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Case Study Questions for Class 9 Maths Chapter 2 Polynomials

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Case Study Questions

Question 1:

On one day, principal of a particular school visited the classroom. Class teacher was teaching the concept of 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.

(i) 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

(ii) 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

(iii) 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

(iv) 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

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

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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

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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)

Mycbseguide: blessing in disguise.

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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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.

CBSE Class 9 All Students can also Download here Class 9 Other Study Materials in PDF Format.

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• CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF 10 February, 2023, 6:20 pm
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• CBSE Class 9th English 2023 : Chapter-wise Competency-Based Test Items with Answer; Download PDF 21 December, 2022, 5:16 pm
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CBSE Important Questions for Class 9 Maths are available in Printable format for Free Download.Here you may find NCERT Important Questions and Extra Questions for Class 9 Mathematics chapter wise with answers also. These questions will act as chapter wise test papers for Class 9 Mathematics. These Important Questions for Class 9 Mathematics are as per latest NCERT and CBSE Pattern syllabus and assure great success in achieving high score in Board Examinations

Total Papers :

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

For Preparation of exams students can also check out other resource material

CBSE Class 9 Maths Sample Papers

CBSE Class 9 Maths Worksheets

CBSE Class 9 Maths Question Papers

CBSE Class 9 Maths Test Papers

CBSE Class 9 Maths Revision Notes

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.

Question Bank of Other Classes

To Prepare better for CBSE paperclass; ?> " title="Download Free CBSE Papers">Ribblu.com brings to you all the previous years papers & worksheets of subject; ?//> for CBSE paperclass; ?>. This CBSE paper and worksheet can be instrumental in students achieving maximum marks in their exams. These Papers and worksheets help students gain confidence and make them ready to face their school examinations. These Papers and worksheets school wise, covers important concepts from an examination perspective. Students and parents can download all the available papers & worksheets directly in the form of PDF. One can use these papers and worksheets to get extensive practice and familiarise themselves with the format of the question paper.

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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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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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Chapter 2 Polynomials Class 9 Maths Important Questions | Free PDF for CBSE 2024-25 Exams

Many important questions are framed in final examinations from Chapter 2 of Mathematics. In this chapter, you will practice various topics like polynomials in one variable, zeros of polynomials, remainder theorem, factorization, and algebraic identities. The PDFs will also cover Class 9 Maths Chapter 2 Most Important Questions for extra help.

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Unit 2: Polynomials

Polynomials 2.1.

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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

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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

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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.

Cuemath is one of the world's leading math learning platforms that offers LIVE 1-to-1 online math classes for grades K-12 . Our mission is to transform the way children learn math, to help them excel in school and competitive exams. Our expert tutors conduct 2 or more live classes per week, at a pace that matches the child's learning needs.

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.

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Polynomials Class 9 Mathematics Notes And Questions

Please refer to Polynomials Class 9 Mathematics notes and questions with solutions below. These revision notes and important examination questions have been prepared based on the latest Mathematics books for  Class 9.  You can go through the questions and solutions below which will help you to get better marks in your examinations.

Class 9 Mathematics Polynomials Notes and Questions

Concept of Polynomials Polynomials

Consider this situation involving trains.

The speed of an express train is ten less than twice that of a passenger train. If each travels for as many hours as its speed, then what is the difference between the distances travelled by them?

Let the speed of the passenger train be x km/hr. Then, travelling time of the train = x hours Distance travelled by it = Speed × Time = x × x = x2 km Now, speed of the express train = (2x − 10) km/hr Its travelling time = (2x − 10) hours Distance travelled by it = (2x − 10) (2x − 10) = (4×2 − 40x + 100) km Thus, required difference = 4×2 − 40x + 100 − x2 = (3×2 − 40x + 100) km

The expression 3×2 − 40x + 100 is an example of a polynomial. Different real-life problems such as the one given above can be expressed in the form of polynomials. Go through this lesson to familiarize yourself with these useful expressions.

Topics to be covered in this lesson:

♦ Identifying polynomials ♦ Constant polynomials ♦ Classification of polynomials according to the number of terms

♦ Did You Know? ♦ Ancient Babylonians developed a unique system to calculate things using formulae. These formulae consisted of letters, mathematical operators (+, −, ×, ÷ ) and numbers. It was this system that led to the development of algebra. The word ‘algebra’ is derived from the Arabic word ‘al-jabr’ meaning ‘the reunion of broken parts’. Another Arabian connection with algebra is the Arab mathematician Muhammad ibn Musa al-Khwarizmi, whose theories greatly influenced this branch of mathematics.

♦ Did You Know? The word ‘polynomial’ is a combination of the Greek words ‘poly’ meaning ‘many’ and ‘nomos’ meaning ‘part or portion’. Thus, a polynomial is an algebraic expression having many parts.

Different Forms of a Polynomial

♦ A polynomial can found and written in different forms. These forms are explained below.

♦ Standard form: If the terms of a polynomial are written in descending order or ascending order of the powers of the variables then the polynomial is said to be in the standard form.

♦ For example, the polynomial 3x + 15×4 − 1 − 13×2 is not in the standard form. It can be written in the standard form as 15×4 − 13×2 + 3x − 1 or −1 + 3x − 13×2+ 15×4.

♦ Index form: Observe the polynomial x6 − 2×4 − 10×3 + 5. In this polynomial, terms having x5, x2 and x are missing. These terms can be added to the polynomial with coefficient 0. Thus, the obtained polynomial will be x6 + 0x5− 2×4 − 10×3 + 0x2 + 0x + 5.

♦ The polynomial obtained on adding the missing terms is said to be in the index form.

♦ Coefficient form: When the coefficients of all the terms of a polynomial are written in a bracket by separating with comma then the polynomial is said to be written in the coefficient form.

♦ It should be noted that if a term is missing then its coefficient is taken as 0. So, it is better to write the given polynomial in the index form before wrting it in the coefficient form.

For example, to write the polynomial x6 − 2×4 − 10×3 + 5 in the coefficient form, we will first write it in the index form as x6 + 0x5− 2×4 − 10×3 + 0x2 + 0x + 5. Now, it can be written in the coefficient form as (1, 0, −2, −10, 0, 0, 5).

Degree of Polynomial More about Polynomials

We know that a polynomial comprises a number of terms, which may have variables or numbers or both. Also, each term can be represented with a variable having some exponent . Exponents of the variables in a given polynomial can be the same or different. Let us consider a polynomial 2×5 + 4×2 + 9. The terms of this polynomial and their exponents are as follows: First term = 2×5; exponent in the first term = 5 Second term = 4×2; exponent in the second term = 2 Third term = 9 = 9×0; exponent in the third term = 0

Note that all the exponents in the above polynomial are different. These exponents help us to identify the degrees of polynomials. Polynomials are categorized based on their degrees.

In this lesson, we will learn about the degrees of polynomials and the classification of polynomials based on the same.

Whiz Kid When a polynomial has an equals sign (=), then it becomes an equation. The maximum number of solutions of an equation is less than or equal to the degree of that equation.

The Degree of a Polynomial in more than one Variable

In case of the polynomials in one variable, the degree of a polynomial is the highest exponent of the variable in the polynomial, but what about the degree of the polynomial in more than one variable?

In this case, the sum of the powers of all variables in each term is obtained and the highest sum among all is the degree of the polynomial.

For example, find the degree of the polynomial 2xy + 3y2z + 4x2yz2 – xyz – 2×3. Let us find the sum of the powers of all variables in each term of this polynomial. Sum of the powers of all variables in the term 2xy = 1 + 1 = 2 Sum of the powers of all variables in the term 3y2z = 2 + 1 = 3 Sum of the powers of all variables in the term 4x2yz2 = 2 + 1 + 2 = 5 Sum of the powers of all variables in the term –xyz = 1 + 1 + 1 = 3 Sum of the powers of all variables in the term –2×3 = 3 Among all the sums, 5 is the highest and thus, the degree of the polynomial 2xy + 3y2z + 4x2yz2 – xyz – 2×3 is 5. Similarly, we can find the degree of any polynomial in more than one variable.

Whiz Kid If all the terms in a polynomial have the same exponent, then the expression is referred to as a homogenous polynomial.

Did You Know? The graphs oflinear polynomials are always straight lines. This is why these polynomials are called ‘linear’ polynomials.

Values of Polynomials at Different Points Value of a Polynomial at Different Points

What do you observe when a ball is dropped from a height? It bounces again and again until it comes to rest after some time. Also, the height to which the ball bounces keeps decreasing and then becomes zero.

Suppose a ball dropped from a certain height h bounces to two-third of that height. If this height of bounce is H, then we can say that H = 2/3 h. In this equality, the value of H changes when the value of h undergoes a change, i.e., the value of H depends upon the value of h.

Polynomials can be described in a similar manner. When the value of the variable in a polynomial changes, the value of the polynomial also undergoes a change. This means that the value of a polynomial depends upon the value of the variable present in it.

In this lesson, we will learn to find the value of a given polynomial at different points.

Finding the Value of a Polynomial at Different Points

Consider the polynomial, p(x) = x2 − 4x + 5. The variable here is x. Hence, for the different values of the variable x, we get different values of the polynomial p(x). Let us find the value of this polynomial at x = 2. We will do so by replacing x with 2 in the given polynomial. p(2) = 22 − 4 × 2 + 5 ⇒ p(2) = 4 − 8 + 5 ⇒ ∴ p(2) = 1 So, the value of the polynomial is 1 when x = 2. Now, let us see what happens on replacing x by −3 in the given polynomial. p(−3) = (−3)2 − 4 (−3) + 5 ⇒ p(−3) = 9 + 12 + 5 ⇒ ∴ p(−3) = 26 The value of the polynomial is different this time. We can see that the value of the polynomial is 26 when x = −3. This shows that the value of the polynomial changes with the change in the value of the variable in it. Similarly, we can find the value of any polynomial for any value of the variable involved.

Did You Know? ♦ The sum of the coefficients of a polynomial p(x) is equal to p(1). For example, let us consider a polynomial, p(x) = x2 − 4x + 5. Here, sum of the coefficients of p(x) = p(1) = 12 − 4(1) + 5 = 2

♦ The constant coefficient of a polynomial p(x) is equal to p(0). Let us consider the same polynomial as above. Here, constant coefficient of p(x) = p(0) = 02 − 4(0) + 5 = 5

Whiz Kid The plotting of the consecutive values of the variable and the polynomial on the coordinate plane gives a curve. Thus, each polynomial represents a curve.

Zeroes of Polynomials Zero Value of a Polynomial

We know that the value of a polynomial differs according to the value of the variable in it. For example, if p(x) is a polynomial with variable x and we put different values of x in p(x), then we will get different values of p(x). In some cases, the value of p(x) can be the same for two or more values of x. Also, in a few cases, the value of p(x) can be zero.

The values of the variable at which a polynomial becomes zero are called zeroes of the polynomial. These values are very special and useful to us. Zeroes of polynomials are used in:

Solving problems related to motion .Solving problems related to path or Focus of a point or geometric figure ♦ Making graphs for economic data In this lesson, we will learn to check whether or not the given values are zeroes of the given polynomials. We will also learn to find the zeroes of different polynomials.

Zeroes or Roots of a Polynomial If the value of a polynomial p(x) at x = a is zero, then a is said to be the zero or root of the polynomial p(x). Let us consider the polynomial p(x) = x2 − 5x + 6 and check its values at x = 1, 2 …. p(1) = 12 − 5 × 1 + 6 = 1 − 5 + 6 = 2 p(2) = 22 − 5 × 2 + 6 = 4 − 10 + 6 = 0 p(3) = 32 − 5 × 3 + 6 = 9 − 15 + 6 = 0 Clearly, p(2) and p(3) are equal to 0; so, x = 2 and x = 3 are the zeroes or roots of the given polynomial.

The value of a constant polynomial can never be zero. Hence, a constant polynomial has no zeroes or roots. For example, p(x) = 8 is a constant polynomial. Let us try to find the roots of this polynomial. On replacing x with any number, we will always get 8. Suppose we replace x with 2. Then, p(2) will still be equal to 8. This will be the case for any value of x.

Did You Know? ♦ The maximum number of roots of a polynomial is less than or equal to the degree of the polynomial. For example, the polynomial x3 − 2x + 10 has a degree 3; so, the number of roots of this polynomial will be 3, 2 or 1. ♦ Every non-constant polynomial has at least one root. ♦ A polynomial can have more than one root.

Whiz Kid The roots of quadratic polynomials of the form ax2 + bx + c can be found by the following formula. x = -b ± √b2 – 4ac/2a For example, the root of x2 + 2x + 1 can be found as follows: x = -2 ± √22 – 4(1)(1)/2(1) = -2± √4-4/2 = -2± √0/2 = -2/2 = -1

Division of Polynomials by Polynomials (Degree 1) Using Long Division Method

Dividing one number by another is something that we know well. For example, let us divide 434 by 9.

In the above division, 434 is the dividend , 9 is the divisor , 48 is the quotient and 2 is the remainder We also know how to represent any division using the division algorithm, which states that:

Dividend = Divisor × Quotient + Remainder Thus, we can write 434 as: 434 = 9 × 48 + 2 We can divide one polynomial by another in the same way as we divide one number by another. In this lesson, we will learn to carry out the division of polynomials and verify the same using the division algorithm. Division Algorithm Polynomials also satisfy the division algorithm. Consider the division of 2×2 − 9x + 4 by x − 2. In this division, we have Dividend = 2×2 − 9x + 4 Divisor = x − 2 Quotient = 2x − 5 Remainder = −6 Now, Divisor × Quotient + Remainder = [(x − 2) (2x − 5)] + (−6) = [x (2x − 5) − 2 (2x − 5)] − 6 = 2×2 − 5x − 4x +10 − 6 = 2×2 − 9x + 4 = Dividend Thus, the given division satisfies the division algorithm, i.e,

Remainder Theorem and Its Application Remainder Theorem

Consider two polynomials p(x) and q(x), where p(x) = 5×4 − 4×2 − 50 and q(x) = x − 2. We know how to divide p(x) by q(x) using the long division method. The result of this division will give the quotient as 5×3 + 10×2 +16x + 32 and the remainder as 14.

The long division method of finding the remainder is quite tedious. There is a simpler way to find the above remainder. This method is generalized in the form of a theorem called the remainder theorem. This theorem helps us find the remainder when a polynomial is to be divided by a linear polynomial.

In this lesson, we will study the remainder theorem and some of its applications in the form of examples.

Understanding the Remainder Theorem

Consider the division of a polynomial p(x) by a polynomial q(x), where p(x) = 5×4 − 4×2 − 50 and q(x) = x − 2. In this case, we have: Dividend = p(x) and divisor = q(x) On dividing p(x) by q(x) using the long division method, we get: Quotient = 5×3 + 10×2 +16x + 32 and remainder = 14 Now, let us find the value of p(x) at x = 2. p(2) = 5 × 24 − 4 × 22 − 50 = 5 × 16 − 4 × 4 − 50 = 80 − 16 − 50 = 14 Note how the value of p(2) is the same as the remainder obtained by the long division of p(x) by q(x). Also observe how x = 2 is a zero of the polynomial q(x).

Thus, if we replace x in the dividend with the zero (or root) of the divisor, then we get the remainder. This method of finding the remainder is called the remainder theorem. It can be stated as follows: For a polynomial p(x) of a degree greater than or equal to 1 and for any real number a, if p(x) is divided by a linear polynomial x − a, then the remainder will be p(a). Proof of the Remainder Theorem

For a polynomial p(x) of a degree greater than or equal to 1 and for any real number a, if p(x) is divided by a linear polynomial x − a, then the remainder will be p(a).

Proof Let p(x) be a polynomial of a degree greater than or equal to 1 and a be any real number. When divided by x − a, let p(x) leave the remainder r(x). Let q(x) be the quotient obtained. Then, p(x) = (x − a) q(x) + r(x), where r(x) = 0 or degree r(x) < degree (x − a) Now, x − a is a polynomial of degree 1; so, either r(x) = 0 or r(x) = constant (since a polynomial of degree less than 1 is a constant). Let r(x) = constant = r (say). Then, p(x) = (x − a) q(x) + r On putting x = a, we get p(a) = (a − a) q(a) + r = 0 × q(a) + r = r Thus, if p(x) is divided by x − a, then the remainder will be p(a).

Notes: 1) If p(x) is divided by x + a, then the remainder will be p(−a). 2) If p(x) is divided by ax − b, then the remainder will be p(b/a) 3) If p(x) is divided by ax + b, then the remainder will be p(-b/a)

Thus, when a = −9 and b = 24, the divisions of x3 + ax2 + bx − 20 by x − 5 and x −3 leave 0 and −2 respectively as the remainders.

Factor Theorem and Its Applications

Factor Theorem

We know the relation between a number and its factor. If we divide 91 by 7, then we get 13 as the quotient and zero as the remainder. In this case, we say that 7 is a factor of 91 as the remainder is zero. Now, if we divide 107 by 9, then we get 11 as the quotient and 8 as the remainder. In this case, we say that 9 is not a factor of 107 as the remainder is not zero. Thus, the relation between a number and its factor is given as follows:

If a number is completely divisible by another number, i.e., the remainder is zero,  then the second number is a factor of the first number.

Similarly, a polynomial p(x) is said to be completely divisible by a polynomial q(x) if we get zero as the remainder on dividing p(x) by q(x). In this case, we say that q(x) is a factor of p(x).

We have studied the remainder theorem that helps us to find the remainder. Similarly, we have a factor theorem that helps us to determine whether or not a polynomial is a factor of another polynomial, without actually performing the division. In this lesson, we will study the factor theorem and solve some problems based on it.

Understanding the Factor Theorem

We can easily determine whether a polynomial q(x) is a factor of a polynomial p(x) without performing the division. This can be done by using the factor theorem, which can be stated as follows:

For a polynomial p(x) of a degree greater than or equal to 1 and for any real number c, i) if p(c) = 0, then x − c will be a factor of p(x) and ii) if x − c is a factor of p(x), then p(c) will be equal to zero. Consider the polynomial, p(x) = x2 − 3x + 2. On putting x = 2 in p(x), we get: p(2) = 22 − 3 × 2 + 2 = 4 − 6 + 2 = 0 Thus, we can say that x − 2 is a factor of p(x), where 2 is a real number.

Proof of the Factor Theorem Statement For a polynomial p(x) of a degree greater than or equal to 1 and for any real number c, i) if p(c) = 0, then x − c will be a factor of p(x) and ii) if x − c is a factor of p(x), then p(c) will be equal to zero.

Proof Let p(x) be a polynomial of a degree greater than or equal to 1 and c be any real number such that p(c) = 0. Let quotient q(x) be obtained when p(x) is divided by x − c. i) p(c) = 0 By the remainder theorem, the remainder obtained is p(c). ⇒ p(x) = (x − c) q(x) + p(c) ⇒ p(x) = (x − c) q(x) [∵ p(c) = 0] ⇒ x − c is a factor of p(x). ii) x − c is a factor of p(x) ⇒ When divided by x − c, p(x) leaves zero as the remainder. However, by the remainder theorem, the remainder obtained is p(c). ⇒ p(c) = 0

Notes 1) x + c will be a factor of p(x) if p(−c) = 0 2) cx − d will be a factor of p(x) if p(d/c) = 0 3) cx + d will be a factor of p(x) if p(-d/c) = 0 4) (x − c) (x − d) will be a factor of p(x) if p(c) = 0 and p(d) = 0

Factorisation of Quadratic Polynomials Using Factor Theorem and Splitting Middle Term

Factorisation of Quadratic Polynomials

We know that 7 × 6 = 42. Here, 7 and 6 are factors of 42. Now, consider the linear polynomials x − 2 and x + 1. On multiplying the two, we get: x (x + 1) − 2 (x + 1) = x2 + x − 2x − 2 = x2 − x − 2, which is a quadratic polynomial. So, x − 2 and x + 1 are factors of the quadratic polynomial x2 − x − 2. A quadratic polynomial can have a maximum of two factors.

In the above example, we found the quadratic polynomial from its two factors. We can also find the factors from the quadratic polynomial. This process of decomposing a polynomial into a product of its factors (which when multiplied give the original expression) is called factorisation.

There are two ways of finding the factors of quadratic polynomials viz., by applying the factor theorem and by splitting the middle term. We will discuss these methods of factorisation in this lesson and also solve some examples based on them.

Factorisation of Quadratic Polynomials Using the Factor Theorem

The factor theorem states that: For a polynomial p(x) of a degree greater than or equal to 1 and for any real number a, if p(a) = 0, then x − a will be a factor of p(x).

Consider the quadratic polynomial, p(x) = x2 − 5x + 6. To find its factors, we need to ascertain the value of x for which the value of the polynomial comes out to be zero. For this, we first determine the factors of the constant term in the polynomial, and then check the value of the polynomial at these points.

In the given polynomial, the constant term is 6 and its factors are ±1, ±2, ±3 and ±6.

Let us now check the value of the polynomial for each of these factors of 6. p(1) = 12 − 5 × 1 + 6 = 1 − 5 + 6 = 2 ≠ 0 Hence, x − 1 is not a factor of p(x). p(2) = 22 − 5 × 2 + 6 = 4 − 10 + 6 = 0 Hence, x − 2 is a factor of p(x). p(3) = 32 − 5 × 3 + 6 = 9 − 15 + 6 = 0 Hence, x − 3 is also a factor of p(x). We know that a quadratic polynomial can have a maximum two factors which are already obtained as: (x − 2) and (x − 3). Thus, the given polynomial = p(x) = x2 − 5x + 6 = (x − 2) (x − 3)

Factorisation of Cubic Polynomial Using Factor Theorem Factorization of Cubic Polynomials

A cubic polynomial can be written as p(x) = ax3 + bx2 + cx + d, where a, b, c and d are real numbers. We cannot factorize a cubic polynomial in the manner in which we factorize a quadratic polynomial. We use a different approach for this purpose.

A cubic polynomial can have a maximum of three linear factors. By knowing one of these factors, we can reduce it to a quadratic polynomial. Thus, to factorize a cubic polynomial, we first find a factor by the hit and trial method or by using the factor theorem, and then reduce the cubic polynomial into a quadratic polynomial. The resultant quadratic polynomial is solved by splitting its middle term or by using the factor theorem.

In this lesson, we will learn how to factorize a cubic polynomial and solve some examples related to the same.

Know More Hit and trial method

Hit and trial method is used to find the factors or roots of a polynomial of degree more than two. In this method, we put some value in the given polynomial to see if it satisfies the polynomial. If it does, then it is the zero of that polynomial. Using this method, we can reduce a polynomial of degree, say n, to a polynomial of degree n − 1.

Using Identity for Square of Sum of Three Terms Algebraic Identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

When we solve an algebraic equation, we get the values of the variables present in it. When an algebraic equation is valid for all values of its variables, it is called an algebraic identity.

So, an algebraic identity is a relation that holds true for all possible values of its variables. We can use algebraic identities to expand, factorise and evaluate various algebraic expressions.

Many algebraic identities are used in mathematics. One such identity is(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx. In this lesson, we will study this identity and solve some examples based on it.

Proof of the Identity Let us prove the identity (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx We can write (x + y + z)2 as : (a + z) 2, where a = x + y = a2 + 2az + z2 [Using the identity (x + y) 2 = x2 + 2xy + y2] = (x + y) 2 + 2(x + y) z + z2 (Substituting the value of a) = x2 + 2xy + y2 + 2xz + 2yz + z2 (Using the identity (x + y) 2 = x2 + 2xy + y2) ∴ (x + y + z) 2 = x2 + y2 + z2 + 2xy + 2yz + 2zx The above identity holds true for all values of the variables present in it. Let us verify this by substituting random values for the variables x, y and z. If x = 2, y = 3 and z = 4, then: (2 + 3 + 4)2 = 22 + 32 + 42 + 2 × 2 × 3 + 2 × 3 × 4 + 2 × 4 × 2 ⇒ 92 = 4 + 9 + 16 + 12 + 24 + 16 ⇒ 81 = 81 ⇒ LHS = RHS Thus, we see that the identity holds true for random values of the variables present in it. Let us now use this identity to expand, factorise and evaluate various algebraic expressions.

Deriving Identity Geometrically

The identity (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx can also be derived with the help of geometrical construction. The steps of construction are as follows:

(1) Draw a square PQRS of side measuring (x + y + z) taking any convenient values of x, y and z.

(2) Mark two points A and B on side PQ such that l(PA) = x and l(AB) = y. Thus, l(BQ) = z. Also, mark two points H and G on side PS such that l(PH) = x and l(HG) = y. Thus, l(GS) = z.

(3) From points A and B, draw segments AF and BE parallel to side PS and intersecting RS at F and E respectively.

(4) From points H and G, draw segments HC and GD parallel to side PQ and intersecting QR at C and D respectively.

From the figure, it can be observed that Area of square PQRS = Sum of areas of squares PAIH, IJKL and KDRE + Sum of areas of rectangles ABJI, BQCJ, JCDK, HILG, LKEF and GLFS ⇒ (x + y + z)2 = (x2 + y2 + z2) + (xy + zx + yz + xy + yz + zx) ⇒ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Using Identities for Cube of Sum or Difference of Two Terms Algebraic Identities: (x + y)3 = x3 + y3 + 3xy (x + y) and (x − y)3 = x3 − y3 − 3xy (x − y)

Consider the number ‘999’. Suppose we have to calculate its cube. One way to find the cube is to multiply 999 by itself three times. However, this method is tedious and, therefore, prone to error.

Here is another way to solve the problem. Let us write 9993 as (1000 − 1)3. We have thus changed the number into the form (x − y)3. Now, the expansion of (x − y)3 will give the cube of 999. The required calculation will be easy since the values of x and y are simple numbers whose multiplication is also simple.

Thus, we see algebraic identities help make calculations simpler and less tedious. In this lesson, we will study the identities (x + y)3 = x3 + y3 + 3xy (x + y) and (x − y)3 = x3 − y3 − 3xy (x − y). We will also solve some examples based on them.

Understanding the Identities We have the two algebraic identities as follows: ♦ (x + y)3 = x3 + y3 + 3xy (x + y) OR (x + y)3 = x3 + y3 + 3x2y + 3xy2 ♦ (x − y)3 = x3 − y3 − 3xy (x − y) OR (x − y)3 = x3 − y3 − 3x2y + 3xy2 The above identities hold true for all values of the variables present in them. Let us verify this by substituting random values for the variables x and y in the first identity. If x = 2 and y = 3, then: (2 + 3)3 = 23 + 33 + 3 × 2 × 3 × (2 + 3) ⇒ 53 = 8 + 27 + 18 × 5 ⇒ 125 = 8 + 27 + 90 ⇒ 125 = 125 ⇒ LHS = RHS Thus, we see that the first identity holds true for random values of the variables present in it. We can prove the same for the second identity as well. Here are some other ways in which the two identities can be represented ♦ x3 + y3 = (x + y)3 − 3xy (x + y) OR x3 + y3 = (x + y) (x2 − xy + y2) ♦ x3 − y3 = (x − y)3 + 3xy (x − y) OR x3 − y3 = (x − y) (x2 + xy + y2)

Proof of the Identities: (x + y)3 = … and x3 + y3 = …

Let us prove the identity (x + y)3 = x3 + y3 + 3x2y + 3xy2 OR (x + y)3 = x3 + y3 + 3xy (x + y) We can write (x + y)3 as: (x + y) (x + y)2 = (x + y) (x2 + 2xy + y2) = x3 + 2x2y + 2xy2 + x2y + 2xy2 + y3 = x3 + y3 + 3x2y + 3xy2 ∴ (x + y)3 = x3 + y3 + 3x2y + 3xy2 ⇒ (x + y)3 = x3 + y3 + 3xy (x + y) … (1) Let us prove the identity x3 + y3 = (x + y)3 − 3xy (x + y) OR x3 + y3 = (x + y) (x2 − xy + y2) We can rewrite equation 1 as: x3 + y3 = (x + y)3 − 3xy (x + y) ⇒ x3 + y3 =(x + y) [(x + y)2 − 3xy] ⇒ x3 + y3 = (x + y) (x2 + 2xy + y2 − 3xy) ⇒ x3 + y3 = (x + y) (x2 − xy + y2)

Proof of the Identities: (x − y)3 = … and x3 − y3 = …

Let us prove the identity (x − y)3 = x3 − y3 − 3x2y + 3xy2 OR (x − y)3 = x3 − y3 − 3xy (x − y) We can write (x − y)3 as: (x − y) (x − y)2 = (x − y) (x2 − 2xy + y2) = x3 − 2x2y + xy2 − x2y + 2xy2 − y3 = x3 − y3 − 3x2y + 3xy2 ∴ (x−y)3 = x3−y3− 3x2y + 3xy2 ⇒ (x − y)3 = x3 − y3 − 3xy (x − y) … (1) Let us prove the identity x3 − y3 = (x − y)3 + 3xy (x − y) OR x3 − y3 = (x − y) (x2 + xy + y2) We can rewrite equation 1 as: x3 −y3 = (x−y)3 + 3xy (x−y) ⇒ x3 − y3= (x − y) [(x − y)2 + 3xy] ⇒ x3 − y3 = (x − y) (x2 − 2xy + y2 + 3xy) ⇒ x3 −y3 = (x−y) (x2 + xy + y2)

Alternate method

On using the identity p3 − q3 = (p − q) (p2 + pq + q2), where p = (2x + 5y) and q = (2x − 5y), we get: (2x + 5y)3 − (2x − 5y)3 = [(2x + 5y) − (2x − 5y)] [(2x + 5y)2 + (2x + 5y) (2x − 5y) + (2x − 5y)2] = (2x + 5y − 2x + 5y) (4×2 + 20xy + 25y2 + 4×2 + 10xy − 10xy − 25y2 + 4×2 − 20xy + 25y2) = 10y (12×2 + 25y2) = 120x2y + 250y3

Solving Problems Using the Identity (x + y + z) (x2 + y2 + z2 − xy − yz − zx) Algebraic Identity: x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx)

Algebraic identities help us solve problems with ease and in minimum time. Say, for example, we need to find the value of (−323 + 153 + 173). One may solve this problem by calculating the cube of each of the given numbers and then adding and subtracting the values so obtained. This method is easy in cases where we are dealing with small numbers. However, when big numbers are involved (as in the present case), this method proves to be tedious.

A simpler and less time-consuming way of solving the above problem is to use an appropriate algebraic identity. In the given expression, we find that −32 + 15 + 17 = 0. So, we need to state an identity under the condition x + y + z = 0.

In this lesson, we will focus on the identity x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) and its expansion under the condition x + y + z = 0. We will also solve examples based on the same.

Understanding the Identity

We have the algebraic identity as follows:

x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) Or x3 + y3 + z3 − 3xyz = 1/2 (x + y + z) [(x − y)2 + (y − z)2 + (z − x)2] The above identity holds true for all values of the variables present in it. Let us verify this by substituting random values for the variables x, y and z. If x = 1, y = 2 and z = 3, then: 13 + 23 + 33 − 3 × 1 × 2 × 3 = (1 + 2 + 3) (12 + 22 + 32 − 1 × 2 − 2 × 3 − 3 × 1) ⇒ 1 + 8 + 27 − 18 = 6 (1 + 4 + 9 − 2 − 6 − 3) ⇒ 18 = 6 × 3 ⇒ 18 = 18 ⇒ LHS = RHS Thus, we see that the identity holds true for random values of the variables present in it.

Proof of the Identity

Let us prove the identity x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) Or x3 + y3 + z3 − 3xyz = 1/2 (x + y + z) [(x − y)2 + (y − z)2 + (z − x)2] We can write x3 + y3 + z3 − 3xyz as: (x3 + y3) + z3 − 3xyz = [(x + y)3 − 3xy(x + y)] + z3 − 3xyz = a3 − 3axy + z3 − 3xyz, where a = x + y = (a3 + z3) − 3axy − 3xyz = (a + z) (a2 − az + z2) − 3xy(a + z) = (a + z) (a2 − az + z2 − 3xy) = (x + y + z) [(x + y)2 − (x + y)z + z2 − 3xy] = (x + y + z) (x2 + y2 + 2xy − zx − yz + z2 − 3xy) = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) ∴ x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) On multiplying and dividing the above expanded form by 2, we get: 1/2 × 2 (x + y + z) (x2 + y2 + z2 − xy − yz − zx) = 1/2 (x + y + z) (2×2 + 2y2 + 2z2 − 2xy − 2yz − 2zx) = 1/2 (x + y + z) (x2 + x2 + y2 + y2 + z2 + z2 − 2xy − 2yz − 2zx) = 1/2 (x + y + z) (x2 + y2 − 2xy + y2 + z2 − 2yz + z2 + x2 − 2zx) = 1/2 (x + y + z) [(x − y)2 + (y − z)2 + (z − x)2] ∴ x3 + y3 + z3 − 3xyz = 1/2 (x + y + z) [(x − y)2 + (y − z)2 + (z − x)2]

Case I of the Identity

A special case for the identity x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) is given below.

Case: When x + y + z = 0 then x3 + y3 + z3 = 3xyz. Proof: We have, x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) On substituting x + y + z = 0, we obtain x3 + y3 + z3 − 3xyz = 0 × (x2 + y2 + z2 − xy − yz − zx) ⇒ x3 + y3 + z3 − 3xyz = 0 ⇒ x3 + y3 + z3 = 3xyz Using this condition, we can factorize and find the values of many complex expressions.

Case II of the Identity

One more special case of the identity x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) is there which is explained below.

Case: When x + y + z ≠ 0 and x3 + y3 + z3 − 3xyz = 0 then x = y = z. Proof: We have, x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) On substituting x3 + y3 + z3 − 3xyz = 0, we obtain 0 = (x + y + z) (x2 + y2 + z2 − xy − yz − zx) ⇒ x2 + y2 + z2 − xy − yz − zx = 0 (x + y + z ≠ 0) ⇒ 1/2 (x + y + z) [(x − y)2 + (y − z)2 + (z − x)2] ⇒ [(x − y)2 + (y − z)2 + (z − x)2] = 0 (x + y + z ≠ 0) Since, the sum of non negative terms such as (x − y)2, (y − z)2 and (z − x)2 is 0, each term is 0. ∴ (x − y)2 = 0, (y − z)2 = 0 and (z − x)2 = 0 ⇒ x − y = 0, y − z = 0 and z − x = 0 ⇒ x = y, y = z and z = x ⇒ x = y = z This condition can be very helpful to factorize and find the values of many complex expressions.

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