case study based questions class 9 maths polynomials

Test: Polynomials- Case Based Type Questions - Class 9 MCQ

10 questions mcq test - test: polynomials- case based type questions, beti bachao, beti padhao (bbbp) is a personal campaign of the government of india that aims to generate awareness and improve the efficiency of welfare services intended for girls. in a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on beti bachao, beti padhao. q. if in the group, there are 10 teachers and 58 girls, then what is the number of boys.

No. of teachers = x + y = 10

⇒ (x + y) 2 = (10) 2

⇒ x 2 + y 2 + 2xy = 100

[Since (a + b) 2 = a 2 + b 2 + 2ab]

No. of students = (x 2 + y 2 ) = 58

⇒ 58 + 2xy = 100

⇒ 2xy = 100 – 58

⇒ xy = 42/2

Now, since (x + y) 3 = [x 3 + y 3 + 3xy(x + y)]

⇒ (10) 3 = [x 3 + y 3 + 3 × 21(10)]

⇒ 1000 = (x 3 + y 3 + 630)

⇒ 1000 – 630 = (x 3 + y 3 )

⇒ (x 3 + y 3 ) = 370

Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls. In a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on Beti Bachao, Beti Padhao. Q. Which is the correct identity?

• A. (a + b) 2 = a 2 + b 2 – 2ab
• B. (a + b) 2 = a 2 + b 2 + 2ab
• C. (a + b) 2 = a 2 – b 2 – 2ab
• D. All are correct.

Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls. In a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on Beti Bachao, Beti Padhao. Q. Which mathematical concept is used here?

• A. Linear equations
• B. Triangles
• C. Polynomials

In mathematics, a polynomial is an expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls.

In a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on Beti Bachao, Beti Padhao.

Q. (x – y) 3 =

(x 2 – y 2 – 3xy(x – y))

(x 3 – y 3 – 2xy(x – y)

(x 3 – y 3 – 3xy(x – y)

(x 3 – y 3 – 3xyx – y)

We know that (x - y) 3 can be written as

(x - y)(x - y)(x - y)

We know that (x - y)(x - y) can be multiplied and written as

= x 2 - xy - yx + y 2 (x - y)

= x 2 - 2xy + y 2 (x - y)

= x 3 - 2x 2 y + xy 2 - yx 2 + 2xy 2 - y 3

= x 3 – y 3 – 2xy(x – y)

Q. Using part (D), find (x 2 – y 2 ) if x – y = 23.

Also, x + y = 10

x 2 – y 2 = (x + y)(x – y)

National Association For The Blind (NAB) aimed to empower and well-inform visually challenged population of our country, thus enabling them to lead a life of dignity and productivity.

Q. Find the amount donated by Ravi.

Q. (x – y) 3 =

• A. x 3 – y 3 – 3xy
• B. x 3 – y 3 – 3xy(x – y)
• C. x 3 – y 3 – 3xy(x + y)
• D. x 3 – y 3

x 2 - xy - yx + y 2 (x - y)

= x 3 – y 3 – 3xy(x – y)

Q. (x + a)(x + b) = x 2 + ................ x + ab

Q. Which mathematical concept is involved in the above situation?

• A. Polynomial
• C. Lines and angles
• D. Triangle

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Important Questions for Polynomials- Case Based Type Questions

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CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF

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CBSE Class 9 Maths exam 2022-23 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions.

Each question has five sub-questions, each followed by four options and one correct answer. Students can easily download these questions in PDF format and refer to them for exam preparation.

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Case Study Questions for Class 9 Maths

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Are you preparing for your Class 9 Maths board exams and looking for an effective study resource? Well, you’re in luck! In this article, we will provide you with a collection of Case Study Questions for Class 9 Maths specifically designed to help you excel in your exams. These questions are carefully curated to cover various mathematical concepts and problem-solving techniques. So, let’s dive in and explore these valuable resources that will enhance your preparation and boost your confidence.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

CBSE Class 9 Maths Board Exam will have a set of questions based on case studies in the form of MCQs. The CBSE Class 9 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out.  CBSE Case Study Questions for Class 9  will provide you with detailed, latest, comprehensive & confidence-inspiring solutions to the maximum number of Case Study Questions covering all the topics from your  NCERT Text Books !

Table of Contents

CBSE Class 9th – MATHS: Chapterwise Case Study Question & Solution

Case study questions are a form of examination where students are presented with real-life scenarios that require the application of mathematical concepts to arrive at a solution. These questions are designed to assess students’ problem-solving abilities, critical thinking skills, and understanding of mathematical concepts in practical contexts.

Chapterwise Case Study Questions for Class 9 Maths

Case study questions play a crucial role in the field of mathematics education. They provide students with an opportunity to apply theoretical knowledge to real-world situations, thereby enhancing their comprehension of mathematical concepts. By engaging with case study questions, students develop the ability to analyze complex problems, make connections between different mathematical concepts, and formulate effective problem-solving strategies.

• Case Study Questions for Chapter 1 Number System
• Case Study Questions for Chapter 2 Polynomials
• Case Study Questions for Chapter 3 Coordinate Geometry
• Case Study Questions for Chapter 4 Linear Equations in Two Variables
• Case Study Questions for Chapter 5 Introduction to Euclid’s Geometry
• Case Study Questions for Chapter 6 Lines and Angles
• Case Study Questions for Chapter 7 Triangles
• Case Study Questions for Chapter 8 Quadilaterals
• Case Study Questions for Chapter 9 Areas of Parallelograms and Triangles
• Case Study Questions for Chapter 10 Circles
• Case Study Questions for Chapter 11 Constructions
• Case Study Questions for Chapter 12 Heron’s Formula
• Case Study Questions for Chapter 13 Surface Area and Volumes
• Case Study Questions for Chapter 14 Statistics
• Case Study Questions for Chapter 15 Probability

The above  Case studies for Class 9 Mathematics will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Studies have been developed by experienced teachers of schools.studyrate.in for benefit of Class 10 students.

• Class 9 Science Case Study Questions
• Class 9 Social Science Case Study Questions

How to Approach Case Study Questions

When tackling case study questions, it is essential to adopt a systematic approach. Here are some steps to help you approach and solve these types of questions effectively:

• Read the case study carefully: Understand the given scenario and identify the key information.
• Identify the mathematical concepts involved: Determine the relevant mathematical concepts and formulas applicable to the problem.
• Formulate a plan: Devise a plan or strategy to solve the problem based on the given information and mathematical concepts.
• Solve the problem step by step: Apply the chosen approach and perform calculations or manipulations to arrive at the solution.
• Verify and interpret the results: Ensure the solution aligns with the initial problem and interpret the findings in the context of the case study.

Tips for Solving Case Study Questions

Here are some valuable tips to help you effectively solve case study questions:

• Read the question thoroughly and underline or highlight important information.
• Break down the problem into smaller, manageable parts.
• Visualize the problem using diagrams or charts if applicable.
• Use appropriate mathematical formulas and concepts to solve the problem.
• Show all the steps of your calculations to ensure clarity.
• Check your final answer and review the solution for accuracy and relevance to the case study.

Benefits of Practicing Case Study Questions

Practicing case study questions offers several benefits that can significantly contribute to your mathematical proficiency:

• Enhances critical thinking skills
• Improves problem-solving abilities
• Deepens understanding of mathematical concepts
• Develops analytical reasoning
• Prepares you for real-life applications of mathematics
• Boosts confidence in approaching complex mathematical problems

Case study questions offer a unique opportunity to apply mathematical knowledge in practical scenarios. By practicing these questions, you can enhance your problem-solving abilities, develop a deeper understanding of mathematical concepts, and boost your confidence for the Class 9 Maths board exams. Remember to approach each question systematically, apply the relevant concepts, and review your solutions for accuracy. Access the PDF resource provided to access a wealth of case study questions and further elevate your preparation.

Q1: Can case study questions help me score better in my Class 9 Maths exams?

Yes, practicing case study questions can significantly improve your problem-solving skills and boost your performance in exams. These questions offer a practical approach to understanding mathematical concepts and their real-life applications.

Q2: Are the case study questions in the PDF resource relevant to the Class 9 Maths syllabus?

Absolutely! The PDF resource contains case study questions that align with the Class 9 Maths syllabus. They cover various topics and concepts included in the curriculum, ensuring comprehensive preparation.

Q3: Are the solutions provided for the case study questions in the PDF resource?

Yes, the PDF resource includes solutions for each case study question. You can refer to these solutions to validate your answers and gain a better understanding of the problem-solving process.

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CBSE Class 9 Mathematics Case Study Questions

Table of Contents

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If you’re looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter. With over 10,000 study notes, solved sample papers and practice questions, it’s got everything you need to ace your exams. Plus, it’s updated regularly to keep you aligned with the latest CBSE syllabus . So why wait? Start your journey to success with myCBSEguide today!

Significance of Mathematics in Class 9

Mathematics is an important subject for students of all ages. It helps students to develop problem-solving and critical-thinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.

CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .

Case studies in Class 9 Mathematics

A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.

Example of Case study questions in Class 9 Mathematics

The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on real-life scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.

The following are some examples of case study questions from Class 9 Mathematics:

Class 9 Mathematics Case study question 1

There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of   XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

• Now he told Raju to draw another line CD as in the figure
• The teacher told Ajay to mark  ∠ AOD  as 2z
• Suraj was told to mark  ∠ AOC as 4y
• Clive Made and angle  ∠ COE = 60°
• Peter marked  ∠ BOE and  ∠ BOD as y and x respectively

Now answer the following questions:

• 2y + z = 90°
• 2y + z = 180°
• 4y + 2z = 120°
• (a) 2y + z = 90°

Class 9 Mathematics Case study question 3

• (a) 31.6 m²
• (c) 513.3 m³
• (b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.

Class 9 Mathematics Curriculum at Glance

At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve real-world problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.

The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.

CBSE Class 9 Mathematics (Code No. 041)

Class 9 Mathematics question paper design

The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problem-solving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.

QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)

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Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.

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14 thoughts on “CBSE Class 9 Mathematics Case Study Questions”

This method is not easy for me

aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10-?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2-?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3-?2 4+ ?2 2+ ?2 d) 2-?3 the identity applied to solve (?10-?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (a-b)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b

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All questions was easy but search ? hard questions. These questions was not comparable with cbse. It was totally wastage of time.

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Can I have more questions without downloading the app.

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CBSE Class 9 Maths Most Important Case Study Based Questions With Solution

Cbse class 9 mathematics case study questions.

In this post I have provided CBSE Class 9 Maths Case Study Based Questions With Solution. These questions are very important for those students who are preparing for their final class 9 maths exam.

All these questions provided in this article are with solution which will help students for solving the problems. Dear students need to practice all these questions carefully with the help of given solutions.

As you know CBSE Class 9 Maths exam will have a set of cased study based questions in the form of MCQs. CBSE Class 9 Maths Question Bank given in this article can be very helpful in understanding the new format of questions for new session.

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Case studies in class 9 mathematics.

The Central Board of Secondary Education (CBSE) has included case study based questions in the Class 9 Mathematics paper in current session. According to new pattern CBSE Class 9 Mathematics students will have to solve case based questions. This is a departure from the usual theoretical conceptual questions that are asked in Class 9 Maths exam in this year.

Each question provided in this post has five sub-questions, each followed by four options and one correct answer. All CBSE Class 9th Maths Students can easily download these questions in PDF form with the help of given download Links and refer for exam preparation.

There is many more free study materials are available at Maths And Physics With Pandey Sir website. For many more books and free study material all of you can visit at this website.

Given Below Are CBSE Class 9th Maths Case Based Questions With Their Respective Download Links.

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• Important Questions with Solutions for CBSE Class 9 Maths Chapter 2 - Polynomials

CBSE Class 9 Maths Chapter-2 Important Questions - Free PDF Download

Mathematics is one of the most important disciplines in today's academic curriculum. Maths is concerned with the operation of a certain job statically and quantitatively. It is a complex subject with several themes, formulae, and theories. As a result, students must focus on the topic in order to progress academically. Students must begin with a strong core understanding in order to deal with the difficulties of mathematics in upper grades.

Class 9 is a vital part of a student's academic career as the first board examinations knock at the door in the upcoming year. Therefore they must have a good understanding of crucial subjects like Maths to secure well in the examinations. Here, Vedantu steps in as an efficient guide to the students to provide a push to score well in their examinations through important questions for class 9 Maths Chapter 2 and solutions framed by professionals. You can also Download NCERT Maths Class 9 to help you to revise the complete Syllabus and score more marks in your examinations. Students can also avail of NCERT Solutions for Class 9 Science from our website. Besides, find NCERT Solutions to get more understanding of various subjects. The solutions are up-to-date and are sure to help in your academic journey.

Download CBSE Class 9 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 9 Maths Important Questions for other chapters:

Important Questions for Class 9 Maths Polynomials

It is the most extensively used subject in science and computer, standing based on logic and reasoning. Therefore, every student must have an in-depth learning of the subject to frame a successful career. Students with a weak core of knowledge in Mathematics may face difficulties to deal with the complex formulas and concepts of Maths, which in return serves as a hurdle to score well in the examination. In class 9, algebra is one of the crucial domains of Maths where maximum students fumble.

An efficient learning algebra and its concepts enable the students to dismantle diplomatic Mathematics problems into more straightforward and convenient processes. Therefordents must practice important questions for class 9  maths chapter 2 to master the topic perfectly. Students who find the topic challenging or get stuck with any concept or problems of algebra many refer to the detailed solutions available on the Internet. Quality study material like important questions for class 9 maths polynomials are also availed for the students to access freely in text and PDF formats.

Class 9 Maths Chapter 2 Important Questions

Students are presented with an extensive view of the algebraic concepts and theories in class 9 Maths Ch 2 important questions. To explore different concepts of the chapter and practice a variety of problems, students must have their hands on important polynomials for class 9. Now, let's discuss some of the details about the chapter:

Polynomials

Polynomials can be quoted as an algebraic expression formed using indeterminates or variables and constants or coefficients. This algebraic expression allows such to perform addition, subtraction, multiplication, positive integer exponentiation of variables. The word polynomial is framed from 'poly' meaning 'many' and 'nominal' meaning 'term,' depicting many terms, which means a polynomial contains many terms but not infinite terms.

A polynomial expression comprises variables like x, y,z, coefficients like 1,2, and exponents like 2 in x². The polynomial function is generally depicted by P(x), where x is the variable. For instance,

P(x) = x² + 7x + 15, here x is the variable and 15 is the constant.

Types of Polynomials

Polynomials are categorised into three groups depending upon the number of terms it comprises of. Here are the types of polynomials.

A monomial is a type of polynomial in algebra consisting of a single non-zero term. A polynomial expression consists of one or more terms. Therefore, every term of a polynomial expression is a monomial. Every numeric value such as 6, 12, 151 is a monomial by itself, whereas the variables can x, y, a can also be included in the list of monomials in algebra. Example of a monomial expression – 7x².

Rules for monomial algebraic expressions:

If a monomial is multiplied by a constant, the output will also be a monomial.

If a monomial is multiplied by a monomial, the result will also be a monomial. For instance, if a monomial three is multiplied by 3, the result 8 is also a monomial.

A binomial is a type of polynomial expression comprising of two non-zero terms. Let's see some examples to make it clear,

7x² + 8y is a binomial expression with two variables.

10x⁴ + 9y is also a binomial expression with two variables.

A trinomial is a type of polynomial expression comprising of three non-zero terms. Let's see some examples to make it clear,

5x²+8x+9 is a trinomial expression with one variables x.

a + b+ c is a trinomial expression with three variables.

7x – 6y + 9z is a trinomial expression with three variables.

Students can explore different questions polynomial types through the class 9 Maths chapter 2 important questions.

Polynomial Theorems

Some of the vital theorems of polynomials are as follows:

Remainder Theorem

The polynomial remainder theorem, also quoted as the little Bezout's theorem, implies that if a polynomial P(x) is divided by any linear polynomial depicted by (x – a), the remainder of the operation will be a constant given by P(a), i.e., r = P(a).

Factor Theorem

The factor theorem implies that if P(x) is a polynomial of degree n > 1, and 'a' is a real number, this portrays that:

If P(x) = 0, then (x – a) is the factor of P(x),

If (x – a) is the factor of P(x), P(x) = 0.

Bezout's Theorem

Bezout's Theorem states that if P(x) = 0, then P(x) gets divided by (x – a), with 'r' as the remainder.

Intermediate Value Theorem

The intermediate value theorem states that when a polynomial function transforms from a negative to a positive value, it must cross the x-axis. In other words, the theorem highlights the properties of continuity of a function.

Fundamental Theorem of Algebra

The fundamental theorem of algebra states that each non-constant single variable that consists of a complex coefficient possess a minimum of one complex root.

Polynomial Equations

A polynomial equation is an algebraic equation comprising of variables with positive integer exponents and constants. A polynomial expression may contain many exponents, and the highest exponent value is termed as the degree of the equation. Let's take an example to make it clear,

ax⁴ + bx² + x + c, is a polynomial expression with degree = 4.

Algebraic identities of polynomials

Identity 1 : (x + z )2 = x² + 2xz + z²

Identity 2 : (y – z) 2 = y² – 2yz + z²

Identity 3 : y² – z² = (y + z) (y – z)

Identity 4 : (x + y) (x + z) = x² + (y + z)x + yz

Important Questions of Polynomials for Class 9

To present the students an insight into the algebraic world, we have highlighted here some of the important questions class 9 Maths chapter 2 , after a proper analysis of sample question papers:

What is a polynomial? Explain with example.

What are the types of polynomial expressions?

Explain the Remainder Theorem with an example.

Prove the Factor theorem of polynomials.

Illustrate Bezout's Theorem, and mention it's importance.

What do you mean by the degree of the polynomial? Explain with examples.

How can we add or subtract polynomials?

Explain the standard form of polynomials.

What do you mean by roots of equations? And how to find them.

Find the roots of polynomial equation, f(x) = x⁴ + 5x² + 7x + 19.

Benefits of Class 9 Maths Ch 2 Important Questions

The students preparing for the boards in the upcoming year must prepare a strong core foundation for developing an in-depth logic and understanding of algebra. Therefore they can blend the benefits by practising the class 9th Maths chapter 2 important questions . Here we have listed some of the fruitfulness of class 9 polynomials important questions:

Students can develop deep learning of the topics by exploring different types of questions presented in the important polynomials for class 9.

Vedantu, with an efficient team of top-notch educators, has carefully designed the questions after proper research and analysis of the past year's question papers and sample test papers.

The important questions of ch 2 Maths class 9 are carefully designed under the CBSE board's rules ’ strict guidance.

To perform well in mathematics, academic success is practice; the students must efficiently practice the polynomials class 9 important questions.

To prevent any issues or mistakes in the important questions for class 9 maths polynomials , expert teachers have reviewed and analysed the papers.

Mathematics is the foundation for logic and reasoning. As a result, in order to grasp the topic's various subjects, students must work with insufficient fundamental Mathematics comprehension and study important polynomial questions for class 9. Students must have a good comprehension of the crucial questions for class 9 mathematics chapter 2 in order to begin a career in science and technology.

Important Related Links for CBSE Class 9

FAQs on Important Questions with Solutions for CBSE Class 9 Maths Chapter 2 - Polynomials

1. Where can I find some important questions for Class 9 Maths Chapter 2?

Ans: Crucial Class 9 Maths Chapter 2 questions titled Polynomials assist students in preparing for the Class 9 Mathematics Test. This will give you a general sense of the types of questions you could encounter in the test and from which chapter. Understanding what to study in a subject makes learning easier and faster since it requires less time. Hence, while preparing for Class 9 Maths Chapter 2 named Polynomials in CBSE Class 9 Mathematics, students in Class 9th are encouraged to answer these key questions.

2. Does Vedantu provide solutions to Class 9 Maths Chapter 2 of NCERT textbook?

Ans: Free PDF download of Important Questions with solutions for CBSE Class 9 Maths Chapter 2 named Polynomials are prepared by Vedantu’s in-house expert Mathematics teachers from the latest edition of NCERT textbooks. You can register online for Maths tuition on Vedantu if you are eager to score more marks in the final examination. You can also Download NCERT Maths Class 9 Solutions in order to help yourself revise the complete syllabus and score more marks in your examinations. All the Class 9th students can also avail of NCERT Solutions for Science from Vedantu website and mobile application very easily which help them to prepare for the exam along with the important questions.

3. Can I download the solutions to Important Questions with solutions for CBSE Class 9 Maths Chapter 2 offered by Vedantu?

Ans: Yes, you can download the solutions to Important Questions with solutions for CBSE Class 9 Maths Chapter 2 offered by Vedantu. These are available on Vedantu’s official website and mobile app at free of cost in PDF format. In that case, you can download the Vedantu app from the Google play store to avail these Important Questions with solutions for CBSE Class 9 Maths Chapter 2 offered by Vedantu.

4. What is taught in Class 9 Maths Chapter 2 of CBSE curriculum?

Ans: Polynomials is the second chapter of Class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses the Polynomials and their applications. The introduction of the chapter includes whole numbers, integers, and rational numbers.

The chapter begins with the introduction of Polynomials in section 2.1 followed by two other very important topics explained in section 2.2 and 2.3.

Section 2.1 - Polynomials in one Variable – This topic discusses the Linear, Quadratic and Cubic Polynomial.

Section 2.2 - Zeros of a Polynomial – This chapter explains that a zero of a polynomial need not be zero and can have more than one zero.

Section 2.3 - Real Numbers and Their Decimal Expansions – Here you will study the decimal expansions of real numbers and understand if it can help in distinguishing between rationals and irrationals.

5. Can I print the MCQ Questions for Chapter 2 of Class 9 Maths with answers?

Ans: Yes, the MCQ Questions and Answers for Chapter 2 of Class 9 Maths are in a downloadable PDF format and can be printed easily. These are available on Vedantu's website and app and you can download them according to your comfort and timing and can print the MCQs for future reference. Studying these important questions will ensure you get a good score in the final exam for Class 9 Maths.

6. Are these free or is there any charge for the MCQ Questions for Chapter 2 of Class 9 Maths with answers?

Ans: Yes, the MCQ Questions and Answers for Chapter 2 of Class 9 Maths are absolutely free and do not carry any hidden charge or cost. You can also download these from the Vedantu app. You can download these anytime according to your comfort so that you can study as and when required. These important questions have been carefully selected by the experts at Vedantu and are guaranteed to help you to score the best.

7. How Many questions are there in NCERT Solutions of Chapter 2 of Class 9 Maths?

Ans: The question break-up of the exercises covered in NCERT Solutions of Chapter 2 of Class 9 Maths are as follows:

Exercise 2.1 includes five questions

Exercise 2.2 includes four questions

Exercise 2.3 includes three questions

8. What are the Important Topics covered in NCERT Solutions of Chapter 2 of Class 9 Maths?

Ans: The major topics covered in NCERT Solutions for Chapter 2 of Class 9 Maths are as follows: 

Polynomials in One Variable

Zeros of a Polynomial

Factorisation of Polynomials

Algebraic Identities

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

Ans: NCERT Solutions for Class 9 Mathematics Chapter 2 will certainly give you an advantage. The NCERT Solutions are created by experts in the field at Vedantu to provide students with effective solutions to problems while explaining their concepts so that they are never stuck with the same issue again in the future. This will ensure that you get the greatest grades and fully comprehend complicated ideas. As a result, you must without a doubt select NCERT Solutions for Chapter 2 of Class 9 Mathematics.

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Maths Topics to be covered for Class 9

• Real Numbers
• Polynomials
• Coordinate Geometry
• Linear Equations in Two Variables
• Introduction to Euclid's Geometry
• Lines and Angles
• Heron’s Formula
• Probability
• Quadrilaterals
• Area of Parallelogram and Triangles
• Constructions
• Surface Areas and Volumes

Structure of CBSE Maths Sample Paper for Class 9 is

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CBSE Class 9 Maths Sample Papers

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Question Bank of Other Subjects of Class 9

Importance of Question Bank for Exam Preparation?

There are many ways to ascertain whether a student has understood the important points and topics of a particular chapter and is he or she well prepared for exams and tests of that particular chapter. Apart from reference books and notes, Question Banks are very effective study materials for exam preparation. When a student tries to attempt and solve all the important questions of any particular subject , it becomes very easy to gauge how much well the topics have been understood and what kind of questions are asked in exams related to that chapter.. Some of the other advantaging factors of Question Banks are as follows

• Since Important questions included in question bank are collections of questions that were asked in previous exams and tests thus when a student tries to attempt them they get a complete idea about what type of questions are usually asked and whether they have learned the topics well enough. This gives them an edge to prepare well for the exam.Students get the clear idea whether the questions framed from any particular chapter are mostly either short or long answer type questions or multiple choice based and also marks weightage of any particular chapter in final exams.
• CBSE Question Banks are great tools to help in analysis for Exams. As it has a collection of important questions that were asked previously in exams thereby it covers every question from most of the important topics. Thus solving questions from the question bank helps students in analysing their preparation levels for the exam. However the practice should be done in a way that first the set of questions on any particular chapter are solved and then solutions should be consulted to get an analysis of their strong and weak points. This ensures that they are more clear about what to answer and what can be avoided on the day of the exam.
• Solving a lot of different types of important questions gives students a clear idea of what are the main important topics of any particular chapter that needs to focussed on from examination perspective and should be emphasised on for revision before attempting the final paper. So attempting most frequently asked questions and important questions helps students to prepare well for almost everything in that subject.
• Although students cover up all the chapters included in the course syllabus by the end of the session, sometimes revision becomes a time consuming and difficult process. Thus, practicing important questions from Question Bank allows students to check the preparation status of each and every small topic in a chapter. Doing that ensures quick and easy insight into all the important questions and topics in each and every individual. Solving the important questions also acts as the revision process.

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CBSE Case Study Questions Class 9 Maths Chapter 2 Polynomials PDF Download

CBSE Case Study Questions Class 9 Maths Chapter 2 Polynomials PDF Download  are very important to solve for your exam. Class 9 Maths Chapter 2 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Case Study Questions Class 9 Maths Chapter 2 Polynomials

• Checkout: Class 9 Science Case Study Questions
• Checkout: Class 9 Maths Case Study Questions

Polynomials Case Study Questions With Answers

Case study questions class 9 maths chapter 2.

Case Study/Passage-Based Questions

Case Study 1. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p(x) = 4x 2 + 12x + 5, which is the product of their individual shares.

Coefficient of x 2 in the given polynomial is (a) 2 (b) 3 (c) 4 (d) 12

Answer: (c) 4

Total amount invested by both, if x = 1000 is (a) 301506 (b)370561 (c) 4012005 (d)490621

Answer: (c) 4012005

The shares of Ankur and Ranjan invested individually are (a) (2x + 1),(2x + 5)(b) (2x + 3),(x + 1) (c) (x + 1),(x + 3) (d) None of these

Answer: (a) (2x + 1),(2x + 5)

Name the polynomial of amounts invested by each partner. (a) Cubic (b) Quadratic (c) Linear (d) None of these

Answer: (c) Linear

Find the value of x, if the total amount invested is equal to 0. (a) –1/2 (b) –5/2 (c) Both (a) and (b) (d) None of these

Answer: (c) Both (a) and (b)

Case Study 2. One day, the principal of a particular school visited the classroom. The class teacher was teaching the concept of a polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked various questions to students. Some of them are given below. Answer them

Which one of the following is not a polynomial? (a) 4x 2 + 2x – 1 (b) y+3/y (c) x 3 – 1 (d) y 2 + 5y + 1

Answer: (b) y+3/y

The polynomial of the type ax 2 + bx + c, a = 0 is called (a) Linear polynomial (b) Quadratic polynomial (c) Cubic polynomial (d) Biquadratic polynomial

Answer: (a) Linear polynomial

The value of k, if (x – 1) is a factor of 4x 3 + 3x 2 – 4x + k, is (a) 1 (b) –2 (c) –3 (d) 3

Answer: (c) –3

If x + 2 is the factor of x 3 – 2ax 2 + 16, then value of a is (a) –7 (b) 1 (c) –1 (d) 7

Answer: (b) 1

The number of zeroes of the polynomial x 2 + 4x + 2 is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (b) 2

Case Study 3. Amit and Rahul are friends who love collecting stamps. They decide to start a stamp collection club and contribute funds to purchase new stamps. They both invest a certain amount of money in the club. Let’s represent Amit’s investment by the polynomial A(x) = 3x^2 + 2x + 1 and Rahul’s investment by the polynomial R(x) = 2x^2 – 5x + 3. The sum of their investments is represented by the polynomial S(x), which is the sum of A(x) and R(x).

Q1. What is the coefficient of x^2 in Amit’s investment polynomial A(x)? (a) 3 (b) 2 (c) 1 (d) 0

Answer: (a) 3

Q2. What is the constant term in Rahul’s investment polynomial R(x)? (a) 2 (b) -5 (c) 3 (d) 0

Answer: (c) 6

Q3. What is the degree of the polynomial S(x), representing the sum of their investments? (a) 4 (b) 3 (c) 2 (d) 1

Answer: (c) 2

Q4. What is the coefficient of x in the polynomial S(x)? (a) 7 (b) -3 (c) 0 (d) 5

Answer: (b) -3

Q5. What is the sum of their investments, represented by the polynomial S(x)? (a) 5x^2 + 7x + 4 (b) 5x^2 – 3x + 4 (c) 5x^2 – 3x + 5 (d) 5x^2 + 7x + 5

Answer: (b) 5x^2 – 3x + 4

Case Study 4. A school is organizing a fundraising event to support a local charity. The students are divided into three groups: Group A, Group B, and Group C. Each group is responsible for collecting donations from different areas of the town.

Group A consists of 30 students and each student is expected to collect ‘x’ amount of money. The polynomial representing the total amount collected by Group A is given as A(x) = 2x^2 + 5x + 10.

Group B consists of 20 students and each student is expected to collect ‘y’ amount of money. The polynomial representing the total amount collected by Group B is given as B(y) = 3y^2 – 4y + 7.

Group C consists of 40 students and each student is expected to collect ‘z’ amount of money. The polynomial representing the total amount collected by Group C is given as C(z) = 4z^2 + 3z – 2.

Q1. What is the coefficient of x in the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 0

Answer: (b) 5

Q2. What is the degree of the polynomial B(y)? (a) 2 (b) 3 (c) 4 (d) 1

Answer: (b) 3

Q3. What is the constant term in the polynomial C(z)? (a) 4 (b) 3 (c) -2 (d) 0

Answer: (c) -2

Q4. What is the sum of the coefficients of the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 17

Answer: (c) 10

Q5. What is the total number of students in all three groups combined? (a) 30 (b) 20 (c) 40 (d) 90

Answer: (c) 40

Hope the information shed above regarding Case Study and Passage Based Questions for Case Study Questions Class 9 Maths Chapter 2 Polynomials with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Maths Polynomials Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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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.

The equation √2x-1 can be written as 0x 2 +√2x-1 [Since 0x 2 is 0]

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

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

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

We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 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

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

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

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

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

(iii) 6x 2 +5x-6 

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

We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -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

We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -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)

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

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+y) 2 –3xy]

Taking (x+y) common ⇒ x 3 +y 3 = (x+y)[(x 2 +y 2 +2xy)–3xy]

⇒ 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–y) 2 +3xy]

Taking (x+y) common ⇒ x 3 −y 3 = (x–y)[(x 2 +y 2 –2xy)+3xy]

⇒ 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)[(3y) 2 –(3y)(5z)+(5z) 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)[(4m) 2 +(4m)(7n)+(7n) 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)[(3x) 2 +y 2 +z 2 –3xy–yz–3xz]

= (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)

⇒ x 3 +y 3 +z 3 –3xyz = (1/2)(x+y+z)[2(x 2 +y 2 +z 2 –xy–yz–xz)]

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

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

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

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

We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20 = 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

We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=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[y(3y+5)–1(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 Class 9 Maths Chapter 2 Polynomials

NCERT solutions for class 9 maths Chapter 2 Polynomials are all about the basics of polynomials like the different types of polynomials, finding roots, or solutions to a polynomial equation. Polynomials are algebraic expressions having one variable or more. These NCERT solutions class 9 maths Chapter 2 also explain the remainder theorem and factor theory of polynomials in detail, the algebraic identities, and polynomials of various degrees.

Class 9 Maths NCERT Solutions Chapter 2 polynomials illustrate the difference between linear, quadratic, and cubic polynomials. Important theorems mentioned are the Remainder theorem and the Factor theorem, which help identify the factors of a polynomial. Students can access the solutions from the pdf links given below and also find some of these in the exercises given below.

• NCERT Solutions Class 9 Maths Chapter 2 Ex 2.1
• NCERT Solutions Class 9 Maths Chapter 2 Ex 2.2
• NCERT Solutions Class 9 Maths Chapter 2 Ex 2.3
• NCERT Solutions Class 9 Maths Chapter 2 Ex 2.4
• NCERT Solutions Class 9 Maths Chapter 2 Ex 2.5

NCERT Solutions for Class 9 Maths Chapter 2 PDF

The exercises related to identifying the type of polynomial, finding the roots or solution of a polynomial equation, and finding factors of the polynomial are available for free pdf download using the four links provided below:

☛ Download Class 9 Maths NCERT Solutions Chapter 2 Polynomials

NCERT Class 9 Maths Chapter 2   Download PDF

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

These fundamental properties and theorems of polynomials form the building blocks for higher mathematics. Thus, it is very important to master the fundamentals by solving many different example exercises using the links provided above. These NCERT Solution exercises will help understand the properties of polynomials better, as well as how to utilize them. Chapter-wise detailed analysis of NCERT Solutions Class 9 Maths Chapter 2 Polynomials is given below.

• Class 9 Maths Chapter 2 Ex 2.1 - 21 Questions
• Class 9 Maths Chapter 2 Ex 2.2 - 22 Questions
• Class 9 Maths Chapter 2 Ex 2.3 - 7 Questions
• Class 9 Maths Chapter 2 Ex 2.4 - 7 Questions
• Class 9 Maths Chapter 2 Ex 2.5 - 41 Questions

☛ Download Class 9 Maths Chapter 2 NCERT Book

Topics Covered: The topics that are covered under the chapter on polynomials include an explanation of polynomials as a special set of algebraic equations, different types of polynomials, solutions of polynomial equations, factor theorem, and remainder theorem. Also, these class 9 maths NCERT solutions Chapter 2 define the algebraic identities , which help in factorizing the algebraic equations.

Total Questions: Class 9 Maths Chapter 2 Polynomials consists of a total of 45 questions, of which 31 are easy, 9 are moderate, and 5 are long answer type questions.

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List of Formulas in NCERT Solutions Class 9 Maths Chapter 2

NCERT solutions class 9 maths Chapter 2 covers lots of important concepts crucial for understanding higher grade maths. By learning to factorize a polynomial expression, one can find the roots of the polynomial equation. This is a relatively simple process that can greatly improve an individual's understanding of polynomial equations. Some important algebraic identities or formulas which help in factorization and are covered in NCERT solutions for class 9 maths chapter 2 are given below.

• ( x + y ) 2 = x 2 + 2xy + y 2
• ( x + y ) 3 = x 3 + y 3 + 3xy (x+y)

Important Questions for Class 9 Maths NCERT Solutions Chapter 2

Video solutions for class 9 maths ncert chapter 2, faqs on ncert solutions class 9 maths chapter 2, how cbse students can utilize ncert solutions class 9 maths chapter 2 effectively.

Algebra forms the basis of higher mathematical studies. Hence, students should focus on the important terms defined in this chapter, like the degree of a polynomial, the difference between constant and variable, to get a clear understanding of the polynomials. This will help them to make their base strong to appear for their board exams and face any kind of difficult questions.

Why are Class 9 Maths NCERT Solutions Chapter 2 Important?

The NCERT Solutions Class 9 Maths Chapter 2 includes a detailed explanation of the remainder and the factor theorem, which hold an important place in algebra. Also, the crucial algebraic identities are discussed in an elaborate manner with plenty of questions to solve for the students. A list of all key equations and concepts is available at the end of the chapter. This is a significant benefit because students can use this list whenever required instead of figuring it out from between the lengthy chapter text. Overall, these solutions cover all of the major concepts, approaches, and formulas, making them of utmost importance for class 9 math students.

How Many Questions are there in NCERT Solutions Class 9 Maths Chapter 2 Polynomials?

Overall the NCERT Solutions Class 9 Maths Chapter 2 has 98 questions that can be categorized as easy, medium, and difficult ones. Roughly 70 questions are straightforward and easy to solve, 20 questions are of medium difficulty level while 8 would require some thinking as they are long-form questions.

What are the Important Topics Covered in NCERT Solutions Class 9 Maths Chapter 2?

The important topics that are covered under the NCERT Solutions Class 9 Maths Chapter 2 include the basic understanding of polynomials, the components of algebraic expressions, and their definitions. The chapter focuses on the types of polynomials and how to solve them, with special emphasis on factor and remainder theorem and the algebraic entities.

What are the Important Formulas in NCERT Solutions Class 9 Maths Chapter 2?

Since the NCERT Solutions Class 9 Maths Chapter 2 covers the polynomials from their basic structure, several definitions of important terms have been explained with their formulas, like the factor and the remainder theorem. But the most important formula would be the algebraic identities as they help in factorization itself. For example, (a + b) 2 = a 2 + 2ab + b 2

Do I Need to Practice all Questions Provided in NCERT Solutions Class 9 Maths Polynomials?

NCERT Solutions Class 9 Maths Polynomials encompass a variety of questions that explore all the algebraic concepts related to polynomials. Hence, it would be good if the students make use of this resource and start practicing by solving the examples first, which will help them in getting an idea of what steps are to be followed when questions related to polynomials are solved.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Ncert solutions for class 9 maths chapter 2 polynomials| pdf download.

• Exercise 2.1 Chapter 2 Class 9 Maths NCERT Solutions
• Exercise 2.2 Chapter 2 Class 9 Maths NCERT Solutions
• Exercise 2.3 Chapter 2 Class 9 Maths NCERT Solutions
• Exercise 2.4 Chapter 2 Class 9 Maths NCERT Solutions
• Exercise 2.5 Chapter 2 Class 9 Maths NCERT Solutions

NCERT Solutions for Class 9 Maths Chapters:

How can i download chapter 2 polynomials class 9 ncert solutions pdf , what do you mean by algebraic expressions, what is the coefficient of a zero polynomial, what is biquadratic polynomial, contact form.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 2 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

Class 9 Maths Chapter 2 Polynomials NCERT Solutions

Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.1

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.5

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials – Topic Discussion

Below we have listed the topics that have been discussed in this chapter. As this is one of the important topics in maths, It comes under the unit – Algebra which has a weightage of 20 marks in class 9 maths board exams.

• Polynomials in one Variable – Discussion of Linear, Quadratic and Cubic Polynomial.
• Zeroes of a Polynomial
• Real Numbers and their Decimal Expansions
• Representing Real Numbers on the Number Line Operations on Real Numbers
• Laws of Exponents for Real Numbers.

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