NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.2) Exercise 2.2

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, which are based on the chapter titled “Polynomials,” underline the fact that this is a crucial chapter included in the Class 9 Mathematics curriculum. To achieve higher results in the Class 9 Mathematics examinations, every student must study these NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. The chapter aids students in grasping the foundational concepts needed to comprehend challenging mathematical themes in higher grades. All of the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, on the Extramarks website and Learning App have been specially crafted by subject-matter experts.

Polynomials in One Variable, the Remainder Theorem, Polynomial Zeroes, Polynomial FactoriSation, and many other themes are covered in NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. A collection of practice questions from Extramarks is available for all the subtopics, along with the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. Students should concentrate on reviewing all of the solutions to the in-text questions that Extramarks provides.

The most crucial concepts are covered in NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. The principles of the subject must be understood by students in order to aid their long-term preparation for competitive tests. While the concepts may initially appear challenging to grasp, as students regularly practice different kinds of problems, it will become simpler for them to find the answers with the help of the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2.

Extramarks provides answers to all textbook questions based on Class 9 Mathematics Chapter 2. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, are offered for practice and reference and are based on a consideration of the nuances of the most recent syllabus. Various educational resources, such as interactive 3D videos, NCERT Exemplars, Extramarks Explainers, etc., are all available on the Extramarks learning portal for students to practice with.

The following are some key ideas to take away from the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2:

  • Assume that f(x) is a polynomial with a degree greater than or equal to 1 and that a is any real number. The Remainder Theorem states this: A division of f(x) by (xa) results in the remainder (a) equaling f(x)
  • Factor Theorem: If “a” is a real number such that f(a)=0 and “f(x)” is a polynomial of degree higher than or equal to one, then (xa) is a factor of f. (x). On the other hand, f(a)=0 if (xa) is a factor of f(x).
  • Polynomial zeros: If f()=0, then a real number is the zero (or root) of the polynomial f(x).
  • There can be a maximum of n roots in a polynomial of degree n.
  • A constant polynomial that is non-zero has no root.
  • The zero polynomial has roots equal to all real numbers.

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 by Extramarks, offer a thorough and step-by-step explanation of each response to the problems in the exercises. These NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 aid students in remembering the approach used to address the various types of questions in the designated exercise. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 are created in accordance with NCERT guidelines to ensure that the entire syllabus is appropriately covered. These NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 are highly beneficial for getting high marks in board examinations.

In the CBSE Class 9 Mathematics curriculum, Polynomials are covered in Chapter 2. It is a highly significant chapter that is organised into eight major major divisions. To master the theme of polynomials, it is highly recommended that students repeatedly study and practice the solutions provided in the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2.

To provide students with a preview of the key themes covered in the chapter on polynomials, which is also the central theme of NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, the list below has been given:

  • Definition
  • 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
  • Monomials
  • Binomials
  • Trinomials
  • Value of a polynomial
  • Zero of a polynomial
  • Operations on polynomials

Ideally, students should optimally utilise NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 in order to be able to tackle challenging polynomial-based mathematical problems with ease.

Students learn about Polynomials in the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. Examinees are made aware of every concept of polynomials, whether it be discovering value, applying algebraic identities, or factorisation of polynomials, by Extramarks’ NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, by Extramarks provide comprehensive explanations of polynomials in a straightforward manner. For all of the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, Extramarks offers students a PDF download option.

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, based on Polynomials answer questions pertaining to the idea of the Polynomial zero, which requires that the polynomial be equated to zero. The value obtained after solving the equation is then referred to as the polynomial’s specific “zero” or root. The solutions to a total of 22 questions have been encapsulated in the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, 16 of which are brief and involve a basic application of logic while the remaining 6 are of medium difficulty.

NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 (Ex 2.2) (include PDF)

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, focused on Polynomials, are a good way to comprehend the fundamentals of polynomials because these solutions are inclusive of straightforward methods that can be practised routinely. Students are advised to regularly study with the aid of NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. The basic themes covered in NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, include inquiries about polynomial degrees and zeros. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, provide thorough examples that will aid students in comprehending the logic behind the questions raised in the exercises.

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 are accessible on the Extramarks website and the Extramarks Learning App. Students just need to register themselves on the Extramarks website or the mobile application to be able to access the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2.

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, are part of the Class 9 Mathematics NCERT Solutions on Extramarks. Students can thus access the Class 9 Mathematics NCERT Solutions to acquire the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, conveniently. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, are therefore extremely useful while doing homework and studying. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, are written in a well-versed manner to make it easy for students’ comprehension.

Verification of zeroes of the specific equation is covered in NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. Finding the functional value of the equation is a major thematic component of the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. Solutions to the following exercises are provided on Extramarks in addition to NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2:

  • Polynomials Exercise 2.1
  • Polynomials Exercise 2.3.
  • Polynomials Exercise 2.4
  • Polynomials Exercise 2.5

Access NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials

Obtaining the expression of polynomial equations is covered in NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. In the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, the first type of question is about determining the value of a polynomial expression with a single variable. Further, the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, also include the method to solve a polynomial equation with double- and triple-degree zeros.

These are some of the reasons why students should refer to NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2:

  • The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, are essential since they contain comprehensive solutions to the problems from Chapter 2, which will provide students with a better knowledge of numbers besides whole numbers.
  • Students will be able to excel in exams if they follow the method of solutions provided in the NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, can also be used for efficient revision during examinations.
  • Students may encounter short-answer or lengthy questions from the types of solutions covered in the NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 in their final exam question papers.

Class 9 Maths Chapter 2 – Exercise 2.2

The fundamental techniques of counting, measuring, and describing the shapes of objects have given rise to Mathematics, the Science of structure, order, and relation. It deals with quantitative calculations and Logical Reasoning. Mathematics has been a necessary complement to the Physical Sciences and Technology since the 17th century, and more recently, it has taken on a similar role in the quantitative components of the Life Sciences.

Mathematics has advanced much beyond simple counting in many cultures, owing to the demands of practical endeavours like business and agriculture. Students who are striving to pursue a career in mathematics are therefore encouraged to seek the guidance of NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 by Extramarks for better results.

What are Polynomials? Constants and Variables

A Polynomial may or may not result from the division of two polynomials. The long division method is employed when a polynomial has multiple terms. The steps of the same are as follows.

  • Students should begin by writing the given polynomial.
  • Then, they are required to divide the terms by the highest power after checking.
  • As the division sign, students ought to use the response from step 2.
  • Students must subtract that now before bringing down the next term.
  • Up until there are no more phrases to carry down, students are required to repeat steps 2 through 4.
  • Students must take note that the complete response, including the remainder, will be in fraction form (last subtract term).

To further explain the definition of polynomials, a polynomial in mathematics is an expression made up of variables, coefficients, and constants that are connected by addition, subtraction, multiplication, and division. The following is a list of the several terminologies connected to polynomials:

  • Terms: A term in an expression can be either a variable, a constant, or a combination of both.
  • Coefficient: A coefficient is a number that is used in the same sentence as a variable.
  • Variable: A variable is a letter that stands in for an expression’s unknowable value.
  • Constant: In an equation, a constant is an integer whose value never changes.

The classification of Polynomials based on Degree in One Variable can be described as follows: The degree of a polynomial can be used to categorise the polynomials in a single variable. Students can take a look at a polynomial’s degree before delving further into its categorisation. The largest power of a polynomial’s variable is referred to as the polynomial’s degree. The highest sum of the different variables in any of the terms if a polynomial has more than one variable.

Algebraic expressions called Polynomials include coefficients and variables. Indeterminates are another name for variables. For polynomial expressions, students can do mathematical operations like addition, subtraction, multiplication, and positive integer exponents but not division by variables. x2+x-12 is an illustration of a polynomial with a single variable. There is no limit to the number of terms that can exist in a polynomial.

The Class 9 Maths Chapter 2 Exercise 2.2 Solutions serve as an excellent guide for Exercise 2.2 Class 9 Maths. Class 9 Maths Chapter 2 Exercise 2.2 is a crucial part of the syllabus, and the solutions to the polynomials in Class 9 Exercise 2.2 are available on the Extramarks learning application and website.

For extensive details on the topic of classification of Polynomials, students are encouraged to acquire the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 on the Extramarks website and learning application. Polynomials can be classified into numerous categories.

  • Zero Polynomial or Constant Polynomial:

The polynomial is known as a zero or constant polynomial if its degree is zero (0). These polynomials solely contain constants. There are no variables in them.

  • Linear Polynomial:

A polynomial is referred to as a linear polynomial if its degree is 1 (one). There is just one solution to the linear polynomial in one variable.

  • Quadratic Polynomial:

A quadratic polynomial is a polynomial with a degree of 2 as its maximum. One-variable quadratic polynomials only have two solutions. Examples of quadratic polynomials in one variable include the following:

  • 9×2 – 10
  • x2 +5x+9
  • m2+25
  • Cubic Polynomial:

A polynomial is referred to as a cubic polynomial if the maximum exponent of any variable in the polynomial is 3, or if the degree of the polynomial is 3. The number of solutions to a cubic polynomial in one variable is three. Examples of cubic polynomials in one variable include the following:

  • 7×3– 21
  • 8×3+2x+9
  • 10m3 + (5/4)

Students can refer to the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, for the definition and extensive detail of a constant. A constant is defined as an entity whose value remains constant regardless of its usage. These values do not change even after the change in expression. Constants are numerical values like a natural number, whole number, real number, integer, or decimal number. The portion of an algebraic equation that only contains integers is known as a constant. These are referred to as constants because their values are unchangeable. It is unambiguous. The constant’s value cannot be altered by any variables in the term. The fact that everyone is aware of the value of the number 8 makes it a constant. It is unchangeable. The second term, 4, is the constant in the formula 3XY – 4 = 2Y. 3XY cannot be a constant because it depends on the values of X and Y, which changes the whole expression.

A variable is an entity whose value isn’t fixed and changes as its usage changes.The letters in an algebraic expression stand in for variables. These terms are not constant; occasionally, their value may change. Variables are represented by the values x and y in the formula 2y+3x=0. Y might be 2, 3, or any other integer, and x could also be any number that the equation accepts. In an algebraic expression, the letters can be replaced by more than one value.

Variables are things whose values are unknown; as a result, they have no numerical value and are denoted by letters of the alphabet such as a, b, c, and so on. Algebraic expressions include constants and variables, which are regarded as the fundamental building blocks of these expressions. Students can also find the detailed explanation of the variable in terms of a Polynomial on the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 on the Extramarks website and mobile application.

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

Before they delve deeper into the chapter it is important that students know a little more about Algebra in general. The subject matter of the Mathematics branch known as Algebra is relations, operations, and their constructions. It is one of the fundamental building blocks of Mathematics and has a plethora of uses in daily life. One of the many different areas of Mathematics is Algebra. It focuses on binary relationships, operations, and Mathematical function structures. A student’s education must include algebra. It aids in the enhancement of a general understanding of other fields of Mathematics such as Calculus, Arithmetic, Geometry, and others.

The Arabic word al-jabr, which means “reunion of broken parts,” is from which the English word Algebra is derived. It is the study of numerous operations and relationships, numerous geometric figure constructions, and the clarification and justification of ideas like polynomials, equations, and algebraic structures.

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, are the solutions to an important exercise on polynomials included in the Class 9 Mathematics curriculum. To achieve better results in the Class 9 Mathematics examination, every student must study these NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2, would aid students in grasping the foundational concepts needed to comprehend more complex mathematical topics. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2 offered on Extramarks have been curated by subject-matter experts.

Here are some points to remember before students solve the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.2:

Class 9 Maths Chapter 2 Exercise 2.1:

The Chapter 2 Polynomials Exercise 2.1 NCERT Solutions for Class 9 Mathematics are available on the Extramarks website and mobile application. Subject specialists at Extramarks have created these NCERT Mathematics solutions to make learning simple for students. As they work through the exercise questions with the assistance of these solutions, students can gain proficiency in various implicit concepts. Exercise 2.1, the first in the series on polynomials, covers the theme of polynomials in one or More Variables. Each response to the questions in the exercises in the Class 9 NCERT textbook is given in-depth, step-by-step explanations. The solutions are always created in accordance with NCERT recommendations in order to properly cover the entire curriculum. These are highly beneficial for getting high marks in board examinations.

Class 9 Maths Chapter 2 Exercise 2.3:

NCERT Solutions Chapter 2 exercise 2.3 in Mathematics for grade 9 is based on the Remainder Theorem, which asserts that if a(x) is a polynomial and has a degree of 1 or greater, and is divided by a linear polynomial in the form of x-b, the remainder will equal a. There are 7 questions in Chapter 2 Exercise 2.3 of the NCERT Solutions for Class 9 Mathematics, 5 of which are short answers and 2 of which are a little longer.

Learn Online with Extramarks’ Online Class

The Extramarks digital learning portal is available to students. With the aid of interesting images and animations, the Learning App’s artificial intelligence-based learning assistant lets the user enjoy a tailored learning experience. The use of interactive learning modules enables students to get all of their questions answered right away. Through visual and interactive learning modules, the education software promises to address all of the students’ problems. By offering homework, projects, essays, quizzes, or tasks, Extramarks also improves the effectiveness of self-guided learning. The NCERT solutions, which are highly popular among many users of the learning platform, are also offered on the android software. Solutions for the textbooks of every academic discipline are available on Extramarks.

Extramarks includes all the entire K–12 curriculum from kindergarten through grade 12. The learning portal is well-known for offering scholarly reference materials for a wide range of subjects, including, Science, Mathematics, Social Studies, English (core and elective), Hindi, Sanskrit, Economics, History, Geography, Political Science, Accounting, Business Studies, Physics, Chemistry, and Biology, as well as Computer Science (C++ and Python), Informatics Practises, Multimedia Web, and Technology. The learning application’s practice and study materials cover all topics and include NCERT Solutions. They are aligned with NCERT books. The educational software offers feedback, scorecards, a summary of the student’s performance, a challenge section where users can participate in quiz competitions, and more. The Extramarks mobile application also provides live online classes for students of all grades and in all subjects.

NCERT Solutions for Class 9

All of the questions in the Class 9 NCERT Mathematics textbook have answers in the NCERT Solutions for Class 9 Mathematics. Students can access the chapter-by-chapter solutions to these chapters in PDF format by clicking on the links on the Extramarks website and mobile application. All of the chapters included in the NCERT textbook for Class 9 are covered in these NCERT Solutions, including the Number System, Coordinate Geometry, Polynomials, Euclidean Geometry, Quadrilaterals, Triangles, Circles, Constructions, Surface Areas and Volumes, Statistics, Probability, etc.

Students can practice all types of questions from the chapters with the aid of the resources provided by Extramarks. Proficient mentors of Extramarks have created the CBSE Class 9 Maths Solutions in a well-structured manner to offer a variety of potential approaches to solving mathematical problems, which would also facilitate a thorough comprehension of the chapter. For their exams, it is advised that the students thoroughly practice all of these solutions. Additionally, these learning resources will assist students in laying the groundwork for more challenging courses.

Students are given additional online learning resources, such as notes, books, question papers, sample problems, worksheets, etc., in addition to NCERT Solutions, which are accessible on Extramarks. These materials were created with consideration for the NCERT and CBSE guidelines. Additionally, it is suggested that students practice the CBSE Class 9 sample papers to get a sense of the final examination’s question format.

The NCERT textbook for Class 9 Mathematics contains 15 chapters. These Class 9 Mathematics NCERT textbook chapters serve as a basis for the academic curriculum of Mathematics in Class 10. Students can conveniently secure access to the PDF versions of learning assets provided by Extramarks, and they can download them as well.

CBSE Study Materials for Class 9

The academic curriculum of Class 9 is considered a crucial milestone in a student’s academic career, as it introduces new and challenging ideas and subjects that will be explored further in subsequent classes. Therefore, in order to succeed in the senior classes, it is preferable to acquire a clear understanding of thematic fundamentals in Class 9 itself. For the aforementioned academic year, Extramarks specifically created scholarly content for CBSE Class 9. The updated term-by-term curriculum, NCERT Solutions, additional problems, NCERT chapter notes, and other crucial e-learning resources have been compiled by Extramarks in order to assist students’ learning effectively. Students can inculcate efficacious time management skills with the aid of these study materials.

CBSE Study Materials 

One of the best-known and most esteemed educational boards in the nation is the Central Board of Secondary Education, or CBSE for short. It is a part of the Union Government of India. Numerous public and private schools that are part of the CBSE board use the NCERT curriculum. For students to learn and prepare for the examination more effectively, Extramarks offers CBSE study materials that are developed in an exciting, engaging, and student-friendly manner. Subject matter specialists have created the study materials with the most recent CBSE curriculum in mind. These digital learning assets, which include the syllabus, books, sample papers, test questions, NCERT solutions, NCERT exemplar solutions, important questions, and CBSE notes, are beneficial for all students.

It is strongly advised that students study on a routine basis. They will be able to finish the curriculum on time if they do this. Additionally, they will have adequate time before the annual examination to review and practice.

Q.1 Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = –1
(iii) x = 2

Ans
(i) Putting x = 0, we get
P(0) = 5(0) – 4(0)2 + 3
= 3
(ii) Putting x = –1, we get
P(–1) = 5(–1) – 4(–1)2 + 3
= – 6
(iii) Putting x = 2, we get
P(2) = 5(2) – 4(2)2 + 3
= – 3

Q.2 Find p(0), p(1) and p(2) for each of the following polynomials: (i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t2 – t3 (iii) p(x) = x3 (iv) p(x) = (x – 1) (x + 1)

Ans
(i) p(y) = y2 – y + 1 Putting y = 0, we get
p(0) = (0)2 – (0) + 1 = 1
Putting y = 1, we get
p(1) = (1)2 – (1) + 1 = 1
Putting y = 2, we get
p(2) = (2)2 – (2) + 1 = 3
(ii) p(t) = 2 + t + 2t2 – t3 Putting t = 0, we get
p(0) = 2 + (0) + 2(0)2 –(0)3 = 2
Putting t = 1, we get
p(1) = 2 + (1) + 2(1)2 –(1)3 = 4
Putting t = 2, we get
p(2) = 2 + (2) + 2(2)2 –(2)3 = 4 (iii) p(x) = x3 Putting x = 0, we get
p(0) = (0)3 = 0
Putting x = 1, we get
p(1) = (1)3 =1
Putting x = 2, we get
p(2) = (2)3 = 8
(iv) p(x) = (x – 1) (x + 1) Putting x = 0, we get
p(0) = (0 – 1) (0 + 1) = –1
Putting x = 1, we get
p(1) = (1 – 1) (1 + 1)= 0
Putting x = 2, we get
p(2) = (2 – 1) (2 + 1)= 3

Q.3

Verify whether the following are zeroes of the polynomial, indicated against them.i px=3x+1,x=−13ii px=5x−π,x=45iii px=x2−1,x=1,−1iv px=(x+1)(x−2),x=−1,2v px=x2,x=0vi px=lx+m,x=–mlvii px=3x2−1,x=−13,23viii px=2x+1,x=12

Ans

i px=3x+1,x=−13Putting x=−13 in p(x), we getp −13=3×−13+1  =0Since, remainder is zero. So, x=−13 is a zero of p(x).ii p x=5x−π,x=45Putting x=45 in p(x), we getp45=5×45−π    =4−π≠0Since, remainder is not zero. So, x=45 is not a zero of p(x).(iii) p(x)=x2−1,x=1,−1Putting x=1 in p(x), we get       p1=(1)2−1  =0Since, remainder is zero. So, x=1 is a zero of p(x).Putting x=−1 in p(x), we get  p−1=(−1)2−1 =0Since, remainder is zero. So, x=−1 is a zero of p(x).(iv) p(x)=(x+1)(x−2),x=−1,2Putting x=1 in p(x), we get   p(−1)=(−1+1)(−1−2)  =0Since, remainder is zero. So, x =1 is a zero of p(x).Putting x=2 in p(x), we get     p(2)=(2+1)(2−2) =0Since, remainder is zero. So, x=2 is a zero of p(x).(v)p(x)=x2,x=0Putting x=0 in p(x), we get      p0=(0)2     =0Since, remainder is zero. So, x=0 is a zero of p(x).vi p x=lx+m,x=−mlPutting x=−ml in p(x), we get   p−ml=l−ml+m            =0Since, remainder is zero. So, x=−ml is a zero of p(x).vii px=3x2−1,x=−13,23Putting x=−13 in p(x), we get   p−13=3−132−1              =0Since, remainder is zero. So, x=−13 is a zero of p(x).Putting x=23 in p(x), we get   p23=3232−1                =3≠0Since, remainder is not zero. So, x=−13 is not a zero of p(x).viii px=2x+1,x=12Putting x =12 in p(x), we get        p12=212+1          =1+1          =2≠0Since, remainder is not zero. So, x=12 is not a zero of p(x).

Q.4

Find the remainder when x3+3x2+3x+1 is divided byi x+1iix−12iii x iv x+π v 5+2x

Ans

(i)  x3+3x2+3x+1 is divided by x+1x2+2x+1x+1x3+3x2+3x+1           ±x3±x2¯                                2x2+3x+1                           ±2x2±2x ¯                                            x+1                                ±x±1¯                            0   ¯Thus, the remainder is 0.Another method:Let      p(x)=x3+3x2+3x+1 and the zero of x+1 is −1.So, p(−1)=(−1)3+3(−1)2+3(−1)+1           =−1+3−3+1=0Thus, the remainder is 0.

(ii)  x3+3x2+3x+1 is divided by x−12x2+72x+194x−12   x3+3x2+3x+1                     ±x3∓12x2¯                                    72x2+3x+1                                 ±72x2∓74x ¯                                                  194x+1                         ±194x∓198¯                                     278¯Thus, the remainder is 278.Another method:Let      p(x)=x3+3x2+3x+1 and the zero of x−12 is 12.So, p(12)=(12)3+3(12)2+3(12)+1          =18+34+32+1          =1+6+12+88=278

Thus, the remainder is 27 8 .

(iii)  x3+3x2+3x+1 is divided by x                     x2+3x+3                 xx3+3x2+3x+1                     ±x3 ¯                          3x2+3x+1                       ±3x2¯                                3x+1                    ±3x¯                           1   ¯

Thus, the remainder is 1. Another method: Let      p( x )= x 3 + 3 x 2 + 3x + 1 and the zero of x is 0. So, p( 0 )= ( 0 ) 3 + 3 ( 0 ) 2 + 3( 0 ) + 1           =0+0+0+1           =1 Thus, the remainder is 1. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyzaiaab2gacaqGHbGaaeyAaiaab6gacaqGKbGaaeyzaiaabkhacaqGGaGaaeyAaiaabohacaqGGaWaaSaaaeaacaaIYaGaaG4naaqaaiaaiIdaaaGaaeOlaaqaamaabmaabaGaamyAaiaadMgacaWGPbaacaGLOaGaayzkaaGaaGPaVlaaykW7caWG4bWaaWbaaSqabeaacaaIZaaaaGqabOGaa8hiaiabgUcaRiaa=bcacaaIZaGaamiEamaaCaaaleqabaGaaGOmaaaakiaa=bcacqGHRaWkcaWFGaGaaG4maiaadIhacaWFGaGaey4kaSIaa8hiaiaaigdacaqGGaGaaeyAaiaabohacaqGGaGaaeizaiaabMgacaqG2bGaaeyAaiaabsgacaqGLbGaaeizaiaabccacaqGIbGaaeyEaiaabccacaqG4baabaGaaCzcaiaaxMaacaWLjaGaaeiEamaaAeaabaGaamiEamaaCaaaleqabaGaaG4maaaakiaa=bcacqGHRaWkcaWFGaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccaWFGaGaey4kaSIaa8hiaiaaiodacaWG4bGaa8hiaiabgUcaRiaa=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bcacaaIZaGaamiEamaaCaaaleqabaGaaGOmaaaakiaa=bcacqGHRaWkcaWFGaGaaG4maiaadIhacaWFGaGaey4kaSIaa8hiaiaaigdacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG6bGaaeyzaiaabkhacaqGVbGaaeiiaiaab+gacaqGMbGaaeiiaiaabIhacaqGGaGaaeyAaiaabohacaqGGaGaaGimaiaab6caaeaacaqGtbGaae4BaiaabYcacaqGGaGaaeiCamaabmaabaGaaGimaaGaayjkaiaawMcaaiabg2da9maabmaabaGaaGimaaGaayjkaiaawMcaamaaCaaaleqabaGaaG4maaaakiaa=bcacqGHRaWkcaWFGaGaaG4mamaabmaabaGaaGimaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiaa=bcacqGHRaWkcaWFGaGaaG4mamaabmaabaGaaGimaaGaayjkaiaawMcaaiaa=bcacqGHRaWkcaWFGaGaaGymaaqaaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGimaiabgUcaRiaaicdacqGHRaWkcaaIWaGaey4kaSIaaGymaaqaaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGymaaqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyzaiaab2gacaqGHbGaaeyAaiaab6gacaqGKbGaaeyzaiaabkhacaqGGaGaaeyAaiaabohacaqGGaGaaeymaiaab6caaaaa@26CC@

(iv)  x3+3x2+3x+1 is divided by x+π                x2+(3−π)x+(3−3π+π2)         x+π  x3+3x2+3x+1                                     ±x3±πx2¯                           (3−π)x2+3x+1                        ±(3−π)x2±π(3−π)x¯                                             (3−3π+π2)x+1                                        ±(3−3π+π2)x±π(3−3π+π2)¯                                                    1−3π+3π2−π3   ¯

Thus, the remainder is ( 1−3π+3 π 2 − π 3 ). Another method: Let      p( x )= x 3 + 3 x 2 + 3x + 1 and the zero of x+π is −π. So, p( −π )= ( −π ) 3 + 3 ( −π ) 2 + 3( −π ) + 1           =− π 3 +3 π 2 −3π+1           =1−3π+3 π 2 − π 3 Thus, the remainder is 1−3π+3 π 2 − π 3 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyzaiaab2gacaqGHbGaaeyAaiaab6gacaqGKbGaaeyzaiaabkhacaqGGaGaaeyAaiaabohacaqGGaWaaeWaaeaacaaIXaGaeyOeI0IaaG4maGGaaiab=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b8aWjaabccacaqGPbGaae4CaiaabccacqGHsislcqWFapaCcaqGUaaabaGaae4uaiaab+gacaqGSaGaaeiiaiaabchadaqadaqaaiabgkHiTiab=b8aWbGaayjkaiaawMcaaiabg2da9maabmaabaGaeyOeI0Iae8hWdahacaGLOaGaayzkaaWaaWbaaSqabeaacaaIZaaaaOGaa4hiaiabgUcaRiaa+bcacaaIZaWaaeWaaeaacqGHsislcqWFapaCaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccaGFGaGaey4kaSIaa4hiaiaaiodadaqadaqaaiabgkHiTiab=b8aWbGaayjkaiaawMcaaiaa+bcacqGHRaWkcaGFGaGaaGymaaqaaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaeyOeI0Iae8hWda3aaWbaaSqabeaacaaIZaaaaOGaey4kaSIaaG4maiab=b8aWnaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiodacqWFapaCcqGHRaWkcaaIXaaabaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpcaaIXaGaeyOeI0IaaG4maiab=b8aWjabgUcaRiaaiodacqWFapaCdaahaaWcbeqaaiaaikdaaaGccqGHsislcqWFapaCdaahaaWcbeqaaiaaiodaaaaakeaacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaabwgacaqGTbGaaeyyaiaabMgacaqGUbGaaeizaiaabwgacaqGYbGaaeiiaiaabMgacaqGZbGaaeiiaiaaigdacqGHsislcaaIZaGae8hWdaNaey4kaSIaaG4maiab=b8aWnaaCaaaleqabaGaaGOmaaaakiabgkHiTiab=b8aWnaaCaaaleqabaGaaG4maaaakiaab6caaaaa@1A

(v)  x3+3x2+3x+1 is divided by 5+2x or 2x+5           12x2+14x+782x+5   x3+3x2+3x+1         ±x3±52x2¯                      12x2+3x+1                  ±12x2±54x¯                           74x+1                               ±74x±358¯                                       −278   ¯

Thus, the remainder is − 27 8 . Another method: Let      p( x )= x 3 + 3 x 2 + 3x + 1 and the zero of 5+2x is − 5 2 . So, p( − 5 2 )= ( − 5 2 ) 3 + 3 ( − 5 2 ) 2 + 3( − 5 2 ) + 1           =− 125 8 + 75 4 − 15 2 +1           = −125+150−60+8 8 =− 27 8 Thus, the remainder is − 27 8 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbbjxAHXgarmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYBg9LrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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bcacqGHRaWkcaWFGaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccaWFGaGaey4kaSIaa8hiaiaaiodacaWG4bGaa8hiaiabgUcaRiaa=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bcacaaIZaWaaeWaaeaacqGHsisldaWcaaqaaiaaiwdaaeaacaaIYaaaaaGaayjkaiaawMcaaiaa=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@0AF1@

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