case study class 9 maths chapter 2

• Polynomials Class 9 Case Study Questions Maths Chapter 2

Last Updated on August 26, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 9 maths. In this article, you will find case study questions for CBSE Class 9 Maths Chapter 2 Polynomials. It is a part of Case Study Questions for CBSE Class 9 Maths Series.

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Case Study Questions on Polynomials

In the current scenario, people use such door whose top half part is made of glass and bottom half part is wooden.

The glass portion of the door is having length and width in the ratio of 5 : 3. The wooden frame around the glass portion adds 11 inches to the total width and 14 inches to the total length. Consider the length of the glass portion as 5 x  inches:

On the basis of the above information, solve the following questions:

Q. 1. Find the total length of the glass portion of the door (in inches) is represented in terms of x .

Q. 2. Find the total width of the glass portion of the door (in inches).

Q. 3. Write the polynomial representation of the area top half part of the door.

Q. 4. Find the zeroes of the polynomial representing the area.

1. The total length of the glass portion in the door is represented by $(5 x+14)$ inches. 2. The total width of the glass portion in the door is $(3 x+11)$ inches. 3. The area of top half part of the door $$ \begin{aligned} & =\text { length } \times \text { width } \\ & =(5 x+14)(3 x+11) \\ =& 15 x^2+55 x+42 x+154 \\ & =15 x^2+97 x+154 \end{aligned} $$ 4. We have area, $p(x)=(5 x+14)(3 x+11)$ For finding zeroes, put $p(x)=0$ \begin{array}{ll} \therefore(5 x+14) (3 x+11)=0 \\ \Rightarrow (5 x+14)=0 \text { or }(3 x+11)=0 \\ \Rightarrow x=\frac{-14}{5} \text { or } x=\frac{-11}{3} \end{array}

Algebraic Identities

Algebraic equations that are true for all values of variables occurring in it. Some useful algebraic identities are:

(i) $(x+y)^2=x^2+2 x y+y^2$ (ii) $(x-y)^2=x^2-2 x y+y^2$ (iii) $x^2-y^2=(x+y)(x-y)$ (iv) $(x+a)(x+b)=x^2+(a+b) x+a b$ (v) $(x+y+z)^2$ =$x^2+y^2+z^2+2 x y+2 y z+2 z x$ (vi) $(x+y)^3$ =$x^3+y^3+3 x y(x+y)$ =$x^3+y^3+3 x^2 y+3 x y^2$ (vii) $(x-y)^3$ =$x^3-y^3-3 x y(x-y)$ =$x^3-y^3-3 x^2 y+3 x y^2$ (viii) $x^3+y^3$ =$(x+y)\left(x^2-x y+y^2\right)$ =$(x+y)^3-3 x y(x+y)$ (ix) $x^3-y^3$ =$(x-y)\left(x^2+x y+y^2\right)$ =$(x-y)^3+3 x y(x-y)$ (x) $\left(x^3+y^3+z^3-3 x y z\right)$ $=\left(x+y+z)(x^2+y^2+z^2-x y-y z-z x\right)$ Note: If $x+y+z=0$, then $x^3+y^3+z^3=3 x y z$

• Heron’s Formula Class 9 Case Study Questions Maths Chapter 10
• Circles Class 9 Case Study Questions Maths Chapter 9
• Quadrilaterals Class 9 Case Study Questions Maths Chapter 8
• Triangles Class 9 Case Study Questions Maths Chapter 7
• Lines and Angles Class 9 Case Study Questions Maths Chapter 6
• Introduction to Euclid’s Geometry Class 9 Case Study Questions Maths Chapter 5
• Linear Equations in Two Variables Class 9 Case Study Questions Maths Chapter 4

Coordinate Geometry Class 9 Case Study Questions Maths Chapter 3

Number systems class 9 case study questions maths chapter 1, topics from which case study questions may be asked.

• Definition of a polynomial in one variable, with examples and counter examples.
• Coefficients of a polynomial
• Terms of a polynomial and zero polynomial.
• Degree of a polynomial.
• Constant, linear, quadratic and cubic polynomials.
• Monomials, binomials, trinomials.
• Factors and multiples.
• Zeros of a polynomial.
• Remainder Theorem with examples.

Linear polynomial can be monomial or binomial. Quadratic polynomial can be monomial, binomial or trinomial. The degree of a zero polynomial is not defined.

Case study questions from the above given topic may be asked.

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Frequently Asked Questions (FAQs) on Polynomials Case Study

Q1: what are polynomials in mathematics.

A1: Polynomials are algebraic expressions that consist of variables and coefficients, connected by addition, subtraction, and multiplication. They can have multiple terms, and the degree of a polynomial is determined by the highest power of the variable in the expression.

Q2: H ow many types of polynomials are there?

A2: Polynomials are classified based on the number of terms they have: Monomial: A polynomial with just one term. Binomial: A polynomial with two terms. Trinomial: A polynomial with three terms Multinomial: A polynomial with more than three terms

Q3: What is the degree of a polynomial?

A3: The degree of a polynomial is the highest power of the variable in the polynomial.

Q4: What are the key concepts covered in Chapter 2 of CBSE Class 9 Maths regarding polynomials?

A4: Chapter 2 of CBSE Class 9 Maths covers concepts such as understanding polynomials and its types. (i) Types of polynomials (ii) Terms and coefficient of polynomials (iii) Zeroes of a polynomial (iv) Division algorithm (v) Remainder theorem (vi) Factor theorem (vii) Factorisation of quadratic polynomial

Q5: What is the difference between a polynomial and a non-polynomial expression?

A5: A polynomial expression consists of variables with non-negative integer exponents and real coefficients, combined using addition, subtraction, and multiplication. Non-polynomial expressions may include variables with negative exponents, fractional exponents, or other operations like division by variables or roots

Q6: What are the common mistakes to avoid when working with polynomials?

A6: Common mistakes include: Misidentifying the degree of the polynomial. Incorrectly applying the distributive property when multiplying polynomials. Forgetting to arrange the polynomial terms in standard form (descending order of exponents). Overlooking the signs while combining like terms. Misapplying the rules of exponents.

Q7: Are there any online resources or tools available for practicing polynomials case study questions?

A7: We provide case study questions for CBSE Class 9 Maths on our website. Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit Physics Gurukul website. they are having a large collection of case study questions for all classes.

Q8: What are the important keywords for CBSE Class 9 Maths Polynomials?

A8: List of important keywords given below – Algebraic Expression: Any expression that contains constants and variables, connected by some or all of the operations +, -, x, ÷. Polynomials: An algebraic expression in which the variables involved have only non-negative integral powers. Polynomials in one Variable: An algebraic expression which consist of only one type of variables in the entire expression. Degree of Polynomial: Highest power of a variable in the polynomial. Constant Polynomial: Polynomial of zero degree. Zero Polynomials: A polynomial consisting of one term, namely zero. Zeroes of a Polynomial: Let p(x) be a polynomial in one variable and ‘a’ be a real number such that the value of polynomial at x=a is zero i.e., p(a) = 0, then ‘a’ is said to be a zero polynomial of p(x). Remainder Theorem: Let p(x) be a polynomial having degree 1 or more than 1 and let ‘a’ be any real number. If p(x) is divided by (x-a), then remainder is p(a).

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CBSE Class 9 Mathematics Case Study Questions

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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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16 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

MATHS PAAGAL HAI

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

Where is search ? bar

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Can I have more questions without downloading the app.

I love math

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I am Riddhi Shrivastava… These questions was very good.. That’s it.. ..

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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 Exam 2024-25 : Skill Subject Sample Papers and Marking Scheme Released; Download PDF

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CBSE Case Study Questions for Class 9 Maths - Pdf PDF Download

CBSE Case Study Questions for Class  9 Maths

CBSE Case Study Questions for Class 9 Maths are a type of assessment where students are given a real-world scenario or situation and they need to apply mathematical concepts to solve the problem. These types of questions help students to develop their problem-solving skills and apply their knowledge of mathematics to real-life situations.

Chapter Wise Case Based Questions for Class 9 Maths

The CBSE Class 9 Case Based Questions can be accessed from Chapetrwise Links provided below:

Chapter-wise case-based questions for Class 9 Maths are a set of questions based on specific chapters or topics covered in the maths textbook. These questions are designed to help students apply their understanding of mathematical concepts to real-world situations and events.

Chapter 1: Number System

• Case Based Questions: Number System

Chapter 2: Polynomial

• Case Based Questions: Polynomial

Chapter 3: Coordinate Geometry

• Case Based Questions: Coordinate Geometry

Chapter 4: Linear Equations

• Case Based Questions: Linear Equations - 1
• Case Based Questions: Linear Equations -2

Chapter 5: Introduction to Euclid’s Geometry

• Case Based Questions: Lines and Angles

Chapter 7: Triangles

• Case Based Questions: Triangles

Chapter 8: Quadrilaterals

• Case Based Questions: Quadrilaterals - 1
• Case Based Questions: Quadrilaterals - 2

Chapter 9: Areas of Parallelograms

• Case Based Questions: Circles

Chapter 11: Constructions

• Case Based Questions: Constructions

Chapter 12: Heron’s Formula

• Case Based Questions: Heron’s Formula

Chapter 13: Surface Areas and Volumes

• Case Based Questions: Surface Areas and Volumes

Chapter 14: Statistics

• Case Based Questions: Statistics

Chapter 15: Probability

• Case Based Questions: Probability

Weightage of Case Based Questions in Class 9 Maths

Why are Case Study Questions important in Maths Class  9?

• Enhance critical thinking:  Case study questions require students to analyze a real-life scenario and think critically to identify the problem and come up with possible solutions. This enhances their critical thinking and problem-solving skills.
• Apply theoretical concepts:  Case study questions allow students to apply theoretical concepts that they have learned in the classroom to real-life situations. This helps them to understand the practical application of the concepts and reinforces their learning.
• Develop decision-making skills:  Case study questions challenge students to make decisions based on the information provided in the scenario. This helps them to develop their decision-making skills and learn how to make informed decisions.
• Improve communication skills:  Case study questions often require students to present their findings and recommendations in written or oral form. This helps them to improve their communication skills and learn how to present their ideas effectively.
• Enhance teamwork skills:  Case study questions can also be done in groups, which helps students to develop teamwork skills and learn how to work collaboratively to solve problems.

In summary, case study questions are important in Class 9 because they enhance critical thinking, apply theoretical concepts, develop decision-making skills, improve communication skills, and enhance teamwork skills. They provide a practical and engaging way for students to learn and apply their knowledge and skills to real-life situations.

Class 9 Maths Curriculum at Glance

The Class 9 Maths curriculum in India covers a wide range of topics and concepts. Here is a brief overview of the Maths curriculum at a glance:

• Number Systems:  Students learn about the real number system, irrational numbers, rational numbers, decimal representation of rational numbers, and their properties.
• Algebra:  The Algebra section includes topics such as polynomials, linear equations in two variables, quadratic equations, and their solutions.
• Coordinate Geometry:  Students learn about the coordinate plane, distance formula, section formula, and slope of a line.
• Geometry:  This section includes topics such as Euclid’s geometry, lines and angles, triangles, and circles.
• Trigonometry: Students learn about trigonometric ratios, trigonometric identities, and their applications.
• Mensuration: This section includes topics such as area, volume, surface area, and their applications.
• Statistics and Probability:  Students learn about measures of central tendency, graphical representation of data, and probability.

The Class 9 Maths curriculum is designed to provide a strong foundation in mathematics and prepare students for higher education in the field. The curriculum is structured to develop critical thinking, problem-solving, and analytical skills, and to promote the application of mathematical concepts in real-life situations. The curriculum is also designed to help students prepare for competitive exams and develop a strong mathematical base for future academic and professional pursuits.

Students can also access Case Based Questions of all subjects of CBSE Class 9

• Case Based Questions for Class 9 Science
• Case Based Questions for Class 9 Social Science
• Case Based Questions for Class 9 English
• Case Based Questions for Class 9 Hindi
• Case Based Questions for Class 9 Sanskrit

Frequently Asked Questions (FAQs) on Case Based Questions for Class 9 Maths

What is case-based questions.

Case-Based Questions (CBQs) are open-ended problem solving tasks that require students to draw upon their knowledge of Maths concepts and processes to solve a novel problem. CBQs are often used as formative or summative assessments, as they can provide insights into how students reason through and apply mathematical principles in real-world problems.

What are case-based questions in Maths?

Case-based questions in Maths are problem-solving tasks that require students to apply their mathematical knowledge and skills to real-world situations or scenarios.

What are some common types of case-based questions in class 9 Maths?

Common types of case-based questions in class 9 Maths include word problems, real-world scenarios, and mathematical modeling tasks.

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FAQs on CBSE Case Study Questions for Class 9 Maths - Pdf

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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.

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

• Class 9 Important Question
• 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.

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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.

CBSE Class 9 Maths Important Questions

Cbse study materials.

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 Class 9 Polynomials

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Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. Answers to each and every question is explained in an easy to understand way, with videos of all the questions.

In this chapter, we will learn

• What is a Polynomial
• What are Polynomials in One Variable
• What are monomials, binomials, trinomials
• What is the Degree of a Polynomial
• What are Linear, Quadratic and Cubic Polynomials
• What is zero of a polynomial (or root of a polynomial)
• Finding zeroes of a polynomial
• Dividing Polynomials and finding remainder
• Finding Remainder using Remainder Theorem
• Checking if it is a factor or not
• Factorising Quadratic and Cubic Polynomials using Factor Theorem
• Solving questions using Algebra Identities (Check full list of Algebra Formulas )

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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 for Class 9 Maths Chapter 2 Polynomials

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

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

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

Download most important questions for class 9 maths chapter – 2 polynomials.

In NCERT Solutions for Class 9, students are introduced to many important topics that will be helpful for those who wish to pursue Mathematics as a subject in higher studies. NCERT Solutions help students to prepare for their upcoming exams by covering the updated CBSE syllabus for 2023-24 and its guidelines.

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

NCERT Class 9 Maths Chapter 2 Polynomials Topics

As this is one of the important Chapters in Class 9 Maths, it comes under the unit – Algebra and has a weightage of 12 marks in the Class 9 Maths CBSE examination. This chapter talks about:

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

Students can refer to the NCERT Solutions for Class 9  while solving exercise problems and preparing for their Class 9 Maths exams.

NCERT Class 9 Maths Chapter 2 – Polynomials Summary

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials is the second chapter of Class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses Polynomials and their applications. The introduction of the chapter includes whole numbers, integers, and rational numbers.

The chapter starts with the introduction of Polynomials in section 2.1, followed by two very important topics in sections 2.2 and 2.3

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

Next, it discusses the following topics:

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

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

• These NCERT Solutions for Class 9 Maths help you solve and revise the updated CBSE syllabus of Class 9 for 2023-24.
• After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
• It follows NCERT guidelines which help in preparing the students accordingly.
• It contains all the important questions from the examination point of view.
• It helps in scoring well in Class 10 CBSE Maths exams.

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

NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials

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List of Exercises in Class 9 Maths Chapter 2 Polynomials

Class 9 Maths Chapter 2 Polynomials contains 5 exercises. Based on the concept of polynomials, each exercise provides a number of questions. Click on the below links to access the exercise-wise NCERT solutions for Class 9 Maths Chapter 2 polynomials.

Exercise 2.1 Solutions 5 Questions

Exercise 2.2 Solutions 4 Questions

Exercise 2.3 Solutions 3 Questions

Exercise 2.4 Solutions 5 Questions

Exercise 2.5 Solutions 16 Questions

Access Answers of NCERT Class 9 Maths Chapter 2 – Polynomials

Exercise 2.1 page: 32.

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

(i) 4x 2 –3x+7

The equation 4x 2 –3x+7 can be written as 4x 2 –3x 1 +7x 0

Since x is the only variable in the given equation and the powers of x (i.e. 2, 1 and 0) are whole numbers, we can say that the expression 4x 2 –3x+7 is a polynomial in one variable.

(ii) y 2 +√2

The equation y 2 + √2 can be written as y 2 + √ 2y 0

Since y is the only variable in the given equation and the powers of y (i.e., 2 and 0) are whole numbers, we can say that the expression y 2 + √ 2 is a polynomial in one variable.

(iii) 3√t+t√2

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

Though t is the only variable in the given equation, the power of t (i.e., 1/2) is not a whole number. Hence, we can say that the expression 3√t+t√2 is not a polynomial in one variable.

The equation y+2/y can be written as y+2y -1

Though y is the only variable in the given equation, the power of y (i.e., -1) is not a whole number. Hence, we can say that the expression y+2/y is not a polynomial in one variable.

(v) x 10 +y 3 +t 50

Here, in the equation x 10 +y 3 +t 50

Though the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression

x 10 +y 3 +t 50 . Hence, it is not a polynomial in one variable.

2. Write the coefficients of x 2 in each of the following:

(i) 2+x 2 +x

The equation 2+x 2 +x can be written as 2+(1)x 2 +x

We know that the coefficient is the number which multiplies the variable.

Here, the number that multiplies the variable x 2 is 1

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

(ii) 2–x 2 +x 3

The equation 2–x 2 +x 3 can be written as 2+(–1)x 2 +x 3

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

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

Hence, the coefficient of x 2 in 2–x 2 +x 3 is -1.

(iii) ( π /2)x 2 +x

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

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

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

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

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

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

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

For example,  3x 35 +5

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

For example,  4x 100

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

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

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

Here, 5x 3 +4x 2 +7x = 5x 3 +4x 2 +7x 1

The powers of the variable x are: 3, 2, 1

The degree of 5x 3 +4x 2 +7x is 3, as 3 is the highest power of x in the equation.

Here, in 4–y 2 ,

The power of the variable y is 2

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

(iii) 5t–√7

Here, in 5t –√7 

The power of the variable t is: 1

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

Here, 3 = 3×1 = 3× x 0

The power of the variable here is: 0

Hence, the degree of 3 is 0.

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

We know that,

Linear polynomial: A polynomial of degree one is called a linear polynomial.

Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial.

Cubic polynomial: A polynomial of degree three is called a cubic polynomial.

The highest power of x 2 +x is 2

The degree is 2

Hence, x 2 +x is a quadratic polynomial

The highest power of x–x 3 is 3

The degree is 3

Hence, x–x 3 is a cubic polynomial

(iii) y+y 2 +4

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

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

The highest power of 1+x is 1

The degree is 1

Hence, 1+x is a linear polynomial.

The highest power of 3t is 1

Hence, 3t is a linear polynomial.

The highest power of r 2 is 2

Hence, r 2 is a quadratic polynomial.

The highest power of 7x 3 is 3

Hence, 7x 3 is a cubic polynomial.

Exercise 2.2 Page: 34

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

(ii) x = – 1

(iii) x = 2

Let f(x) = 5x−4x 2 +3

(i) When x = 0

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

(ii) When x = -1

f(x) = 5x−4x 2 +3

f(−1) = 5(−1)−4(−1) 2 +3

(iii) When x = 2

f(2) = 5(2)−4(2) 2 +3

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

(i) p(y)=y 2 −y+1

p(y) = y 2 –y+1

∴ p(0) = (0) 2 −(0)+1 = 1

p(1) = (1) 2 –(1)+1 = 1

p(2) = (2) 2 –(2)+1 = 3

(ii) p(t)=2+t+2t 2 −t 3

p(t) = 2+t+2t 2 −t 3

∴ p(0) = 2+0+2(0) 2 –(0) 3 = 2

p(1) = 2+1+2(1) 2 –(1) 3 =2+1+2–1 = 4

p(2) = 2+2+2(2) 2 –(2) 3 =2+2+8–8 = 4

(iii) p(x)=x 3

∴ p(0) = (0) 3 = 0

p(1) = (1) 3 = 1

p(2) = (2) 3 = 8

(iv) P(x) = (x−1)(x+1)

p(x) = (x–1)(x+1)

∴ p(0) = (0–1)(0+1) = (−1)(1) = –1

p(1) = (1–1)(1+1) = 0(2) = 0

p(2) = (2–1)(2+1) = 1(3) = 3

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

(i) p(x)=3x+1, x = −1/3

For, x = -1/3, p(x) = 3x+1

∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0

∴ -1/3 is a zero of p(x).

(ii) p(x) = 5x–π, x = 4/5

For, x = 4/5, p(x) = 5x–π

∴ p(4/5) = 5(4/5)- π = 4-π

∴ 4/5 is not a zero of p(x).

(iii) p(x) = x 2 −1, x = 1, −1

For, x = 1, −1;

p(x) = x 2 −1

∴ p(1)=1 2 −1=1−1 = 0

p(−1)=(-1) 2 −1 = 1−1 = 0

∴ 1, −1 are zeros of p(x).

(iv) p(x) = (x+1)(x–2), x =−1, 2

For, x = −1,2;

p(x) = (x+1)(x–2)

∴ p(−1) = (−1+1)(−1–2)

= (0)(−3) = 0

p(2) = (2+1)(2–2) = (3)(0) = 0

∴ −1, 2 are zeros of p(x).

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

For, x = 0 p(x) = x 2

p(0) = 0 2 = 0

∴ 0 is a zero of p(x).

(vi) p(x) = lx +m, x = −m/ l

For, x = -m/ l ; p(x) = l x+m

∴ p(-m/ l) = l (-m/ l )+m = −m+m = 0

∴ -m/ l is a zero of p(x).

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

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

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

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

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

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

For, x = 1/2 p(x) = 2x+1

∴ p(1/2) = 2(1/2)+1 = 1+1 = 2≠0

∴ 1/2 is not a zero of p(x).

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

(i) p(x) = x+5 

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

(ii) p(x) = x–5

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

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

p(x) = 2x+5

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

(iv) p(x) = 3x–2 

p(x) = 3x–2

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

(v) p(x) = 3x 

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

(vi) p(x) = ax, a≠0

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

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

p(x) = cx + d

∴ x = -d/c is a zero polynomial of the polynomial p(x).

Exercise 2.3 Page: 40

1. Find the remainder when x 3 +3x 2 +3x+1 is divided by

∴ Remainder:

p(−1) = (−1) 3 +3(−1) 2 +3(−1)+1

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

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

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

p(0) = (−π) 3 +3(−π) 2 +3(−π)+1

= −π 3 +3π 2 −3π+1

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

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

Let p(x) = x 3 −ax 2 +6x−a

p(a) = (a) 3 −a(a 2 )+6(a)−a

= a 3 −a 3 +6a−a = 5a

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

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

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

= (-343-147)/9

= -490/9 ≠ 0

∴ 7+3x is not a factor of 3x 3 +7x

Exercise 2.4 Page: 43

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

(i) x 3 +x 2 +x+1

Let p(x) = x 3 +x 2 +x+1

p(−1) = (−1) 3 +(−1) 2 +(−1)+1

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

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

Let p(x)= x 4 +x 3 +x 2 +x+1

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

= 1−1+1−1+1

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

(iii) x 4 +3x 3 +3x 2 +x+1 

Let p(x)= x 4 +3x 3 +3x 2 +x+1

The zero of x+1 is -1.

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

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

(iv) x 3 – x 2 – (2+√2)x +√2

Let p(x) = x 3 –x 2 –(2+√2)x +√2

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

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

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

(i) p(x) = 2x 3 +x 2 –2x–1, g(x) = x+1

p(x) = 2x 3 +x 2 –2x–1, g(x) = x+1

∴ Zero of g(x) is -1.

p(−1) = 2(−1) 3 +(−1) 2 –2(−1)–1

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

(ii) p(x)=x 3 +3x 2 +3x+1, g(x) = x+2

p(x) = x 3 +3x 2 +3x+1, g(x) = x+2

∴ Zero of g(x) is -2.

p(−2) = (−2) 3 +3(−2) 2 +3(−2)+1

= −8+12−6+1

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

(iii) p(x)=x 3 –4x 2 +x+6, g(x) = x–3

p(x) = x 3 –4x 2 +x+6, g(x) = x -3

∴ Zero of g(x) is 3.

p(3) = (3) 3 −4(3) 2 +(3)+6

= 27−36+3+6

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

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

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

By Factor Theorem

⇒ (1) 2 +(1)+k = 0

⇒ 1+1+k = 0

(ii) p(x) = 2x 2 +kx+ √2

⇒ 2(1) 2 +k(1)+√2 = 0

⇒ 2+k+√2 = 0

⇒ k = −(2+√2)

(iii) p(x) = kx 2 – √ 2x+1

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

⇒ k(1) 2 -√2(1)+1=0

(iv) p(x)=kx 2 –3x+k

⇒ k(1) 2 –3(1)+k = 0

⇒ k−3+k = 0

4. Factorise:

(i) 12x 2 –7x+1

Using the splitting the middle term method,

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

12x 2 –7x+1= 12x 2 -4x-3x+1

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

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

(ii) 2x 2 +7x+3

We have to find a number whose sum = 7 and product = 2×3 = 6

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

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

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

(iii) 6x 2 +5x-6 

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

6x 2 +5x-6 = 6x 2 +9x–4x–6

= 3x(2x+3)–2(2x+3)

= (2x+3)(3x–2)

(iv) 3x 2 –x–4 

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

3x 2 –x–4 = 3x 2 –4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

5. Factorise:

(i) x 3 –2x 2 –x+2

Let p(x) = x 3 –2x 2 –x+2

Factors of 2 are ±1 and ± 2

p(x) = x 3 –2x 2 –x+2

p(−1) = (−1) 3 –2(−1) 2 –(−1)+2

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

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x 2 –3x+2) = (x+1)(x 2 –x–2x+2)

= (x+1)(x(x−1)−2(x−1))

= (x+1)(x−1)(x-2)

(ii) x 3 –3x 2 –9x–5

Let p(x) = x 3 –3x 2 –9x–5

Factors of 5 are ±1 and ±5

By the trial method, we find that

So, (x-5) is factor of p(x)

p(x) = x 3 –3x 2 –9x–5

p(5) = (5) 3 –3(5) 2 –9(5)–5

= 125−75−45−5

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

(x−5)(x 2 +2x+1) = (x−5)(x 2 +x+x+1)

= (x−5)(x(x+1)+1(x+1))

= (x−5)(x+1)(x+1)

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

Let p(x) = x 3 +13x 2 +32x+20

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

So, (x+1) is factor of p(x)

p(x)= x 3 +13x 2 +32x+20

p(-1) = (−1) 3 +13(−1) 2 +32(−1)+20

= −1+13−32+20

Now, Dividend = Divisor × Quotient +Remainder

(x+1)(x 2 +12x+20) = (x+1)(x 2 +2x+10x+20)

= (x+1)x(x+2)+10(x+2)

= (x+1)(x+2)(x+10)

(iv) 2y 3 +y 2 –2y–1

Let p(y) = 2y 3 +y 2 –2y–1

Factors = 2×(−1)= -2 are ±1 and ±2

So, (y-1) is factor of p(y)

p(y) = 2y 3 +y 2 –2y–1

p(1) = 2(1) 3 +(1) 2 –2(1)–1

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

Now, Dividend = Divisor × Quotient + Remainder

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

= (y−1)(2y(y+1)+1(y+1))

= (y−1)(2y+1)(y+1)

Exercise 2.5 Page: 48

1. Use suitable identities to find the following products:

(i) (x+4)(x +10) 

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

(x+4)(x+10) = x 2 +(4+10)x+(4×10)

= x 2 +14x+40

(ii) (x+8)(x –10)     

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

= x 2 +(8−10)x–80

= x 2 −2x−80

(iii) (3x+4)(3x–5)

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

= 9x 2 +3x(4–5)–20

= 9x 2 –3x–20

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

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

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

2. Evaluate the following products without multiplying directly:

(i) 103×107

103×107= (100+3)×(100+7)

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

Here, x = 100

We get, 103×107 = (100+3)×(100+7)

= (100) 2 +(3+7)100+(3×7)

= 10000+1000+21

(ii) 95×96  

95×96 = (100-5)×(100-4)

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

We get, 95×96 = (100-5)×(100-4)

= (100) 2 +100(-5+(-4))+(-5×-4)

= 10000-900+20

(iii) 104×96

104×96 = (100+4)×(100–4)

Here, a = 100

We get, 104×96 = (100+4)×(100–4)

= (100) 2 –(4) 2

3. Factorise the following using appropriate identities:

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

9x 2 +6xy+y 2 = (3x) 2 +(2×3x×y)+y 2

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

Here, x = 3x

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

(ii) 4y 2 −4y+1

4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1

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

Here, x = 2y

4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1 2

= (2y–1)(2y–1)

(iii)  x 2 –y 2 /100

x 2 –y 2 /100 = x 2 –(y/10) 2

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

Here, x = x

= (x–y/10)(x+y/10)

4. Expand each of the following using suitable identities:

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

(ii) (2x−y+z) 2

(iii) (−2x+3y+2z) 2

(iv) (3a –7b–c) 2

(v) (–2x+5y–3z) 2

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

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

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

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

(ii) (2x−y+z) 2  

Here, x = 2x

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

= 4x 2 +y 2 +z 2 –4xy–2yz+4xz

Here, x = −2x

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

= 4x 2 +9y 2 +4z 2 –12xy+12yz–8xz

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

Here, x = 3a

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

= 9a 2 + 49b 2 + c 2 – 42ab+14bc–6ca

Here, x = –2x

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

= 4x 2 +25y 2 +9z 2 – 20xy–30yz+12zx

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

Here, x = (1/4)a

y = (-1/2)b

(i) 4x 2 +9y 2 +16z 2 +12xy–24yz–16xz

(ii) 2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz

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

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

= (2x+3y–4z) 2

= (2x+3y–4z)(2x+3y–4z)

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

2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz

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

= (−√2x+y+2√2z) 2

= (−√2x+y+2√2z)(−√2x+y+2√2z)

6. Write the following cubes in expanded form:

(i) (2x+1) 3

(ii) (2a−3b) 3

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

(iv) (x−(2/3)y) 3

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

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

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

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

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

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

= 8a 3 –27b 3 –18ab(2a–3b)

= 8a 3 –27b 3 –36a 2 b+54ab 2

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

(iv)  (x−(2/3)y) 3

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

7. Evaluate the following using suitable identities: 

(ii) (102) 3

(iii) (998) 3

We can write 99 as 100–1

(99) 3 = (100–1) 3

= (100) 3 –1 3 –(3×100×1)(100–1)

= 1000000 –1–300(100 – 1)

= 1000000–1–30000+300

We can write 102 as 100+2

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

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

We can write 99 as 1000–2

(998) 3 =(1000–2) 3

=(1000) 3 –2 3 –(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992

8. Factorise each of the following:

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

(ii) 8a 3 –b 3 –12a 2 b+6ab 2

(iii) 27–125a 3 –135a +225a 2    

(iv) 64a 3 –27b 3 –144a 2 b+108ab 2

(v) 27p 3 –(1/216)−(9/2) p 2 +(1/4)p

The expression, 8a 3 +b 3 +12a 2 b+6ab 2 can be written as (2a) 3 +b 3 +3(2a) 2 b+3(2a)(b) 2

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

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

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

The expression, 8a 3 –b 3 −12a 2 b+6ab 2 can be written as (2a) 3 –b 3 –3(2a) 2 b+3(2a)(b) 2

8a 3 –b 3 −12a 2 b+6ab 2 = (2a) 3 –b 3 –3(2a) 2 b+3(2a)(b) 2

= (2a–b)(2a–b)(2a–b)

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

(iii) 27–125a 3 –135a+225a 2  

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

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

= (3–5a)(3–5a)(3–5a)

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

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

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

=(4a–3b)(4a–3b)(4a–3b)

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

(v) 27p 3 – (1/216)−(9/2) p 2 +(1/4)p

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

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

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

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

Taking x = 3p and y = 1/6

= (3p–1/6) 3

= (3p–1/6)(3p–1/6)(3p–1/6)

(i) x 3 +y 3 = (x+y)(x 2 –xy+y 2 )

(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 )

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

⇒ x 3 +y 3 = (x+y) 3 –3xy(x+y)

⇒ x 3 +y 3 = (x+y)(x 2 +y 2 –xy)

(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 ) 

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

⇒ x 3 −y 3 = (x–y) 3 +3xy(x–y)

⇒ x 3 +y 3 = (x–y)(x 2 +y 2 +xy)

10. Factorise each of the following:

(i) 27y 3 +125z 3

(ii) 64m 3 –343n 3

The expression, 27y 3 +125z 3 can be written as (3y) 3 +(5z) 3

27y 3 +125z 3 = (3y) 3 +(5z) 3

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

= (3y+5z)(9y 2 –15yz+25z 2 )

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

64m 3 –343n 3 = (4m) 3 –(7n) 3

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

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

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

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

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

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

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

12. Verify that:

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

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

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

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

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

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

Then, x 3 +y 3 +z 3 -3xyz = (0)(x 2 +y 2 +z 2 –xy–yz–xz)

⇒ x 3 +y 3 +z 3 –3xyz = 0

⇒ x 3 +y 3 +z 3 = 3xyz

Hence Proved

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

(i) (−12) 3 +(7) 3 +(5) 3

(ii) (28) 3 +(−15) 3 +(−13) 3

Let a = −12

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

Here, −12+7+5=0

(−12) 3 +(7) 3 +(5) 3 = 3xyz

= 3×-12×7×5

(28) 3 +(−15) 3 +(−13) 3

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

Here, x+y+z = 28–15–13 = 0

(28) 3 +(−15) 3 +(−13) 3 = 3xyz

= 0+3(28)(−15)(−13)

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

(i) Area: 25a 2 –35a+12

(ii) Area: 35y 2 +13y–12

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

25a 2 –35a+12 = 25a 2 –15a−20a+12

= 5a(5a–3)–4(5a–3)

= (5a–4)(5a–3)

Possible expression for length  = 5a–4

Possible expression for breadth  = 5a –3

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

35y 2 +13y–12 = 35y 2 –15y+28y–12

= 5y(7y–3)+4(7y–3)

= (5y+4)(7y–3)

Possible expression for length  = (5y+4)

Possible expression for breadth  = (7y–3)

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

(i) Volume: 3x 2 –12x

(ii) Volume: 12ky 2 +8ky–20k

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

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x–4)

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

12ky 2 +8ky–20k = 4k(3y 2 +2y–5)

= 4k(3y 2 +5y–3y–5)

= 4k(3y+5)(y–1)

Possible expression for length = 4k

Possible expression for breadth = (3y +5)

Possible expression for height = (y -1)

Disclaimer:

Dropped Topics –  2.4 Remainder theorem.

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CBSE Class 9 Maths 30 Most Important Case Study Questions with Answers

Cbse class 9 maths 30 most important case study questions with answers download here free in pdf format..

CBSE Class 9 Maths exam 2023 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 webpage 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. Candidates can easily download these questions in PDF format and refer to them for exam preparation 2023.

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• JEE Main 2023 April 6 Shift 1 Chemistry Question Paper with Solutions April 7, 2023

Class 9 Maths Case Study Questions Chapter 4 Linear Equations in two variables

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Case study Questions in Class 9 Mathematics Chapter 4  are very important to solve for your exam. Class 9 Maths Chapter 4 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  Class 9 Maths Case Study Questions Chapter 4 Linear Equations in two variables

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

In CBSE Class 9 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Linear Equations in two variables Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 4 Linear Equations in two variables

Case Study/Passage-Based Questions

Case Study 1: Deepak bought 3 notebooks and 2 pens for Rs. 80. His friend Ram said that the price of each notebook could be Rs. 25. Then three notebooks would cost Rs.75, the two pens would cost Rs.5 and each pen could be Rs. 2.50. Another friend Ajay felt that Rs. 2.50 for one pen was too little. It should be at least Rs. 16. Then the price of each notebook would also be Rs.16

Lohith also bought the same types of notebooks and pens as Aditya. He paid 110 for 4 notebooks and 3 pens. Later, Deepak guesses the cost of one pen is Rs. 10 and Lohith guess the cost of one notebook is Rs. 30.

(i) Form the pair of linear equations in two variables from this situation by taking the cost of one notebook as Rs. x and the cost of one pen as Rs. y. (a) 3x + 2y = 80 and 4x + 3y = 110 (b) 2x + 3y = 80 and 3x + 4y = 110 (c) x + y = 80 and x + y = 110 (d) 3x + 2y = 110 and 4x + 3y = 80

Answer: (a) 3x + 2y = 80 and 4x + 3y = 110

(ii) Which is the solution satisfying both the equations formed in (i)? (a) x = 10, y = 20 (b) x = 20, y = 10 (c) x = 15, y = 15 (d) none of these

Answer: (b) x = 20, y = 10

(iii) Find the cost of one pen? (a) Rs. 20 (b) Rs. 10 (c) Rs. 5 (d) Rs. 15

Answer: (b) Rs. 10

(iv) Find the total cost if they will purchase the same type of 15 notebooks and 12 pens. (a) Rs. 400 (b) Rs. 350 (c) Rs. 450 (d) Rs. 420

Answer: (d) Rs. 420

(v) Find whose estimation is correct in the given statement. (a) Deepak (b) Lohith (c) Ram (d) Ajay

Answer: (a) Deepak

Case Study 2: In the below given layout, the design and measurements have been made such that area of two bedrooms and Kitchen together is 95 sq. m.

(i) The area of two bedrooms and kitchen are respectively equal to (a) 5x, 5y (b) 10x, 5y (c) 5x, 10y (c) x, y

(ii) Find the length of the outer boundary of the layout. (a) 27 m (b) 15 m (c) 50 m (d) 54 m

(iii) The pair of linear equations in two variables formed from the statements are (a) x + y = 13, x + y = 9 (b) 2x + y = 13, x + y = 9 (c) x + y = 13, 2x + y = 9 (d) None of the above

(iv) Which is the solution satisfying both the equations formed in (iii)? (a) x = 7, y = 6 (b) x = 8, y = 5 (c) x = 6, y = 7 (d) x = 5, y = 8

(v) Find the area of each bedroom. (a) 30 sq. m (b) 35 sq. m (c) 65 sq. m (d) 42 sq. m

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Mathematics Chapter 4 Linear Equations in two variables with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Maths Linear Equations in two variables 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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