NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 Polynomials

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Mathematics is one of the core subjects that students who are in Class 11 and who have taken up science have. Mathematics at the secondary level gets very advanced, and students need to be more careful when they are approaching the chapters. The ideas and concepts that are explained in these chapters are difficult, and it is not easy for students to grasp them and get a hold of them in one go.. The problems that are present in the Mathematics textbook are very conceptual and analytical, and students have to be very patient as they are solving these problems. Students who transition to the higher secondary level from the Class 10 level know very well the change in scale and the advanced level of difficulty that comes  with this. Students are generally scared and overwhelmed by the scale of the syllabus, but teachers always advise that if students take their academics seriously and focus on revising old chapters with the same emergence with which they make progress with the syllabus, then there is nothing for students to be very scared of. Mostly all the ideas that are discussed in the higher secondary level of Mathematics are echoed and repeated in the other science subjects that students have besides mathematics like Physics and Chemistry. Ideas and concepts like Trigonometry, Calculus, Coordinate geometry and other  are present in almost every other science subject. Students must therefore take mathematics seriously if they want to do well in their CBSE exams.Students are given their final grade based on the aggregate of all the subjects that they appear for in their exams, and therefore they cannot divide their attention to a subject,  in a uniform way. Students are implored by teachers that they must seriously attempt to finish the syllabus with sufficient time left for them to revise the chapters. Although, the scale of the syllabus is so vast that it gets difficult for students to make progress with the syllabus while also revising and revisiting thechapters that they have already done once. As a result, Extramarks’ teachers have worked tirelessly to develop the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3.

In an effort to provide comprehensive solutions to all NCERT questions, Extramarks has released the NCERT Solutions for Class 10 Maths, Chapter 2, Exercise 2.3.The NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 provide solutions to every unsolved question that is given in the NCERT textbook. The solutions that are provided in the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 are provided by teachers who are highly qualified and have years of experience in teaching Class 10 students for their Class 10 boards. The entire Class 10 curriculum is very important because it helps students to form the foundation for all the advanced concepts that they are going to learn in future classes, soit is imperative for students to have a very good grasp of the ideas discussed in Class 10. Therefore, NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 is one of the most efficient resources that a Class 10 student can have at their disposal. The answers in NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 are easy to read and extremely well-detailed solutions for the Class 10 NCERT textbook. NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 provides step-by-step solutions, to every unsolved problem. The solutions provided in NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 are meticulously written, making sure that whoever accesses the solutions no matter their academic background, can easily understand the solutions. Students who have used the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 have all shared their positive experiences with them and how it has helped them in situations when they needed help quickly.

Generally, students in their  CBSE 10 class have many subjects to attend to and study, and therefore they are already under a lot of stress. Students in Class 10 are appearing for their first Boards and therefore, the anticipation of the exam builds intensely, making it difficult for students to concentrate. Therefore, in such times of stress, the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 provide great guidance to students. The solutions in NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 are  constantly revised and updated because the guidelines that CBSE and the NCERT follow are constantly changing.  Given that status, it is highly inefficient if the solutions in NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 are not in sync with the question paper pattern. When students practise the unsolved numerical problems through the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 it provides students with a very solid foundation of the kind of questions that one can expect in the CBSE Class 10 Board exams. A regular revision of the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 helps students understand the correct wayfor CBSE to approach a problem. A mathematical problem can be solved by many different methods, but students must solve the problem with the right methodology, which CBSE approves of. When students regularly utilise the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 they can easily see for themselves the accurate way to approach a problem. Teachers have all assured students that these solutions are very safe to use and that theyhave nothing to fear. The NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 are prepared and released after being intently scrutinized. The solutions are highly reliable and devoid of any errors.

Teachers have suggested that the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 can be used dynamically. Students who have the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 in their possession can use them in any way they desire. Although teachers in their years of experience, have observed a distinct pattern in which students can use the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3. One of the primary ways the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 can be used is when students are solving unsolved mathematical problems for the first time. When students are solving an exercise for the first time, they must go through the theory of the chapter in great depth. They must also refer to the solved problems that are provided in the textbook before they start referring to NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3. When Class 10 CBSE students start solving an exercise for the first time, they come across many doubts. Teachers always encourage students to address their doubts and never be daunted by them. Doubts are extremely important to track because they give students a very clear understanding of the topics that need to be addressed. When students introspect on their mistakes, they realise the inconsistencies in their understanding of the chapter. Therefore, NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 helps students  look deeply into their doubts so that they can reach the core of their doubts. The NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 help students in multiple ways, andone of the most important is that when students come across doubts,, they generally wait for their teachers to solve them. Although waiting for their teachers is highly counterintuitive when students are aiming to finish the syllabus. The time that is wasted while waiting for help is highly uncalled for. Students can help themselves whenever they want when they have NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 at their disposal.

Class 10 Maths Chapter 2 Exercise 2.3 is very important among the other exercises in the chapter, and therefore it must be taken care of with diligence.

Students are often seen to refer to the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 while they are revising old chapters after making considerable progress with the syllabus. Students are always told that when they are revising the old chapters, they must do it first by themselves. It is only when they come across a mistake or a doubt that they cannot solve by themselves that they can refer to NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3. Often the solutions in NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 suffice perfectly to help out these students. The solutions are provided by highly qualified teachers, and therefore students are always under the supervision of highly qualified individuals. The NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 can be easily accessed through the Extramarks website as well as their mobile applications.

Class 10 Maths Ch 2 Ex 2.3 discusses quadratic polynomials which are polynomials with the coefficient of 2.

Click on the link below to access the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3.

H2 – NCERT Solutions for Class 10 Maths Chapter 2 Polynomials (Ex 2.3) Exercise 2.3

The solutions to Maths Class Chapter 2 exercise 2.3 can be found on the Extramarks website for free

NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 provides the solution to the second chapter of the Class 10 CBSE curriculum. The third exercise in this chapter discusses quadratic polynomials. The NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 provide very crisp and easy solutions to this chapter, and therefore their ease of access  makes them popular among students.

H2 – NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.3

NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 provide comprehensive solutions to the second chapter of the Class 10 CBSE mathematics curriculum called Polynomials. The main ideas that are discussed in this chapter are – Constant Polynomial, Linear Polynomial, Quadratic Polynomial and Cubic Polynomial. The other ideas that the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 help students  get a good understanding of are calculating the value of a polynomial, the zero of a polynomial, the Graph of a polynomial and the significance of the zeros of quadratic equations.

H3 – NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.3

NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 prepares students in the best possible way because polynomials are an extremely important chapter in the entire curriculum. This chapter is one of the most important building blocks for the advanced concepts that a student would be learning in the later classes. NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3 has therefore become one of the most crucial resources that students in their Class 10 CBSE use.

H3 – NCERT Solutions Class 10 Maths All Chapters

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H3 – NCERT Solutions Class 10 Maths Chapter 2 Exercises

Click on the link below to access the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.3.

H3 – Importance of Polynomials

Many students, after finishing their boards, often during their transition to Class 11, choose science as their field of study.These students have to have this chapter very clear because, based on this chapter, various advanced concepts will be further explicated in the later classes. As a result, the NCERT Solutions for Class 10 Maths, Chapter 2, Exercise 2.3 help students prepare for higher secondary school.

Q.1

Divide the polynomial p(x) by the polynomial g(x)and find the quotient and remainder in each of the following: (i) p(x)=x3−3x2+5x−3, g(x)=x2−2 (ii) p(x)=x4−3x2+4x+5, g(x)=x2+1−x (iii) p(x)=x4−5x+6, g(x)=2−x2 (i) p(x)=x3−3x2+5x−3, g(x)=x2−2x2−2x−3x3−3x2+5x−3               x3             −2x−               +¯               −3x2+7x−3               −3x2           +6         +                    −7x−9Quotient=x−3Remainder=7x−9(ii) p(x)=x4−3x2+4x+5, g(x)=x2+1−xx2+1−xx2+x−3x4−3x2+4x+5                     x4+   x2−x3−    −         + ¯                     x3−4x2+4x+5                     x3−x2    +x  −    +          −         ¯                   −3x2+3x+5                   −3x2+3x−3+         −        +¯                      8Quotient=x2+x−3Remainder=8(iii) p(x)=x4−5x+6, g(x)=2−x22−x2−x2−2x4−5x+6              x4−2 x2−    +               ¯              2x2−5x+6              2x2            −4    −                +    ¯            −5x+10Quotient=−x2−2Remainder=−5x+10

Q.2

Check whether the first polynomial is a facator of the second polymomial by dividing the second polynomial by the first polynomial: (i)  t2−3, 2t4+3t3−2t2−9t−12 (ii) x2+3x+1, 3x4+5x3−7x2+2x+2 (iii) x3−3x+1, x5−4x3+x2+3x+1 (i)  t2−3, 2t4+3t3−2t2−9t−12t2−32t2+3t+42t4+3t3−2t2−9t−12            2t4            −6t2−                +                            ¯           3t3+4t2−9t−12           3t3             −9t−                 +¯                   4t2−12                   4t2−12           −      +¯                        0Remainder = 0So, t2−3 is a factor of 2t4+3t3−2t2−9t−12.(ii) x2+3x+1, 3x4+5x3−7x2+2x+2x2+3x+13x2−4x+23x4+5x3−7x2+2x+2                       3x4+9x3+3x2−       −         −                             ¯                        −4x3−10x2+2x+2                        −4x3 −12x2−4x +          +            +               ¯                             2x2+6x+2                             2x2+6x+2   −        −         −¯                                     0Remainder = 0So, x2+3x+1 is a factor of 3x4+5x3−7x2+2x+2.(iii) x3−3x+1, x5−4x3+x2+3x+1x3−3x+1x2−1x5−4x3+x2+3x+1                        x5−3x3+x2−    +         −                             ¯                               −x3             +3x+1                               −x3             +3x−1       +                  −       +      ¯                                             2Remainder = 2So, x3−3x+1 is not a factor of x5−4x3+x2+3x+1.

Q.3

Obtain all other zeroes of 3x4+6x3−2x2−10x−5,if two of its zeroes are 53 and−53 . p(x)=3x4+6x3−2x2−10x-5The two zeroes of p(x) are 53 and-53 . Therefore, x−53 and x+53 are factors of p(x).Also,x−53 x+53  = x2−53and so x2−53 is a factor of p(x).Now,                3x2+6x+3x2−53 3x4+6x3−2x2−10x−5               3x4             −5x2−                  +                            ¯                6x3+3x2−10x−5               6x3              −10x−                   +                ¯                     3x2             −5                     3x2             −5−                  +¯                                 03x4+6x3−2x2−10x−5=x2−533x2+6x+3                                                    =3x2−53x2+2x+1                                                    =3x2−53x+1(x+1)Equating x2−53x+1(x+1) equal to zero, we getthe zeroes of the given polynomial.Hence, the zeroes of the given polynomial are 53 , −53 ,  1and −1.

Q.4

On dividing x3−3x2+x+2 by a polynomial g(x),the quotient and remainder were x−2 and −2x+4,respectively. Find g(x). Dividend = x3−3x2+x+2Quotient = x−2Remainder = −2x+4Divisor = g(x) = ?We know that,      Dividend=Divisor×Quotient+Remainder∴    x3−3x2+x+2=g(x)×( x−2)+(−2x+4)⇒g(x)×( x−2)=x3−3x2+x+2+2x−4⇒g(x)=(x3−3x2+3x−2)( x−2)x−2x2−x+1x3−3x2+3x−2             x3−2x2−    +                          ¯                  −x2+3x−2                  −x2+2x      +         −              ¯                   x−2                   x−2     −     +¯                   0∴     g(x)=x2−x+1

Q.5

Give examples of polynomials p(x),g(x),q(x) and r(x)which satisfy the division algorithm and (i) deg p(x)=deg q(x) (ii) deg q(x)=deg r(x) (iii) deg r(x)=0 (i) deg p(x)=deg q(x)if divisor is constant then Degree of quotient = degree of dividend Let dividend p(x)=2x2−2x+14and    divisor  g(x)=2Then, wer have              quotient  q(x)=x2−x+7and remainder  r(x)=0We find that deg p(x)=deg q(x)=2Let us check for division algorithm.             dividend=divisor×quotient + remainderor 2x2−2x+14=2(x2−x+7)+0= 2x2−2x+14Thus, the division algorithm is satisfied.(ii) deg q(x)=deg r(x)Let us divide x3+x by x2. Clearly we have,dividend p(x)=x3+x   divisor  g(x)=x2   quotient  q(x)=xand remainder  r(x)=xHere, deg q(x)=deg r(x)=1Let us check for division algorithm.             dividend=divisor×quotient + remainderor x3+x=x2⋅x+x=x3+x Thus, the division algorithm is satisfied.(iii) deg r(x)=0Degree of remainder will be zero if the remainder is constant.Let us divide x2+1 by x. Clearly we have,dividend p(x)=x2+1   divisor  g(x)=x   quotient  q(x)=xand remainder  r(x)=1Here, deg r(x)=0Let us check for division algorithm.             dividend=divisor×quotient + remainderor x2+1=x⋅x+1=x2+1 Thus, the division algorithm is satisfied.

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