CBSE NCERT Solutions

NCERT and CBSE Solutions for free

## Class 9 Mathematics Assignments

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

## Mathematics Class 9 Assignments Pdf Download

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

## Topicwise Assignments for Class 9 Mathematics Download in Pdf

More assignments for class 9 mathematics.

## Advantages of Class 9 Mathematics Assignments

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

## Frequently Asked Questions by Class 9 Mathematics students

At https://www.cbsencertsolutions.com, we have provided the biggest database of free assignments for Mathematics Class 9 which you can download in Pdf

We provide here Standard 9 Mathematics chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Mathematics in Grade 9, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

We have provided here topic-wise Mathematics Grade 9 question banks, revision notes and questions for all difficult topics, and other study material.

We have provided the best collection of question bank and practice tests for Class 9 for all subjects. You can download them all and use them offline without the internet.

## Related Posts

## Class 9 Mathematics Areas of Parallelogram and Triangle Assignments

## Class 9 English Assignments

## Class 9 Mathematics Frequency Distribution Assignments

- School & Boards
- College Admission
- Govt Jobs Alert & Prep
- Current Affairs
- GK & Aptitude
- Class 9 Maths NCERT Solutions

## NCERT Solutions for Class 9 Maths (PDF): Updated for 2022-2023

Ncert solutions for class 9 maths are available here for download in pdf. the chapter-wise solutions are best to prepare for the annual assessment in class 9 maths exam 2022-23..

NCERT Solutions for Class 9 Maths: Check the best explained NCERT Solutions for Class 9 Maths below. Download the chapter-wise solutions in PDF. These solutions have been updated for the latest edition of the NCERT Maths Book and are explained in the simplest manner. Therefore, following the Class 9th Maths NCERT Solutions by Jagran Josh will help students clear the concepts and score well in Class 9 Maths Exam 2022-23.

Also Read| CBSE Class 9 Maths Revised Syllabus 2022-2023

Links to Download Class 9 Maths NCERT Solutions are provided below:

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

Chapter 15: Probability

In Class 9, students are introduced to several new and complex concepts which are important to lay a strong foundation for higher classes. However, it also becomes difficult for students to grasp difficult topics and clear the concepts. But if you have good and reliable study resources then it becomes easier to understand the concepts and logic. NCERT Solutions by Jagran Josh form the best resource to remove all your doubts and clear the concepts. All these NCERT solutions have been explained by the subject matter experts to provide accurate and detailed solutions to class 9 students. These NCERT Solutions are prepared for the latest NCERT book for the new academic session 2021-22.

- Each solution is framed in a way to involve the key concepts.
- Solutions have been rechecked/revised by subject experts.
- Chapter-wise solutions are available for free download.

Also, check Class 9 Mathematics NCERT Exemplar Problems

When working with Mathematics problems, it is quite essential to focus on understanding the process and logic that is involved. This helps you understand how you should approach similar problems in school tests or annual examinations. Following the NCERT solutions provided above in this article will help to cover the complete syllabus in an organised manner.

Follow the Class 9 Maths NCERT Book:

Setting up a strong foundation of concepts during Class 9 can be much effective in cracking the biggest test of a student’s life, i.e., the Class 10 th Board Exam. For this, following the NCERT textbooks is considered as the best practice as these books help to develop a strong foundation of all the concepts.

Importance of Class 9 Maths NCERT Textbooks and NCERT Solutions

CBSE affiliated schools follow NCERT textbooks as questions asked in CBSE Class 9th annual examination are based on the NCERT textbooks only. The contents of the NCERT textbooks i.e., concepts, definitions, theorems, etc., are presented in the precise form. You will find here to the point detail on all topics and concepts. Facts and examples given in NCERT textbooks are universally accepted and it is one of the main reasons why CBSE prescribes NCERT textbooks.

NCERT textbooks play an essential role to get a hold of every topic if studied thoroughly. With simple and easy language, they help to lay a better foundation of all tough and advanced level concepts. To achieve the maximum marks in exams, all students need to read the NCERT books thoroughly and solve the exercise questions given at the end of each chapter. This will surely help them to get the desired results in the exams.

Also, check:

NCERT Books for Class 9 - All Subjects

Get here latest School , CBSE and Govt Jobs notification in English and Hindi for Sarkari Naukari and Sarkari Result . Download the Jagran Josh Sarkari Naukri App .

- CBSE Date Sheet 2024
- CBSE 10th Date Sheet 2024
- CBSE 12th Date Sheet 2024
- CBSE Class 10 Social Science Syllabus 2023-24
- CBSE Class 12 Maths Syllabus 2023-24
- AP TET Answer Key 2024
- SSC CPO Notification 2024
- SSC JE Notification 2024
- Kerala SET Result 2024
- CUET UG Registration 2024

## Trending Categories

- CBSE Class 9 Study Material
- NCERT Solutions for Class 9-12
- CBSE Study Material
- NCERT Solutions for Class 9
- CBSE Class 9

## Latest Education News

BPSC TRE 3 Admit Card 2024: bpsc.bih.nic.in से डाउनलोड करें बिहार शिक्षक परीक्षा एडमिट कार्ड, 16 मार्च की परीक्षा स्थगित

SSC CPO Syllabus 2024 and Exam Pattern for Paper 1, 2 & PET/PST, Download PDF

People with the highest brain power can spot the hotel owner in the vintage graphic within 5 seconds. Can you?

PDUSU Result 2024 OUT at shekhauni.ac.in; Direct Link to Download UG 1st Semester Marksheet

UPPSC Admit Card 2024: कब आएगा यूपी पीसीएस प्रीलिम्स एडमिट कार्ड, इन स्टेप्स से कर सकेंगे प्राप्त

JEE Main Paper 2 Result 2024 Out, Download BArch, BPlanning Paper 2 Scorecard at jeemain.nta.ac.in

JEE Main Paper 2 Result 2024 Declared: How to Check Session 1 NTA Scorecards for BArch, BPlan at jeemain.nta.ac.in, Check session 2 link here

JEE Main Paper 2 Result 2024 Out, Download BArch, BPlan Scorecard at jeemain.nta.ac.in

WBPSC SI Admit Card 2024 Out at psc.wb.gov.in, Direct Link to Hall Ticket, Check Exam Pattern

TSPSC Exam Date 2024 Out for 2734 Group 1, 2 And 3 Posts, Here's Notice Download Link

Genius IQ Test: You have a high IQ if you can find the height of the glass in 10 seconds!

99% of People Failed To Find 7 Cats Hidden In This Optical Illusion Within 7 Seconds. Test Your Skills!

BPSC TRE 3.0 Admit Card 2024: Important Notice Released at bpsc.bih.nic.in

UPPSC Admit Card 2024 Soon: What are the Steps to Download PCS Hall Ticket

BPSC TRE 3 Admit Card Release Date 2024 Announced at bpsc.bih.nic.in, Check Details Here

GK Quiz on Football: Are You a Football Know-It-All? Take this Quiz!

Only highly attentive can spot 3 differences between the boy and rose pictures in 12 seconds.

IB ACIO Tier 1 Result 2024: Check How to Download, Tier 2 Details

BPSC Headmaster Syllabus 2024: PDF Download For Important Topics, Check Exam Pattern

Optical Illusion Eye Test: Find a dog hiding in the garden in 9 seconds!

## Chapter 2 – Real and Complex Numbers

Exercise 2.1, exercise 2.2, exercise 2.3, exercise 2.4, exercise 2.5, exercise 2.6, review exercise, multiple choice questions, this post has 24 comments.

Assalam-o-Alaikum please check ex 2.4 Q1 it’s wrong according to question given on book please reply me

Thanks for correction. we have make it correct.

is there any way, i can download these notes

Yes, you can. By click top right corner arrow button. And download it.

Exercise 2.1 question 2 part 1 is wrong please check

Too 👍👍👍👍👍 good ots really help me in any problem but ex 2.4 q 1 Is wrong I hope u will ✔️✔️check it

yeah. we have make corrections in it.

Thanks for correction.

there is some mistakes in values like in excersice 2.4,question 1,part 1

You’ve given review ex in place of ex 2.1… Kindly provide ex 2.1 plz……☺

Ok, I will make it.

well described

Rashid Aziz MSC Mathematics

Nice notes😇

Thanks So much

There is something wrong in question no 6 part 6 of ex 2.6 check it ok

Kindly mention the mistake.

The question is 1/2i not 1/2

I’ve corrected the mistake. Thanks for the correction.

Kindly recheck the McQs answer

Thank u so much for giving notes. Is there any teacher who taught theorems

It is very good link But this link can’t show the Example …They show only exercises

I solved a question with the method used in question seven part one of exercise 2.6 and my maths teacher did not accepted it 😔 in a test

## Leave a Reply Cancel reply

Talk to our experts

1800-120-456-456

- NCERT Solutions for Class 9 Maths Chapter 2: Polynomials - Exercise 2.2
- NCERT Solutions

## Important Topics under NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.2)

Chapter 2 of the class 9 maths syllabus is on Polynomials. It is a very important chapter that is covered in class 9 and is divided into 8 major topics. Students are advised to learn from ad practice the solved questions time and again to be able to master the concept of polynomials.

The following list has been provided to give the students a glimpse at the important topics that are covered under the chapter on Polynomials.

Classification of polynomials

Degree of a polynomial

Types of polynomial based on the degree

Constant polynomial

Linear polynomial

Quadratic polynomial

Cubic polynomial

Types of polynomial based on terms

Value of a polynomial

Zero of a polynomial

Operations on polynomials

## Importance of Polynomials in Class 9

Polynomials are expressions having more than two algebraic terms. They can also be defined as the sum of several terms where the same variable/variables has/have different powers.

Polynomials are practically a language used in most mathematical expressions and are used to represent appropriate relations between different variables or numbers. Students are encouraged to gather as much information from this chapter to be able to solve tricky problems based on polynomials easily in exams.

NCERT Solutions for Class 9 Maths Chapter 2 teach you about polynomials. Vedantu’s NCERT Solutions for Class 9 Maths Chapter 2 PDF makes you aware of every aspect of Polynomials - be it finding value, using algebraic identities or factorization of polynomials. Our Exercise 2.2 Class 9 Maths solution explains everything about polynomials in an easy to understand way. Vedantu provides students with a Free PDF download option for all the NCERT Book Solutions of updated CBSE textbooks . Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 9 Science , Maths solutions, and solutions of other subjects that are available on Vedantu only.

## Access NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials

Exercise 2.2

1. Find the value of the polynomial $\text{5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$ at

i) $\text{x = 0}$

Ans: Let $\text{p}\left( \text{x} \right)\text{ = 5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$

We will simply put the value of $\text{x}$ in the given polynomial.

$\Rightarrow \text{p}\left( \text{0} \right)\text{ = 5}\left( \text{0} \right)\text{ - 4}\left( {{\text{0}}^{\text{2}}} \right)\text{ + 3}$

$\Rightarrow \text{p}\left( \text{0} \right)\text{ = 3}$

Hence, the value of the polynomial at $\text{x = 0}$ is $\text{3}$.

ii) $\text{x = -1}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = 5}\left( \text{-1} \right)\text{ - 4}\left[ {{\left( \text{-1} \right)}^{\text{2}}} \right]\text{ + 3}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = - 5 - 4 + 3}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = - 6}$

Hence, the value of the polynomial at $\text{x = -1}$ is $\text{-6}$.

iii) $\text{x = 2}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 5}\left( \text{2} \right)\text{ - 4}\left( {{\text{2}}^{\text{2}}} \right)\text{ + 3}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 10 - 16 + 3}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = - 3}$

Hence, the value of the polynomial at $\text{x = 2}$ is $\text{-3}$.

2. Find $\text{p}\left( \text{0} \right)$, $\text{p}\left( \text{1} \right)$ and $\text{p}\left( \text{2} \right)$ for each of the following polynomials:

i) $\text{p}\left( \text{y} \right)\text{ = }{{\text{y}}^{\text{2}}}\text{ - y + 1}$

$\text{p}\left( \text{0} \right)\text{ = }{{\text{0}}^{\text{2}}}\text{ - 0 + 1}$.

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 1}$.

$\text{p}\left( \text{1} \right)\text{ = }{{\text{1}}^{\text{2}}}\text{ - 1 + 1 = 1}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 1}$.

$\text{p}\left( \text{2} \right)\text{ = }{{\text{2}}^{\text{2}}}\text{ - 2 + 1}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4 - 2 + 1}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 3}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 3}$.

ii) $\text{p}\left( \text{t} \right)\text{ = 2 + t + 2}{{\text{t}}^{\text{2}}}\text{ - }{{\text{t}}^{\text{3}}}$

$\text{p}\left( \text{0} \right)\text{ = 2 + 0 + 2}\left( {{\text{0}}^{\text{2}}} \right)\text{ - }\left( {{\text{0}}^{\text{3}}} \right)\text{ = 2}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 2}$.

$\text{p}\left( \text{1} \right)\text{ = 2 + 1 + 2}\left( {{\text{1}}^{\text{2}}} \right)\text{ - }\left( {{\text{1}}^{\text{3}}} \right)\text{ = 4}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 4}$.

$\text{p}\left( \text{2} \right)\text{ = 2 + 2 + 2}\left( {{\text{2}}^{\text{2}}} \right)\text{ - }\left( {{\text{2}}^{\text{3}}} \right)$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4 + 8 - 8}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 4}$.

iii) $\text{p}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$

$\text{p}\left( \text{0} \right)\text{ = }{{\text{0}}^{\text{3}}}\text{ = 0}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 0}$.

$\text{p}\left( \text{1} \right)\text{ = }{{\text{1}}^{\text{3}}}\text{ = 1}$

$\text{p}\left( \text{2} \right)\text{ = }{{\text{2}}^{\text{3}}}\text{ = 8}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 8}$.

iv) $\text{p}\left( \text{x} \right)\text{ =}\left( \text{x - 1} \right)\left( \text{x + 1} \right)$

$\text{p}\left( \text{0} \right)\text{ =}\left( \text{0 - 1} \right)\left( \text{0 + 1} \right)\text{ = -1}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = -1}$.

$\text{p}\left( \text{1} \right)\text{ =}\left( \text{1 - 1} \right)\left( \text{1 + 1} \right)\text{ = 0}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 0}$.

$\text{p}\left( \text{2} \right)\text{ =}\left( \text{2 - 1} \right)\left( \text{2 + 1} \right)$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = }\left( \text{1} \right)\left( \text{3} \right)\text{ = 3}$

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

i) $\text{p}\left( \text{x} \right)\text{ = 3x + 1}$,$\text{x = -}\frac{\text{1}}{\text{3}}$

Ans: We are given: $\text{x = -}\frac{\text{1}}{\text{3}}$. If it is the zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = 3x + 1}$, then $\text{p}\left( \text{-}\frac{\text{1}}{\text{3}} \right)$ should be $\text{0}$.

$\text{p}\left(\text{-}\frac{\text{1}}{\text{3}}\right)\text{=3}\left(\text{-}\frac{\text{1}}{\text{3}} \right)\text{ + 1 = -1 + 1 = 0}$

Hence, we can say that $\text{x = -}\frac{\text{1}}{\text{3}}$ is a zero of the given polynomial.

ii) $\text{p}\left( \text{x} \right)\text{= 5x - }\!\!\pi\!\!\text{ }$ ,$\text{x = }\frac{\text{4}}{\text{5}}$

Ans: We are given: $\text{x = }\frac{\text{4}}{\text{5}}$. If it is the zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = 5x - }\text{ }\!\!\pi\!\!\text{ }$, then $\text{p}\left( \frac{\text{4}}{\text{5}} \right)$ should be $\text{0}$.

$\text{p}\left( \frac{\text{4}}{\text{5}} \right)\text{ = 5}\left( \frac{\text{4}}{\text{5}} \right)\text{ - 3}\text{.14 = 4 - 3}\text{.14 }\ne \text{ 0}$

Hence, we can say that $\text{x = }\frac{\text{4}}{\text{5}}$ is not a zero of the given polynomial.

iii) $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{-1, x = 1, -1}$

Ans: We are given: $\text{x = 1}$ and $\text{x = -1}$ .

If they are zeros of polynomial $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{-1}$ , then $\text{p}\left( \text{1} \right)$ and $\text{p}\left( \text{-1} \right)$ should both be $\text{0}$.

$\text{p}\left( \text{1} \right)\text{ = }{{\left( \text{1} \right)}^{\text{2}}}\text{-1 = 0}$

$\text{p}\left( \text{-1} \right)\text{ = }{{\left( \text{-1} \right)}^{\text{2}}}\text{-1 = 0}$

Hence, we can say that $\text{x = 1}$ and $\text{x = -1}$ are zeroes of the given polynomial.

iv) $\text{p(x) = (x+1)(x-2), x = -1, 2}$

Ans: We are given: $\text{x = -1}$ and $\text{x = 2}$.

If they are zeroes of the polynomial $\text{p}\left( \text{x} \right)\text{ = }\left( \text{x+1} \right)\left( \text{x-2} \right)$ , then $\text{p}\left( \text{-1} \right)$ and $\text{p}\left( \text{2} \right)$ should be $\text{0}$.

$\text{p}\left( \text{-1} \right)\text{ = }\left( \text{-1+1} \right)\left( \text{-1-2} \right)\text{ = }\left( \text{0} \right)\left( \text{-3} \right)\text{ = 0}$

$\text{p}\left( \text{2} \right)\text{ = }\left( \text{2+1} \right)\left( \text{2-2} \right)\text{ = }\left( \text{3} \right)\left( \text{0} \right)\text{ = 0}$

Hence, we can say that $\text{x = -1}$ and $\text{x = 2}$ are zeroes of the given polynomial.

v) $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{, x = 0}$

Ans: We are given $\text{x = 0}$.

If it is a zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{2}}}$ , then $\text{p}\left( \text{0} \right)$ should be $\text{0}$.

$\text{ p}\left( \text{0} \right)\text{ = }{{\left( \text{0} \right)}^{\text{2}}}\text{ = 0}$

Hence, we can say that $\text{x = 0}$ is a zero of the given polynomial.

vi) $\text{p(x) = lx + m, x = -}\frac{\text{m}}{\text{l}}$

Ans: We are given: $\text{x = -}\frac{\text{m}}{\text{l}}$.

If it is a zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = lx + m}$ , then $\text{p}\left( \text{-}\frac{\text{m}}{\text{l}} \right)$ should be $\text{0}$ .

Here, $\text{p}\left( \text{-}\frac{\text{m}}{\text{l}} \right)\text{ = l}\left( \text{-}\frac{\text{m}}{\text{l}} \right)\text{ + m = -m + m = 0}$

Hence, we can say that $\text{x = -}\frac{\text{m}}{\text{l}}$ is a zero of the given polynomial.

vii)$\text{p(x)=3}{{\text{x}}^{\text{2}}}\text{-1,x=-}\frac{\text{1}}{\sqrt{\text{3}}}\text{,}\frac{\text{2}}{\sqrt{\text{3}}}$

Ans: We are given:$\text{x = }\frac{\text{-1}}{\sqrt{\text{3}}}$ and $\text{x = }\frac{\text{2}}{\sqrt{\text{3}}}$.

If they are zeroes of the polynomial $\text{p}\left( \text{x} \right)\text{ = 3}{{\text{x}}^{\text{2}}}\text{-1}$, then $\text{p}\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)$and $\text{p}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)$ should be $\text{0}$ .

$\text{p}\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)\text{ = 3}{{\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)}^{\text{2}}}\text{-1 = 3}\left( \frac{\text{1}}{\text{3}} \right)\text{-1 = 1-1 = 0}$

$\text{p}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)\text{ = 3}{{\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)}^{\text{2}}}\text{-1 = 3}\left( \frac{\text{4}}{\text{3}} \right)\text{-1 = 4-1 = 3}$

Hence, we can say that $\text{x = }\frac{\text{-1}}{\sqrt{\text{3}}}$ is a zero of the given polynomial.

However, the value of $\text{x = }\frac{\text{2}}{\sqrt{\text{3}}}$ is not a zero of the given polynomial.

viii) $\text{p(x) = 2x+1, x = }\frac{\text{1}}{\text{2}}$

Ans: We are given: $\text{x = }\frac{\text{1}}{\text{2}}$.

If it is a zero of polynomial $\text{p}\left( \text{x} \right)\text{ = 2x+1}$ , then $\text{p}\left( \frac{\text{1}}{\text{2}} \right)$ should be $\text{0}$

Here, $\text{p}\left( \frac{\text{1}}{\text{2}} \right)\text{ = 2}\left( \frac{\text{1}}{\text{2}} \right)\text{+1 = 1+1 = 2}$.

So, we get the value $\text{p}\left( \frac{\text{1}}{\text{2}} \right)\ne \text{0}$ .

Hence, we can say that $\text{x = }\frac{\text{1}}{\text{2}}$ is not a zero of the given polynomial.

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

i) $\text{p}\left( \text{x} \right)\text{ = x+5}$

Ans: If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{x+5 = 0}$

$\Rightarrow \text{x = -5}$

Therefore, $\text{x = -5}$ is a zero of the given polynomial.

ii) $\text{p}\left( \text{x} \right)\text{ = x-5}$

$\Rightarrow \text{x-5 = 0}$

$\Rightarrow \text{x = 5}$

Therefore, $\text{x = 5}$ is a zero of the given polynomial.

iii) $\text{p}\left( \text{x} \right)\text{ = 2x+5}$

$\Rightarrow \text{2x+5 = 0}$

$\Rightarrow \text{x = -}\frac{\text{5}}{\text{2}}$

Therefore, $\text{x = -}\frac{\text{5}}{\text{2}}$ is a zero of the given polynomial.

iv) $\text{p}\left( \text{x} \right)\text{ = 3x-2}$

$\Rightarrow \text{3x-2 = 0}$

$\Rightarrow \text{x = }\frac{\text{2}}{\text{3}}$

Therefore, $\text{x = }\frac{\text{2}}{\text{3}}$ is a zero of the given polynomial.

v) $\text{p}\left( \text{x} \right)\text{ = 3x}$

$\Rightarrow \text{3x = 0}$

$\Rightarrow \text{x = 0}$

Therefore, $\text{x = 0}$ is a zero of the given polynomial.

vi) $\text{p}\left( \text{x} \right)\text{ = ax, a}\ne \text{0}$

$\Rightarrow \text{ax = 0}$

It is also given that $\text{a}$ is non-zero.

vii) $\text{p}\left( \text{x} \right)\text{ = cx+d, c}\ne \text{0, c, d}$ are real numbers.

$\Rightarrow \text{cx+d = 0}$

It is also given that $\text{c}$ is non-zero.

$\Rightarrow \text{x = -}\frac{\text{d}}{\text{c}}$

Therefore, $\text{x = -}\frac{\text{d}}{\text{c}}$ is a zero of the given polynomial.

## Class 9 Maths Chapter 2 - Exercise 2.2

Vedantu’s Class 9 Maths Chapter 2 Exercise 2.2 includes the following problems from polynomial-

Finding the value of polynomial when the value of the variable is randomly fixed by the teacher,For example, if I ask you to find the value of 2x + 3x when I have randomly set the value of X as 5, you will put the value - 5 in place of X and then multiply 2 and 3 with 5.

Finding whether a particular value of the variable will result in the polynomial to be equated as For example, if X equals to 5 in the equation 2X - 10, then the X = 5 is a zero of the given polynomial.

Finding which value of the variable will result in the polynomial to be equated as 0. For example, if I want to make the value of 3X+1-7 then I have to assign the value of X as 2.

## What are Polynomials? Constants and Variables

Before discussing polynomials, it is necessary that you have a clear idea about variables and exponents. Consider the following equation -

2x + 3x = 10

Here we don't know at first what X is. When we solve the equation, we come to the conclusion that here X is 2. So, what X does here is it answers the ‘what’ of a question. In this equation, the question that is implied is - what should we multiply 2 and 3 with so that the addition of the product results in 10?

Again when we are confronted with the following equation -

2x + 3x = 20

the result of X changes. So, the value of X varies the situation.

This is what we call variable. This variable is generally denoted by x,y,z etc.

Again, if you look at the above two examples, the values of 2 and 3 never change. In both situations, the values of 2 and 3 remain fixed. These are called constants.

Now, look at this,

Here the value of 2 increases exponentially ( 2*2*2). The power 3 is known as an exponent.

If you look at the algebraic expression - 2x + 3x - written above, you will see there are two parts in it - 2x and 3x. These parts are called terms.

Polynomial is a mathematical expression that has multiple terms and consists of constants, variables and non-negative or non-fractionated exponents. The polynomial will only involve addition, subtraction or multiplication.

So 2x + 3x is a polynomial in the variable X.

Again, 2³ + X³ + 3² is a polynomial.

But X + 1∕X is not a polynomial since 1/x can be written as . As said earlier, a polynomial does not allow negative exponents.

Expressions that have √X are not polynomials too since √X can be written as and we know that this is not accepted by polynomials.

If you look at the expression 2x + 3x, the numbers before the variables are called coefficients.

## Class 9 Maths Chapter 2 – Other Exercises (2.1 and 2.3)

Here are a few other exercises that you will find in the NCERT Solutions for Class 9 Maths Chapter 2 PDF –

Exercise 2.1

In this exercise of Class 9 Maths Chapter 2, students will be introduced to the concept of polynomials. They will obtain an indent knowledge regarding polynomials in one variable and how to solve the associated sums through shortcut techniques. Here are some important questions for exams that are provided in the exercise.

Question 1 : Finding the polynomials among the expressions in one variable and stating reasons for the answers.

Question 2 : Finding the coefficients of x2 in the provided expressions.

Question 3 : Examples of a binomial of degree 35 and monomial of degree 100.

Question 4: Determining the degree of polynomials.

Question 5 : Classifying linear, quadratic and cubic polynomial expressions.

Exercise 2.3

This chapter is a continuation of Exercise 2.2 Class 9 of NCERT maths book. Students will get in-depth knowledge about the Factor Theorem by solving all questions provided in this exercise step by step.

Question 1 : Finding the remainder when divided by expressions provided.

Question 2: Determining remainders.

Question 3: Checking if provided expressions are factors.

## How Can Vedantu Help You?

Vedantu's Polynomials Class 9 solutions will help you understand all the topics included in the Polynomials chapter. NCERT guidelines specifically advise teachers to build the fundamental knowledge of the subjects in the students. Vedantu follows this NCERT advice thoroughly. That is why you will find not only solutions to the questions on polynomials but also the explanations on the logic behind our solutions. For example - in question number 2 of Ex 2.1 of our Polynomials Class 9 NCERT Solutions when we are asked to find the coefficients of given polynomials we showed how has an invisible 1 before it. So the coefficient of X² is 1.

Our answers in Class 9th Maths NCERT Solutions Chapter 2 include every step. No step has been skipped. You will not struggle to understand how we reached the result from a given equation. So in question 1(i) of Ex 2.2, we showed how we arrived at the answer 3 by solving the equation step by step. This will not only help the students to understand the granularities in the chapter but will also help them to get good marks.

Every Polynomial formula Class 9 has been used in our solution. The students will get to learn all the aspects of the chapter. Reading our PDF, a student can answer any question from this chapter with confidence.

The Polynomials Class 9 PDF is written in a lucid language. Apart from the Maths expressions, we have used simple, easy to understand words so that students can understand the solutions. Even the math expressions are adequately explained wherever deemed fit.

The Class 9 Maths Chapter 2 PDF has accurate solutions to every polynomials problem. The answers are written by expert teachers who know how to solve a Maths problem so that students understand the process and even the examiners get impressed by the detailed process. These teachers are well aware of NCERT guidelines and follow them to write their answers. There is no unnecessary information in our solutions.

Vedantu’s Polynomial Class 9 PDF is neatly organised. The solutions are written in an uncluttered, easy to understand way. That is why in question 1 of Ex2.3, you will find that we have used red ink to indicate the cutting off of -2x² and +2x². This visually pleasurable way of reading the solutions will help the students to maintain their focus on the solutions.

You can use our NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 to learn all the intricacies of polynomials. These solutions can provide you with everything that you need to excel in the chapter. The answers in our PDF are not written by part-time bloggers, but they are written by professional teachers who took care to make the solutions as helpful as possible.

The Class 9 Maths Chapter 2 Exercise 2.2 Solutions are absolutely free of cost. You don’t need to pay a single penny to download our PDF. You can simply enter your email id and we will send the download link to your email. Vedantu is not motivated by profit; we are genuinely interested in making the learning process of students better. We believe every Indian student has the right to get a quality education.

## Learn Online with Vedantu’s Online Class

Our PDF on polynomials is exhaustive and can help you in finding solutions to all the questions asked in the textbook. Maths needs explanation - countless of times. You might want to ask questions about a particular solution in our PDF. So to help you, Vedantu holds online classes where you can ask questions directly to the teachers. This interaction is particularly necessary for understanding Maths chapters. The schedule of the classes is announced beforehand. We have helpful videos too which will help you understand the solutions in a much more easy and efficient way.

Our online teachers are experts in their subjects and come from respectable institutions like IITs. These teachers know how to teach Maths to students in such a way that students do not get bored. The teachers are aided by simultaneous images, videos and slides. This visual way of teaching ensures easy retention in students.

With our Master Class, you can experience a whole new learning experience. We are offering free seats in our master class for a limited period of time. These masterclasses can enhance your knowledge in ways you cannot fathom. We do not employ cheap tricks and tips. Our main focus is on building the basic knowledge of the students. Our Master Class is not just for advanced students, intermediate students can get benefit from the Master Classes as they will shed fear for Maths and grow a love for the subject.

Vedantu also has its app. This application will help the students to stay in the learning mode 24*7. The app, as of now, boasts of 1 million-plus downloads on Play Store.

Vedantu is here to help the students move forward with confidence. With Vedantu, you have a golden opportunity to make the pillars of your knowledge strong.

NCERT Solutions for Class 9 Maths

Chapter 1 - Number System

Chapter 3 - Coordinate Geometry

Chapter 4 - Linear Equations in Two Variables

Chapter 5 - Introductions to Euclid's Geometry

Chapter 6 - Lines and Angles

Chapter 7 - Triangles

Chapter 8 - Quadrilaterals

Chapter 9 - Areas of Parallelogram and Triangles

Chapter 10 - Circles

Chapter 11 - Constructions

Chapter 12 - Heron's formula

Chapter 13 - Surface area and Volumes

Chapter 14 - Statistics

Chapter 15 - Probability

## FAQs on NCERT Solutions for Class 9 Maths Chapter 2: Polynomials - Exercise 2.2

1. Give a brief description of the topic.

Polynomial is acquired from the word “poly” which means “many” and the word “nominal” refers to “term”. In Math subject, a polynomial expression consists of variables which are also known as coefficients and indeterminates. The coefficients require the operations of addition, subtraction, non-negative integer exponents of variables and multiplication.

Polynomials are utterances with one or more phrases with a non-zero coefficient. A polynomial can also have one or more than one number of terms. In the form of a polynomial, each utterance in it is known as a term. Suppose x 2 + 4x+ 2 is polynomial, then the expressions x 2 , 4x, and 2 are the terms of the polynomial. Each term of the polynomial has a coefficient. For example, if 4x+1 is the polynomial, then the coefficient of x is 4.

2. What are the important terms of a polynomial?

A term might either be a variable or a single digit or it also could be a sequence of the variable with digits.

The degree of the polynomial is the most distinguished power of the variable in a polynomial.

A polynomial of degree 1 will be termed as a linear polynomial.

A polynomial of degree 2 will be described as a quadratic polynomial.

A polynomial of degree 3 is termed as a cubic polynomial.

A polynomial of 1 term is termed as a monomial.

A polynomial of 2 terms is termed as binomial.

A polynomial of 3 terms is termed as a trinomial.

A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0, where a is also termed as root of the equation p(x) = 0.

A linear polynomial in one variable holds an individual zero, a polynomial of a non-zero constant has no zero, and every real character is a zero of the zero polynomial.

3. What are the topics that are covered in this chapter?

Polynomial is an algebraic expression which includes constants, variables and exponents. It is the expression where the variables have just certain elemental powers. The topics that are covered in this chapter are:

Introduction

Polynomials in One Variable

Zeros of Polynomials

Remainder Theorem

Factorisation of Polynomials

Algebraic Identities

Polynomial expressions are algebraic equations which will have two or more terms with the equal variables of different exponents. This is one of the notable topics for class 9 Mathematics, which students require to learn in order to gain profound knowledge about complex algebraic expressions. Mathematics is one of those subjects that serve as a foundation of numbers, analysis, figures and logic.

4. Are the answers provided by Vedantu, sufficient to attain accurate marks?

Our answers are made to the point and they are drafted to aid you from the exam portion of the view. Answers to the exercising questions are certainly given with examples and they are 100% curate. Our answers will make your learning easy for the exam as they are fitted to be compatible with the tips given with the assistance of using CBSE maths Syllabus and NCERT Book.

Our answers will assist you in advancing a conceptual basis with all the principal ideas in a comprehensible language. The exercising covers all of the vital topics and subtopics of the chapter which might occur for your Class 9 maths exams. You also can clear all of your doubts from here and in this manner, you can definitely perform well in your Class 9 maths exam.

5. Do I need to practice all the questions given in the NCERT Solutions for Class 9 Maths Chapter 2?

The NCERT textbook is the best textbook to prepare for the final exams as it is where most of the questions are asked in the final exam. The exercises and the examples are explained in a simpler manner for all students to understand. Important concepts are also explained in the PDFs. It is important that you practice all the questions and concepts while preparing for the exam. You should not leave any questions. Being thorough with all the questions will help you score good marks in the final exam. You can visit the Vedantu website or download the Vedantu app to access these resources at free of cost.

6. How many exercises are there in this chapter?

Chapter 2 of Maths from the NCERT textbook has three exercises- 2.1, 2.2 and 2.3. Each exercise and its solution is explained step by step in the NCERT Solutions for Class 9 Maths Chapter 2. Not a single question or step has been skipped where they are explained. The examples from within the chapter are also explained. The solutions have been written by professionals with the aim of making it easier for all the students depending on their calibre to understand.

7. How many questions are there in each exercise?

There are a total of three exercises in the Chapter 2 of Class 10 NCERT Maths textbook. Exercise 2.1 has five questions that cover introduction to polynomials and linear and quadratic linear expressions. Exercise 2.2 has four questions that mainly concentrate on finding the value of the polynomials. The last exercise is 2.3 which focuses on the factor theorem. The questions ask to find the remainders and their division.

8. What are constants and variables?

Constants are the numeric values attached to a variable (x,y) in any equation. They stay the same and do not change. They help you determine the value of the variable. Variables are the X or Y attached to a constant whose value needs to be determined. It may vary in different situations. It is a result of the constants and the answer on the other side that helps determine the value of the variables.

9. Is Class 9 Maths NCERT Solutions enough to revise for the final exam?

The NCERT Solutions for each subject and chapter provides summaries and revision notes for the students to refer to and prepare from. These are sufficient as they cover every aspect of the chapter and do not exclude any topic. The students can take the help of the Class 9 Maths NCERT Solutions to prepare for the exam and revise by going through the revision notes provided. The formulas and concepts are explained in an easy and systematic manner.

## NCERT Solutions for Class 9

Cbse study materials for class 9, cbse study materials.

## Mathematics Part II (Solutions) Solutions for Class 9 Math

- Textbook Solutions
- class 9 math

Mathematics Part II (Solutions) Solutions are considered an extremely helpful resource for exam preparation. Meritnation.com gives its users access to a profuse supply of Mathematics Part II (Solutions) questions and their solutions. MAHARASHTRA Board Class 9 math Mathematics Part II (Solutions) Solutions are created by experts of the subject, hence, sure to prepare students to score well. The questions provided in Mathematics Part II (Solutions) Books are prepared in accordance with MAHARASHTRA Board, thus holding higher chances of appearing on MAHARASHTRA Board question papers. Not only do these Mathematics Part II (Solutions) Solutions for Class 9 math strengthen students’ foundation in the subject, but also give them the ability to tackle different types of questions easily.

Our MAHARASHTRA Board Class 9 math textbook solutions give students an advantage with practical questions. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.

- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 1 - Basic concepts in Geometry
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 2 - Parallel Lines
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 3 - Triangles
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 4 - Constructions of Triangles
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 5 - Quadrilaterals
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 6 - Circle
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 7 - Co-ordinate Geometry
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 8 - Trigonometry
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 9 - Surface Area and Volume

## NCERT Solutions for Class 9 Maths Exercise 9.2 Circles

NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Circles updated for new academic session 2023-24. Students can get here Class 9 Maths Ex. 9.2 in Hindi and English Medium based on rationalised syllabus and new NCERT textbooks for CBSE and State Boards 2023-24.

## Class 9 Maths Exercise 9.2 NCERT Solutions

- Class 9 Maths Exercise 9.2 in English
- Class 9 Maths Exercise 9.2 in Hindi
- Class 9 Maths Chapter 9 in PDF
- Class 9 Maths Chapter 9 Solutions
- Class 9 Mathematics NCERT Solutions
- Class 9 all Subjects NCERT Solutions

## Class 9 Maths Exercise 9.2 Questions in Detail

- Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
- If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
- If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
- If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD.
- Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
- A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Before venturing into next chapter we should know that we need to learn theorems from NCERT solutions Maths for class 9th states that chords of equal length of a given circle or of a congruent circles are equidistant from the center (or centers). This is a simple theorem and you can verify by actually drawing a circle and drawing two chords of same lengths.

Theorem says, Chords which are equidistant from the center are equal in length. This is just the converse theorem 6. It is not a difficult theorem but slightly more in difficulty level compared to theorem 6.

Now based on these two theorems exercise 9.2 contains 6 questions. Out of which the first 3 are seemingly easy and the last three needs little practice and application as well.

The above theorems are based on some simple observations. For example – if you want to draw a shortest line from a particular point to a straight line, then you should draw a perpendicular from that point that straight line.

Practically speaking suppose you are crossing a road and there is a vehicle coming from one side. Then it is better to let the vehicle go by.

But if you stuck in between then the shortest to route cross will be the perpendicular distance will be one end to another. Thus we see that geometry have practice application in life.

It is suggested that students revise the old theorems as much as possible because to prove the new theorems i.e., the theorems under consideration we should be familiar with the theorems learnt in the past.

Copyright 2024 by Tiwari Academy | A step towards Free Education

- NCERT Solutions
- NCERT Class 9
- NCERT 9 Maths
- Chapter 2: Polynomial
- Exercise 2.1

## NCERT Solutions for class 9 Maths Chapter 2- Polynomials Exercise 2.1

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.1 are provided here. These NCERT Maths solutions are prepared by our subject experts that make it easy for students to learn. The students can use it for reference while solving the exercise problems. The first exercise in NCERT Class 9 Maths Solutions Chapter 2 , Polynomials – Exercise 2.1, discusses Polynomials in one or more variables.

The solutions provide detailed and step-wise explanations of each answer to the questions given in the exercises in the NCERT textbook for Class 9. The NCERT solutions are always prepared by following the guidelines so that students can cover the whole syllabus accordingly. These are very helpful in scoring well in board examinations.

## NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials Exercise 2.1

carouselExampleControls111

Previous Next

## Access Answers to Maths NCERT Class 9 Chapter 2 – Polynomials Exercise 2.1

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 powers 10, 3, and 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.

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

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.

The coefficients of x 2 in (π/2)x 2 +x is π/2.

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

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

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 of 35 is called a binomial of degree 35.

E.g., 3x 35 +5

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

E.g., 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.

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.

the degree is 2

Hence, y+y 2 +4is 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.

## Access Other Exercise Solutions of Class 9 Maths Chapter 2 – Polynomials

Exercise 2.2 Solutions 4 Questions

Exercise 2.3 Solutions 3 Questions

Exercise 2.4 Solutions 5 Questions

Exercise 2.5 Solutions 16 Questions

## Key Advantages of NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials Exercise 2.1

- The NCERT Class 9 Maths Solutions help students solve and revise all questions of Exercise 2.1.
- After going through the stepwise solutions given by our subject expert teachers, students will be able to get more marks.
- They help to do very well in Maths Board exams.
- They follow NCERT guidelines which help in preparing the students accordingly.

## Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Its very help full 👍

- Share Share

## Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Counselling

## IMAGES

## VIDEO

## COMMENTS

All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Class 9. Students can click on the links below and download all Pdf Assignments for Mathematics class 9 for free. All latest Kendriya Vidyalaya Class 9 Mathematics Assignments with Answers and test papers are given below.

These NCERT Solutions for Class 9 cover all the topics included in the NCERT textbook, like Number System, Coordinate Geometry, Polynomials, Euclid's Geometry, Quadrilaterals, Triangles, Circles, Constructions, Surface Areas and Volumes, Statistics, Probability, etc. With the help of these Solutions of NCERT Books for Class 9 Maths, students ...

NCERT Maths Class 9 Textbook Solutions is solved by expert teachers provide you a strong foundation in the subject Maths. The 9th CBSE Maths Solutions are solved keeping various parameters in mind such as stepwise marks, formulas, mark distribution, etc., This in turn, helps you not to lose even a single mark.

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. (ii) 4-y2. Solution: The highest power of the variable in a polynomial is the degree of the polynomial. Here, in 4-y 2, The power of the variable y is 2.

Step 2: Verify your answers and learn about the Exam Pattern. Step 3: Revise the syllabus and use class 9 Maths NCERT Solution as a Revision Aid. Step 4: Make Effective Self Study in 9th Maths using NCERT Solutions. Step 5: Confined to new rationalised NCERT Textbooks and solve it using NCERT Solutions.

6 exercises are included in the NCERT Solutions for Class 9 Maths Chapter 1 which are Exercise 1.1 to 1.6. Most of the questions are application based and a few are memory-based questions to test students' application skills and cognizance. Chapter 1 - Number System Exercises in PDF Format. Exercise 1.1.

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2 provide a detailed and step-wise explanation of each answer to the questions given in the exercises in the NCERT textbook for Class 9. These solutions help the students memorising the method in which the different types of questions in this exercise are solved.

NCERT solutions for class 9 Maths are exclusively updated for the current academic session 2022-23. Get precise and detailed NCERT Solutions for all chapters of Class 9 Maths in PDF.

CBSE Class 9 Mathematics Chapter 2 Polynomials solutions include a substantial number of solved questions that span the complete syllabus in the form of graded exercises and step-by-step explanations. Vedantu's goal is to clarify the chapter's key subject and to help students build problem-solving abilities. Class:

NCERT Solutions App. Chapter-wise NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 solved by Expert Teachers as per NCERT (CBSE) Book guidelines. Class 9 Maths Chapter 2 Polynomials Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

on January 18, 2024, 9:04 AM. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials in Hindi and English Medium revised and modified for academic year 2023-24. The solutions offered by Tiwari Academy often aim to simplify complex mathematical concepts, making it easier for students to understand and grasp the subject matter.

Download Freeilm.com Android App. Download Class 9 Maths, Chapter 2 Notes, Real and Complex Numbers that contains Solutions of All Exercises, Review Exercises, MCQ's in PDF for free.

Exercise 2.3. This chapter is a continuation of Exercise 2.2 Class 9 of NCERT maths book. Students will get in-depth knowledge about the Factor Theorem by solving all questions provided in this exercise step by step. Question 1: Finding the remainder when divided by expressions provided.

Get NCERT solutions for class 9 for all subjects: Maths, Science, Social Science, English & Hindi. We provide chapter-wise NCERT Solutions for class 9 for all NCERT books. NCERT Solutions for class 9 provide you with a quick way to complete your homework. All solutions are prepared by experts and easy-to-understand.

These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students' understanding. Mathematics Part 2 Solutions Textbook Solutions for Class 9 MATH. Homework Help with Chapter-wise solutions and Video explanations.

NCERT Class 9 Solutions for Maths, Science and Social Science. The NCERT Book is a good resource for students. It explains each chapter in a simple and easily understandable way. At the end of each chapter, there are exercise problems which contain questions based on it. These questions are provided to check whether students have understood the ...

Here is the list of Extra Questions for Class 9 Maths with Answers based on latest NCERT syllabus prescribed by CBSE. Chapter 1 Number Systems Class 9 Extra Questions. Chapter 2 Polynomials Class 9 Extra Questions. Chapter 3 Coordinate Geometry Class 9 Extra Questions. Chapter 4 Linear Equations for Two Variables Class 9 Extra Questions.

NCERT Solutions App. Chapter-wise NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 solved by Expert Teachers as per NCERT (CBSE) Book guidelines. Class 9 Maths Chapter 2 Polynomials Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

Class 9 Maths Exercise 9.2 Questions in Detail. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Ex 2.1 Class 9 Maths Question 2. (i) The given polynomial is 2 + x 2 + x. The coefficient of x 2 is 1. (ii) The given polynomial is 2 - x 2 + x 3. The coefficient of x 2 is -1. (iii) The given polynomial is π 2x2 + x. The coefficient of x 2 is π 2. (iv) The given polynomial is √2 x - 1. The coefficient of x 2 is 0.

These NCERT Maths solutions are created by our subject experts, who make it easy for students to learn. The students use NCERT Class 9 Maths Solutions for reference while solving the exercise problems. The fifth exercise in Polynomials- Exercise 2.5 discusses the Algebraic Identities. The experts provide a detailed and stepwise explanation of ...

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