polynomial, algebraic identities, etc., are explained in this chapter. These concepts form the basis of higher mathematics so you must have a good knowledge of these concepts. For example, factorization is one of the most basic concepts of algebra, and you will find its application in other chapters as well.

You may expect sums for finding the degree of polynomials to carry fewer marks in the examination, and mostly these sums are in the compulsory part of the question paper. The sums of the NCERT maths book of class 9 are solved in a step-by-step manner on Vedantu and you may refer to it for a better understanding of all concepts of ‘Polynomials’.

7. How do I identify the degree of a polynomial?

The degree of polynomials in one variable is the highest power of the variable in the algebraic expression.

For example, X^2+5X+3. The degree of this equation is 2.

8. Are NCERT Solutions for Class 9 Maths Chapter 2 difficult to learn?

No, it is easy to score high in your exams if you regularly practice with Vedantu’s NCERT solutions for Class 9 Maths Chapter 2. The solutions are formulated by experienced subject matter experts who understand the CBSE curriculum. The solutions teach you how to present your answers in exams. Every minute detail is covered in a simple way for students to grasp the concepts easily. The solutions are updated according to the latest guidelines of the CBSE board.

9. What is a Polynomial?

Polynomials are expressions that contain one or more terms with coefficients that are not zero. Polynomial terms can be variables, constants, or both. A polynomial's exponents should always be whole numbers. The degree of a polynomial is the highest power in the polynomial. Polynomials are another way to represent real numbers. Further information may be found in Vedantu's NCERT answers for Class 9 Mathematics Chapter 2.

10. What is the Remainder Theorem?

When a polynomial p(x) is divided by the linear polynomial (x-a), then the remainder comes out to be p(a). The degree of the polynomial should be greater than one. For more details, refer to Vedantu’s NCERT solutions for Class 9 Maths Chapter 2. The solution PDFs and other study materials such as important questions and revision notes can also be downloaded from the Vedantu app as well for free of cost. 

11. What are the conditions for an expression to be Polynomial?

The following conditions are important to consider an expression as a polynomial:

All the coefficients should be real numbers.

The exponent should not be a negative number. 

For more details, refer to Vedantu’s NCERT solutions for Class 9 Maths Chapter 2. The solution PDFs and other study materials such as important questions and revision notes can also be downloaded from the Vedantu app as well for free of cost. 

NCERT Solutions for Class 9 Maths

Ncert solutions for class 9.

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NCERT Exemplar Class 9 Maths Chapter 2 Polynomials

April 10, 2019 by Bhagya

NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths . Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials.

NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials

Exercise 2.1

Question 2. √2 is a polynomial of degree (a) 2 (b) 0 (c) 1 (d)½ Solution: (b) √2 = -√2x°. Hence, √2 is a polynomial of degree 0, because exponent of x is 0.

Question 3. Degree of the polynomial 4x 4 + 0x 3 + 0x 5 + 5x + 7 is (a) 4 (b) 5 (c) 3 (d) 7 Solution: (a) 4 4x 4 + 0x 3 + 0x 5  + 5x + 7 = 4x 4 + 5x + 7 As we know that the degree of a polynomial is equal to the highest power of variable x. Here, the highest power of x is 4. Therefore, the degree of the given polynomial is 4.

Question 4. Degree of the zero polynomial is (A) 0 (B) 1 (C) Any natural number (D) Not defined Solution: (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x 5 etc. Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial.

Question 5. If p (x) = x 2 – 2√2x + 1, then p (2√2) is equal to (a) 0 (b) 1 (c) 4√2 (d) 8 √2 +1 Solution: (b) Given, p(x) = x 2 – 2√2x + 1 …(i) On putting x = 2√2 in Eq. (i), we get P(2√2) = (2√2) 2 – (2√2)(2√2) + 1 = 8 – 8 + 1 = 1

Question 6. The value of the polynomial 5x – 4x 2 + 3, when x = – 1 is (a)-6 (b) 6 (c) 2 (d) -2 Solution: (a) Let p (x) = 5x – 4x 2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1) 2 + 3= -5 – 4 + 3 = -6

Question 7. If p (x) = x + 3, then p(x) + p(- x) is equal to (a) 3 (b) 2x (c) 0 (d) 6 Solution: (d) Given p(x) = x + 3, put x = -x in the given equation, we get p(-x) = -x + 3 Now, p(x) + p(-x) = x + 3 + (-x) + 3 = 6

Question 8. Zero of the zero polynomial is (a) 0 (b) 1 (c) any real number (d) not defined Solution: (c) Zero of the zero polynomial is any real number. e.g., Let us consider zero polynomial be 0(x – k), where k is a real number For determining the zero, put x – k = 0 ⇒ x = k Hence, zero of the zero polynomial be any real number.

Question 9. Zero of the polynomial p(x) = 2x + 5 is (a) -2/5 (b) -5/2 (c) 2/5 (d) 5/2 Solution: (b) Given, p(x) = 2x + 5 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 ⇒ -5/2 Hence, zero of the polynomial p(x) is -5/2.

Question 10. One of the zeroes of the polynomial 2x 2 + 7x – 4 is (a) 2 (b) ½ (c) -1 (d) -2 Solution: (b) Let p (x) = 2x 2 + 7x – 4 = 2x 2 + 8x – x – 4 [by splitting middle term] = 2x(x+ 4)-1(x + 4) = (2x-1)(x+ 4) For zeroes of p(x), put p(x) = 0 ⇒ (2x -1) (x + 4) = 0 ⇒ 2x – 1 = 0 and x + 4 = 0 ⇒ x = ½ and x = -4 Hence, one of the zeroes of the polynomial p(x) is ½.

Question 11. If x 51  + 51 is divided by x + 1, then the remainder is (a) 0 (b) 1 (c) 49 (d) 50 Solution: (d) Let p(x) = x 51 + 51 . …(i) When we divide p(x) by x+1, we get the remainder p(-1) On putting x = -1 in Eq. (i), we get p(-1) = (-1) 51 + 51 = -1 + 51 = 50 Hence, the remainder is 50.

Question 12. If x + 1 is a factor of the polynomial 2x 2 + kx, then the value of k is (a) -3 (b) 4 (c) 2 (d)-2 Solution: (c) Let p(x) = 2x 2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2 – k = 0 ⇒ k = 2 Hence, the value of k is 2.

Question 13. x + 1 is a factor of the polynomial (a) x 3 + x 2 – x + 1 (b) x 3 + x 2 + x + 1 (c) x 4 + x 3 + x 2 + 1 (d) x 4 + 3x 3 + 3x 2 + x + 1 Solution: (b) Let assume (x + 1) is a factor of x 3  + x 2  + x + 1. So, x = -1 is zero of x 3  + x 2  + x + 1 (-1) 3  + (-1) 2 + (-1) + 1 = 0 ⇒ -1 + 1 – 1 + 1 = 0 ⇒ 0 = 0 Hence, our assumption is true.

Question 14. One of the factors of (25x 2 – 1) + (1 + 5x) 2 is (a) 5 + x (b) 5 – x (c) 5x -1 (d) 10x Solution: (d) Now, (25x 2 -1) + (1 + 5x) 2 = 25x 2 -1 + 1 + 25x 2 + 10x [using identity, (a + b) 2 = a 2 + b 2 + 2ab] = 50x 2 + 10x = 10x (5x + 1) Hence, one of the factor of given polynomial is 10x.

Question 15. The value of 249 2 – 248 2 is (a) 1 2 (b) 477 (c) 487 (d) 497 Solution: (d) Now, 249 2 – 248 2 = (249 + 248) (249 – 248) [using identity, a 2 – b 2 = (a – b)(a + b)] = 497 × 1 = 497.

Question 16. The factorization of 4x 2 + 8x+ 3 is (a) (x + 1) (x + 3) (b) (2x + 1) (2x + 3) (c) (2x + 2) (2x + 5) (d) (2x – 1) (2x – 3) Solution: (b) Now, 4x 2 + 8x + 3= 4x 2 + 6x + 2x + 3 [by splitting middle term] = 2x(2x + 3) + 1 (2x + 3) = (2x + 3) (2x + 1)

Question 17. Which of the following is a factor of (x+ y) 3 – (x 3 + y 3 )? (a) x 2 + y 2 + 2 xy (b) x 2 + y 2 – xy (c) xy 2 (d) 3xy Solution: (d) Now, (x+ y)3 – (x 3 + y 3 ) = (x + y) – (x + y)(x 2 – xy + y 2 ) [using identity, a 3 + b 3 = (a + b)(a 2 – ab + b 2 )] = (x + y)[(x + y) 2 -(x 2 – xy + y 2 )] = (x+ y)(x 2 + y 2 + 2xy – x 2 + xy – y 2 ) [using identity, (a + b) 2 = a 2 + b 2 + 2 ab)] = (x + y) (3xy) Hence, one of the factor of given polynomial is 3xy.

Question 18. The coefficient of x in the expansion of (x + 3) 3 is (a) 1 (b) 9 (c) 18 (d) 27 Solution: (d) Now, (x + 3) 3 = x 3 + 3 3 + 3x (3)(x + 3) [using identity, (a + b) 3 = a 3 + b 3 + 3ab (a + b)] = x 3  + 27 + 9x (x + 3) = x 3 + 27 + 9x 2  + 27x Hence, the coefficient of x in (x + 3) 3 is 27.

Question 21. If a + b + c =0, then a 3  + b 3  + c 3 is equal to (a) 0 (b) abc (c) 3abc (d) 2abc Solution: (d) Now, a 3  + b 3  + c 3  = (a + b + c) (a 2 + b 2 + c 2 – ab – be – ca) + 3abc [using identity, a 3  + b 3  + c 3 – 3 abc = (a + b + c)(a 2 + b 2 + c 2 – ab – be – ca)] = 0 + 3abc [∴ a + b + c = 0, given] a 3  + b 3 + c 3 = 3abc

Exercise 2.2

(ii) Polynomial Each exponent of the variable x is a whole number.

(iii) Not polynomial Because exponent of the variable x is 1/2, which is not a whole number.

(iv) Polynomial Because each exponent of the variable x is a whole number.

(v) Not polynomial Because one of the exponents of the variable x is -1, which is not a whole number.

(vi) Not polynomial Because given expression is a rational expression, thus, not a polynomial.

(vii) Polynomial Because each exponent of the variable a is a whole number.

Question 2. Write whether the following statements are True or False. Justify your answer. (i) A binomial can have atmost two terms (ii) Every polynomial is a binomial (iii) A binomial may have degree 5 (iv) Zero of a polynomial is always 0 (v) A polynomial cannot have more than one zero (vi) The degree of the sum of two polynomials each of degree 5 is always 5. Solution: (i) False Because a binomial has exactly two terms.

(ii) False Because every polynomial is not a binomial. e.g., (a)x² + 4x + 3 [polynomial but not a binomial] (b) x² + 5 [polynomial and also a binomial]

(iii) True Because a binomial is a polynomial whose degree is a whole number which is greater than or equal to one. Therefore, a binomial may have degree 5.

(iv) False Because zero of a polynomial can be any real number e.g., for p(x) = x – 1, zero of p(x) is 1, which is a real number.

(v) False Because a polynomial can have any number of zeroes. It depends upon the degree of the polynomial. e.g. for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and -2.

(vi) False Because the sum of any two polynomials of same degree has not always same degree. e.g., Let f(x) = x 5 + 2 and g(x) = -x 5 + 2x 2 ∴ Sum of two polynomials, f(x) + g(x) = x 5  + 2 + (-x 5 + 2x 2 ) = 2x 2  + 2, which is not a polynomial of degree 5.

Exercise 2.3

Question 1. Classify the following polynomials as polynomials in one variable, two variables etc. (i) x 2 + x + 1 (ii) y 3 – 5y (iii) xy + yz + zx (iv) x 2 – Zxy + y 2 + 1 Solution: (i) Polynomial x 2  + x + 1 is a one variable polynomial, because it contains only one variable i.e., x. (ii) Polynomial y 3  – 5y is a one variable polynomial, because it contains only one variable i.e., y. (iii) Polynomial xy + yz + zx is a three variables polynomial, because it contains three variables x, y and z. (iv) Polynomial  x 2  –  2xy + y 2  + 1 is a two variables polynomial, because it contains two variables x and y.

Question 2. Determine the degree of each of the following polynomials. (i) 2x – 1 (ii) -10 (iii) x 3  – 9x + 3x 5 (iv) y 3 (1 – y 4 ) Solution: (i) Degree of polynomial 2x – 1 is 1, Because the maximum exponent of x is 1. (ii) Degree of polynomial – 10 or – 10x° is 0, because the exponent of x is 0. (iii) Degree of polynomial x3 – 9x + 3x 5 is 5, because the maximum exponent of x is 5. (iv) Degree of polynomial y 3 (1 – y 4 ) or y 3 – y 7 is 7, because the maximum exponent of y is 7.

(ii) The coefficient of x 3 in given polynomial is 1/5

(iii) The coefficient of x 6 in given polynomial is -1.

(iv) The constant term in given polynomial is 1/5

(iii) We have, (x – 1) (3x – 4) = 3x 2 – 7x + 4 ∴ Coefficient of x² in 3x² – 7x + 4 is 3.

(iv) We have, (2x – 5) (2x² – 3x + 1) = 2x (2x² – 3x + 1) – 5(2x² – 3x + 1) = 4x³ – 6x² + 2x – 10x² + 15x – 5 = 4x³ – 16x² + 17x – 5 Coefficient of x² in 4x³ – 16x² + 17x – 5 is -16.

Question 5. Classify the following as  constant, linear, quadratic and cubic polynomials: (i) 2 – x² + x³ (ii) 3x³ (iii) 5t – √7 (iv) 4 – 5y² (v) 3 (vi) 2 + x (vii) y³ – y (viii) 1 + x + x² (ix) t² (x) √2x – 1 Solution: (i) Polynomial 2 – x² + x³ is a cubic polynomial, because its degree is 3. (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. (iii) Polynomial 5t – √7 is a linear polynomial, because its degree is 1. (iv) Polynomial 4 – 5y² is a quadratic polynomial, because its degree is 2. (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. (vi) Polynomial 2 + x is a linear polynomial, because its degree is 1. (vii) Polynomial y³ – y is a cubic polynomial, because its degree is 3. (viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2. (ix) Polynomial t² is a quadratic polynomial, because its degree is 2. (x) Polynomial √2x – 1 is a linear polynomial, because its degree is 1.

Question 6. Give an example of a polynomial, which is (i) monomial of degree 1. (ii) binomial of degree 20. (iii) trinomial of degree 2. Solution: (i) The example of monomial of degree 1 is 3x. (ii) The example of binomial of degree 20 is 3x 20  + x 10 (iii) The example of trinomial of degree 2 is x 2  – 4x + 3

Question 7. Find the value of the polynomial 3x 3 – 4x 2 + 7x – 5, when x = 3 and also when x = -3. Solution: Let p(x) =3x 3  – 4x 2  + 7x – 5 At x= 3, p(3) = 3(3) 3 – 4(3) 2 + 7(3) – 5 = 81 – 36 + 21 – 5 ∴ P( 3) = 61 At x = -3, p(-3)= 3(-3) 3 – 4(-3) 2 + 7(-3) – 5 = -81 – 36 – 21 – 5 = -143 ∴ p(-3) = -143 Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

Question 9. Find p(0), p( 1) and p(-2) for the following polynomials (i) p(x) = 10x – 4x 2 – 3 (ii) p(y) = (y + 2)(y – 2) Solution: (i) Given, polynomial is p(x) = 10x – 4x 2  – 3 On putting x = 0, 1 and – 2, respectively in Eq. (i), we get p(0) = 10(0) – 4(0) 2 – 3 = 0 – 0 – 3 = -3 p(1) = 10 (1) – 4 (1 ) 2 – 3 = 10 – 4 – 3= 10 – 7 = 3 and p(-2) =10 (-2) -4 (-2) 2 – 3 = -20- 4 × 4 – 3 = -20 – 16 – 3 = -39 Hence, the values of p(0), p(1) and p(-2) are respectively, -3, 3 and – 39.

(ii) Given, polynomial isp(y) = (y+2)(y-2) On putting y =0,1 and -2, respectively in Eq. (i), we get p(0) = (0+2)(0-2) = -4 p(1) = (1 + 2)(1-2) = 3 x (-1) = -3 and p(-2) = (-2 + 2)(-2 -2) = 0 (-4) = 0 Hence, the values of p(0), p(1) and p(-2) are respectively, -4, -3 and 0.

Question 10. Verify whether the following are true or false. (i) -3 is a zero of at – 3 (ii) -1/3 is a zero of 3x + 1 (iii) -4/5 is a zero of 4 – 5y (iv) 0 and 2 are the zeroes of t 2 – 2t (v) -3 is a zero of y 2 + y – 6 Solution: (i) False Put x – 3 = 0 ⇒ x = 3 Hence, zero of x – 3 is 3.

(ii) True Put 3x + 1 = 0 ⇒ x = -1/3 Hence, zero of 3x + 1 is -1/5.

(iii) False Put 4 – 5y = 0 ⇒ y = 4/5 Hence, zero of 4 – 5y is 4/5.

(iv) True Put t² – 2t = 0 ⇒ t(t – 2) = 0 ⇒ t = 0 and t – 2 = 0 ⇒ t = 0 and t = 2 Hence, the zeroes of t² – 2t are 0 and 2.

(v) True Put y² + y – 6 = 0 ⇒ y² + 3y – 2y – 6 = 0 ⇒ y(y + 3) – 2(y + 3) = 0 = (y-2)(y + 3) = 0 ⇒ y – 2 = 0 and y + 3 = 0 ⇒ y = 2 and y = -3 Hence, the zeroes of y² + y – 6 are 2 and – 3.

Question 11. Find the zeroes of the polynomial in each of the following, (i) p(x)= x – 4 (ii) g(x)= 3 – 6x (iii) q(x) = 2x – 7 (iv) h(y) = 2y Solution: (i) Given, polynomial is p(x) = x – 4 For zero of polynomial, put p(x) = 0 ∴ x – 4 = 0 ⇒ x = 4 Hence, zero of polynomial is 4.

(ii) Given, polynomial is g(x) = 3 – 6x For zero of polynomial, put g(x) = 0 ∴ 3 – 6x = 0 ⇒ x = 3/6 = 1/2. Hence, zero of polynomial is X

(iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0 ∴ 2x – 7 = 0 ⇒ 2x = 7 ⇒ x = 7/2 Hence, zero of polynomial q(x) is 7/2

(iv) Given polynomial h(y) = 2 y For zero of polynomial, put h(y) = 0 ∴ 2y = 0 ⇒ y = 0 Hence, the zero of polynomial h(y) is 0.

Question 12. Find the zeroes of the polynomial p(x) = (x – 2) 2  – (x + 2) 2 . Solution: Given, polynomial is p(x) = (x – 2) 2  – (x + 2) 2 For zeroes of polynomial, put p(x) = 0 ∴ (x – 2) 2  – (x + 2) 2  = 0 ⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0 [using identity, a 2  – b 2 = (a – b)(a + b)] ⇒ (2x)(-4) = 0 ⇒ -8x = 0 ⇒ x = 0 Therefore zero of the polynomial is p(x) is 0.

Question 14. By remainder theorem, find the remainder when p(x) is divided by g(x) (i) p(x) = x 3  – 2x 2  – 4x – 1, g(x) = x + 1 (ii) p(x) = x 3 – 3x 2 + 4x + 50, g(x) = x – 3 (iii) p(x) = x 3  – 12x 2 + 14x – 3, g(x) = 2x – 1 – 1 (iv) p(x) = x 3  – 6x 2  + 2x – 4, g(x) = 1 – (3/2) x Solution: (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 Here, zero of g(x) is -1. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1) ∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1 = -1 – 2 + 4 – 1 = 0 Therefore, remainder is 0.

(ii) We have, p(x) = x³ – 3x² + 4x + 50 and g(x) = x – 3 Here, zero of g(x) is 3. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) ∴ p(3) = (3)³ – 3(3)² + 4(3) + 50 = 27 – 27 + 12 + 50 = 62 Therefore, remainder is 62.

Question 15. Check whether p(x) is a multiple of g(x) or not (i) p(x) = x 3 – 5x 2 + 4x – 3,  g(x) = x – 2. (ii) p(x) = 2x 3 – 11x 2 – 4x + 5,  g(x) = 2x + 1 Solution: (i) We have, g(x) = x – 2 Here, zero of g(x) is 2. Now, p(x) = x 3 – 5x 2 + 4x – 3 ∴ p(2) = (2) 3 – 5(2) 2 + 4(2) – 3 = 8 – 20 + 8 – 3 = – 7 Since, remainder ≠ 0, then p(x) is not a multiple of g(x).

Since, remainder ≠ 0, then p(x) is not a multiple of g(x).

Question 16. Show that, (i) x + 3 is a factor of 69 + 11x – x 2 + x 3 (ii) 2x – 3 is a factor of x + 2x 3 -9x 2 +12 Solution: (i) Let p(x) = x 3 – x 2 + 11x + 69 x + 3 is a factor of p(x) if p(-3) = 0 Now, p(-3) = (-3) 3 – (-3) 2 + 11(-3) + 69 = – 27 – 9 – 33 + 69 = 0 Therefore, (x + 3) is a factor of p(x).

Question 17. Determine which of the following polynomial has x – 2 a factor (i) 3x 2 + 6x – 24 (ii) 4x 2 + x – 2 Solution: (i) Let p(x) = 3x 2 + 6x – 24  … (1) Substituting x = 2 in (1), we get p(2) = 3(2) 2 + 6(2) – 24 = 12 + 12 – 24 = 0 Hence, x – 2 is a factor of p(x).

(ii)Let p(x) = 4x 2 + x – 2 … (2) Substituting x = 2 in (2), we get p(2) = 4(2) 2 + 2 – 2 = 16 ≠ 0 Hence, x – 2 is not a factor of p(x).

Question 18. Show that p-1 is a factor of p 10 -1 and also of p 11  -1. Solution: Let g (p) = p 10  -1  …(1) and h(p) = p 11  -1  …(2) On putting p = 1 in Eq. (i), we get g(1) = 1 10  -1 = 1 – 1 = 0 Hence, p-1 is a factor of g(p). Again, putting p = 1 in Eq. (2), we get h (1) = (1) 11  -1 = 1 – 1 = 0 Hence, p -1 is a factor of h(p).

Question 19. For what value of m is x 3 -2mx 2 +16 divisible by x + 2? Solution: Let p(x) = x 3  – 2mx 2  + 16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2) 3 – 2m(-2) 2 + 16 = 0 ⇒ -8 – 8m + 16 = 0 ⇒ 8m = 8 m = 1 Hence, the value of m is 1.

Question 20. If x + 2a is a factor of a 5  – 4a 2 x 3 + 2x + 2a + 3, then find the value of a. Solution: Let p(x) =a 5  – 4a 2 x 3  + 2x + 2a + 3 Since, x + 2a is a factor of p(x), then put p(-2a) = 0 ∴ (-2a) 5  – 4a 2 (-2a) 3 + 2(-2a) + 2a + 3 = 0 ⇒ -32a 5 + 32a 5 – 4a + 2a + 3 = 0 ⇒ -2a + 3 = 0 2a = 3 a = 3/2. Hence, the value of a is 3/2.

Question 22. If x + 1 is a factor of ax 3 + x 2 – 2x + 4o – 9, find the value of a. Solution: Let p(x) = ax 3 + x 2 – 2x + 4a – 9 Since, x + 1 is a factor of p(x), then p(-1) = 0 a(- 1) 3 + (- 1) 2 – 2(-1) + 4a – 9 = 0 ⇒ -a + 1 + 2 + 4a – 9 = 0 ⇒ 3a = 6 ⇒ a = 2

Question 23. Factorise: (i) x 2 + 9x + 18 (ii) 6x 2 + 7x – 3 (iii) 2x 2 – 7x – 15 (iv) 84 – 2r – 2r 2 Solution: (i) We have, x 2 + 9x +18 = x 2 + 6x + 3x +18 = x(x + 6) + 3(x + 6) = (x + 3) (x + 6) We have, 6x 2 + 7x – 3 = 6x 2 + 9x – 2x – 3 = 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3) We have, 2x 2 – 7x – 15 = 2x 2 – 10x + 3x -15 = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) We have, 84 – 2r – 2r 2 = – 2 (r 2 + r – 42) = -2(r 2 + 7r – 6r – 42) = -2[r(r + 7) -6(r + 7)] = 2(6 – r)(r + 7) or 2(6 – r) (7 + r)

Question 24. Factorise: (i) 2x 3 – 3x 2 – 17x + 30 (ii) x 3 -6x 2 +11x-6 (iii) x 3 + x 2 -4x-4 (iv) 3x 3 -x 2 -3x+1 Solution: (i) We have, 2X 3 – 3x 2 – 17x + 30 = 2x 3 – 4x 2 + x 2 – 2x – 15x + 30 = 2x 2 (x – 2) + x(x – 2) – 15(x – 2) = (x – 2) (2x 2 + x – 15) = (x – 2) (2x 2 + 6x – 5x – 15) = (x – 2) [2x(x + 3) – 5(x + 3)] = (x – 2)(x + 3)(2x – 5)

(ii)We have, x 3 – 6x 2 + 11x – 6 = x 3 – x 2 – 5x 2 + 5x + 6x – 6 = x 2 (x – 1) – 5x (x – 1) + 6(x – 1) = (x -1) (x 2 – 5x + 6) = (x – 1) (x 2 – 3x – 2x + 6) = (x – 1) [x(x – 3) – 2(x – 3)] = (x-1)(x-2)(x-3)

(iii) We have, x 3 + x 2 – 4x – 4 = x 2 (x + 1) – 4(x + 1) = (x + 1)(x 2 – 4) = (x + 1) (x – 2) (x + 2)[∴ a 2 – b 2 = (a – b) (a + b)]

(iv) We have, 3x 3 – x 2 – 3x + 1 = 3x 3 – 3x 2 + 2x 2 – 2x – x + 1 = 3x 2 (x – 1) + 2x(x – 1) -1(x – 1) = (x – 1)(3x 2 + 2x – 1) = (x – 1) (3x 2 + 3x – x – 1) = (x – 1) [3x(x + 1) – 1(x + 1)] = (x – 1) (x + 1)(3x -1)

Question 25. Using suitable identity, evaluate the following: (i) 103 3 (ii) 101 × 102 (iii) 999 2 Solution: (i) We have, 103 3 = (100 + 3) 3 = (100) 3 + (3) 3 + 3(100)(3)(100 + 3) [∴ (a + b) 3 = a 3 + b 3 + 3ab(a + b)] = 1000000 + 27 + 900(103) = 1000027 + 92700 = 1092727

(ii) We have, 101 × 102 = (100 + 1) (100 + 2) = (100) 2 + (1 + 2)100 + (1)(2) [∴ (x + a)(x + b) = x 2 + (a + b)x + ab] = 10000 + 300 + 2 = 10302

(iii) We have, (999) 2 = (1000 -1) 2 = (1000) 2 + (1) 2 – 2(1000)(1) [∴ (a – b) 2 = a 2 + b 2 – 2ab] = 1000000 + 1 – 2000 = 998001

Question 27. Factorise the following: (i) 9x 2  – 12x + 3 (ii) 9x 2  – 12x + 4 Solution: (i) We have, 9x 2 – 12x + 3 = 3(3x 2 – 4x + 1) = 3(3x 2 – 3x – x + 1) = 3[3x(x – 1) – 1(x – 1)] = 3(3x – 1)(x – 1)

(ii) We have, 9x 2 – 12x + 4 = (3x) 2 – 2 × 3x × 2 + (2) 2 = (3x – 2) 2                 [∴ a 2 – 2ab + b 2 = (a – b)²] = (3x – 2) (3x – 2)

Question 28. Expand the following: (i) (4o – b + 2c) 2 (ii) (3o-5b-c) 2 (ii) (-x + 2y – 3z) 2 Solution: We know that (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx (i) We have, (4a – b + 2c) 2 = (4a) 2 + (-b) 2  + (2c) 2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) = 16a 2 + b 2 + 4c 2 – 8ab – 4bc + 16ac

(ii)We have, (3a – 5b – cf = (3a) 2 + (-5b² + (- c²) + 2(3a)(-5b) + 2(-5b)(-c) + 2(-c)(3a) = 9a 2 + 25b 2 + c 2  – 30ab + 10bc – 6ac

(iii) We have, (- x + 2y – 3z) 2 = (- x) 2 + (2y) 2  + (-3z) 2 + 2(-x)(2y) + 2(2y)(- 3z) + 2(- 3z)(- x) = x 2 + 4y 2 + 9z 2 – 4xy – 12yz + 6xz

Question 29. Factorise the following: (i) 9x 2 + 4y 2 +16z 2 +12xy-16yz-24xz (ii) 25x 2 + 16y 2 + 4Z 2 – 40xy +16yz – 20xz (iii) 16x 2 + 4)^ + 9z 2 -^ 6xy – 12yz + 24xz Solution: (i) We have, 9x 2 + 4y 2 + 16z 2 + 12xy – 16yz – 24xz = (3x) 2 + (2y) 2 + (-4z) 2 + 2(3x)(2y) + 2(2y)(-4z) + 2(-4z)(3x) = (3x + 2y – 4z) 2 [∴ a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = (a + b + c) 2 ] = (3x + 2y- 4z) (3x + 2y – 4z)

(ii)We have, 25X 2 + 16y 2 + 4z 2 – 40xy + 16yz – 20xz =(-5x) 2 + (4y) 2 + (2z) 2 + 2(-5x)(4y) + 2(4y)(2z) + 2(2z)(-5x) = (- 5x + 4y + 2z) 2 [∴ a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = (a + b + c) 2 ] = (- 5x + 4y + 2z)(- 5x + 4y + 2z)

(iii) We have, 16x 2 + 4y 2 + 9z 2 – 16xy – 12yz + 24xz = (4x) 2 + (- 2y) 2 + (3z) 2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x) = (4x – 2y + 3z) 2 [∴ a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = (a + b + c) 2 ] = (4x – 2y + 3z)(4x -2y + 3z)

Question 30. lf a+b+c=9 and ab + bc + ca = 26, find a 2 + b 2 + c 2 . Solution: We have, a + b + c = 9 ⇒ (a + b + c) 2 = (9) 2 [Squaring on both sides] ⇒ a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = 81 ⇒ a 2 + b 2 + c 2 + 2 (ab + bc + ca) = 81 ⇒ a 2 + b 2 + c 2 + 2(26) = 81 [∴ ab + bc + ca = 26] ⇒ a 2 + b 2 + c 2 = 81 – 52 = 29

Question 34. Factorise: (i) 1 + 64x 3 (ii) a 3  – 2√2b 3 Solution: (i) We have, 1 + 64x 3 = (1) 3 + (4x) 3 = (1 + 4x)[(1) 2 – (1)(4x) + (4x) 2 ] [∴ a 3 + b 3 = (a + b)(a 2 – ab + b 2 )] = (1 + 4x)(1 – 4x + 16x 2 )

(ii)  We have, a 3 – 2√2b 3 = (a) 3 – (√2b) 3 = (a – √2b)[a 2 + a( √2b) + (√2b) 2 ] [∴ a 3 – b 3 = (a – b)(a 2 + ab + b 2 )] = (a – √2b)(a 2 + √2ab + 2b 2 )

Question 35. Fi nd (2x – y + 3z) (4x 2 + y 2 + 9z 2  + 2xy + 3yz – 6xz). Solution: We have, (2x – y + 3z) (4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) = 2x(4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) – y(4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) + 3z(4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) = 8x³ + 2xy 2 + 18xz 2 + 4x 2 y + 6xyz – 12x 2 z – 4x 2 y – y 3 – 9yz 2 – 2xy 2 – 3y 2 z + 6xyz + 12x 2 z + 3y 2 z + 27z 3 + 6xyz + 9yz 2 – 18xz 2 = 8X 3 – y 3 + 27z 3 + 18xyz

(ii) We have, (0.2) 3 – (0.3) 3 + (0.1) 3 = (0.2) 3 + (- 0.3) 3 + (0.1) 3 Since, 0.2 – 0.3 + 0.1 = 0, ∴ (0.2)³ + (-0.3)³ + (0.1) 3 = 3(0.2) (-0.3) (0.1) [ ∴ If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc] = -0.018

Question 38. Without finding the cubes, factorise (x- 2y) 3 + (2y – 3z) 3 + (3z – x) 3 . Solution: we see that (x – 2y) +(2y – 3z)+ (3z – x) = 0 Therefore, (x – 2y)³ + (2y – 3z)³ + (3z – x)³ = 3(x – 2y)(2y – 3z)(3z – x). If a + b + c = 0, then a 3  + b 3  + c 3 = 3abc

Question 39. Find the value of (i) x 3 + y 3 – 12xy + 64,when x + y = -4. (ii) x 3 – 8y 3 – 36xy-216,when x = 2y + 6. Solution: (i) Since, x + y + 4 = 0, then x 3 + y 3 + (4) 3 = 3xy(4) [∴ If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc] ⇒ x 3 + y 3 + 64 = 12xy ⇒ x 3 + y 3  – 12xy + 64 = 0

(ii) Since, x – 2y – 6 = 0, then x³ + (-2y) 3 + (-6) 3 = 3x(-2y)(-6) [∴ If a + b + c = 0, then a 3 + b 3 + c 3 3abc] ⇒ x 3 – 8y 3 – 216 = 36xy ⇒ x 3 – 8y 3 – 36xy – 216 = 0

Question 40. Give possible expression for the length and breadth of the rectangle whose area is given by 4a 2 + 4a – 3. Solution: Given, area of rectangle = (Length) × (Breadth) = 4a 2 + 4a – 3 = 4a 2 + 6a – 2a – 3 = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3

Exercise 2.4

Question 1. If the polynomials az 3 + 4z 2 + 3z – 4 and z 3  – 4z + o leave the same remainder when divided by z – 3, find the value of a. Solution: Let p 1 (z) = az 3  + 4z 2  + 3z – 4 and p 2 (z) = z 3  – 4z + o When we divide p 1 (z) by z – 3, then we get the remainder p,(3). Now, p 1 (3) = a(3)3 + 4(3)2 + 3(3) – 4 = 27a+ 36+ 9-4= 27a+ 41 When we divide p 2 (z) by z-3 then we get the remainder p 2 (3). Now, p 2 (3) = (3) 3 -4(3)+a = 27-12 + a = 15+a According to the question, both the remainders are same. p 1 (3) = p 2 (3) 27a + 41 = 15 + a 27a – a = 15 – 41 . 26a = 26 ∴ a = -1

Question 2. The polynomial p{x) = x 4  – 2x 3  + 3x 2  – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also, find the remainder when p(x) is divided by x + 2. Solution: We have, p(x) = x 4 – 2x 3 + 3x 2 – ox + 3a – 7 Since p(x) is divided by x + 1, then remainder is p(-1). p(-1) = (-1) 4 – 2(-1) 3 + 3(-1) 2 – a(-1) + 3a – 7 = 1+ 2 + 3 + o + 3a – 7 = 4a – 1 p(- 1) = 19          (Given) ⇒ 4a – 1 = 19 ⇒ 4a = 20 ∴ a = 5 Thus, required polynomial, p(x) = x 4 – 2x 3 + 3x 2 – 5x + 3(5) – 7 = x 4 – 2x 3 + 3x 2 – 5x + 8 Now, this is divided by x + 2, then remainder is p(-2). p(- 2) = (- 2) 4 – 2(- 2) 3 + 3(- 2) 2 – 5(- 2) + 8 = 16 + 16 + 12 + 10 + 8 = 62

Question 4. Without actual division, prove that 2x 4 – Sx 3 + 2x 2 – x + 2 is divisible by x 2 – 3x + 2. [Hint: Factorise x 2 – 3x + 2] Solution: Let p(x) = 2x 4 – 5x 3 + 2x 2 – x + 2 Now, x 2 -3x + 2 = x 2  – 2x – x + 2 = (x-2)(x- 1) Hence, zeroes of x 2 – 3x + 2 are 1 and 2. ⇒ p(x) is divisible by x 2 – 3x + 2 i.e., divisible by x – 1 and x – 2, if p(1) = 0 and p(2) = 0 Now, p(1) = 2(1) 4 – 5(1) 3 + 2(1) 2 -1 + 2 = 2 – 5 + 2 – 1 + 2 = 6 – 6 = 0 and p( 2) = 2(2) 4 – 5(2) 3 + 2(2) 2 -2 + 2 = 32 – 40 + 8 = 40 – 40 = 0 Hence, p(x) is divisible by x 2 – 3x + 2.

Question 6. Multiply x 2 + 4y 2 + z 2 + 2xy + xz – 2yz by (-z + x-2y). Solution: We have, (x 2 + 4y 2 + z 2 + 2xy + xz – 2yz)(- z + x – 2y) = x 2 (- z + x – 2y) + 4y 2 (- z + x – 2y) + z 2 (- z + x – 2y) + 2xy(- z + x – 2y) + xz(- z + x – 2y) – 2yz (- z + x – 2y) = -x 2 z + x 3 – 2 x 2 y – 4y 2 z + 4xy 2 – 81y 3 – z 3 + xz 2 – 2yz 2 – 2xyz + 2x 2 y – 4xy 2 – xz 2 + x 2 z – 2xyz + 2yz 2 – 2xyz + 4y 2 z = x 3 – 8y 3 – z 3 – 6xyz

Question 8. If a + b + c = 5 and ab + bc + ca = 10, then prove that a 3 + b 3 + c 3   – 3abc = -25. Solution: We have, a + b + c = 5,ab + bc + ca = 10 Since (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca), then (5) 2  = a 2 + b 2 + c 2 + 2(10) ⇒ a 2  + b 2  + c 2  = 25 – 20 ⇒ a 2 + b 2 + c 2  = 5     … (i) L.H.S. = a 3 + b 3 + c 3 – 3abc = (a + b + c)(a 2 + b 2 + c 2 – ab – be – ca) = (5) [5 – (ab + be + ca)]             [From (i)] = 5(5 -10) = 5(-5) = – 25 = R.H.S.

Question 9. Prove that (a +b +c) 3 -a 3 -b 3  – c 3 =3(a +b)(b +c)(c +a). Solution: L.H.S. = [(a + b + c) 3 – a 3 ] – (b 3 + c 3 ) = (a + b + c – a)[(a + b + c) 2 + a 2 + (a + b + c)a]  – [(b + c) (b 2 + c 2 – be)] x 3 – y 3 = (x – y)(x 2 + y 2 + xy) and x 3 + y 3 = (x + y)(x 2 + y 2 – xy) = (b + c)[a 2 + b 2 + c 2  + 2 ab + 2 bc + 2 ca + a 2 + a 2 + ab + ac] – (b + c)(b 2 + c 2 – bc) = (b + c)[3a 2 +3ab + 3ac + 3bc] = (b + c)[3(a 2 + ab + ac + bc)] = 3 (b + c)[a(a + b) + c(a + b)] = 3(a + b)(b + c)(c + a) = R.H.S.

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