# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4

The National Council of Educational Research and Training (NCERT) was founded in order to assist and advise the Central and State Governments on policies and programmes for the improvement of high-quality school education. It was established to enable and encourage the diverse cultural practises that are practised throughout the nation as a whole, as well as to support a unified educational system for the country with a national character. The basic goals of NCERT and its units are to conduct, promote, and coordinate research in school education-related disciplines. NCERT offers information about the globe and covers all significant topics. Students have moved away from the traditional approach to learning. It has precise content that students can readily understand. Additionally, it organises pre-service and in-service teacher training programmes, supports creative teaching methods, and collaborates and networks with other educational institutions, including NGOs and state education departments. The Central Board of Secondary Education advises students to use the NCERT textbooks. In addition to emphasising the benefits of learning from NCERT textbooks, CBSE claims that studying from NCERT can considerably improve students’ test performance.

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**CBSE Class 10 NCERT Solutions for Maths Chapter 2 Polynomials Exercise 2.4**

The NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 are offered to help students with their studies. These solutions are created by Extramarks’ subject specialists to help Class 10 board examination students study effectively. These experts produce the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 so that students can quickly and easily tackle the NCERT problems. They make sure that students can quickly learn from this and also pay attention to how easily they can understand the subject. It consists of additional chapter-related questions for practice. The NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 present a series of questions, each with a detailed answer from the experts. Following NCERT criteria, the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 are created with the intention of covering the entire curriculum. These are highly beneficial for getting good examination results.

The Class 10th Math Exercise 2.4 is the fourth exercise in Class 10 Mathematics Chapter 2. In Class 9, Polynomials are introduced, but in Class 10, they are covered in greater detail. Students will be able to get additional marks after reading the detailed solutions provided by Extramarks subject-matter experts. It adheres to the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4, which aid in the students’ appropriate preparation. It contains all the crucial questions from the perspective of the examination.It helps students achieve high marks in mathematics board examinations. Students can complete and review all of the questions in the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 using these NCERT Solutions.

**NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4**

Mathematics is an important aspect of human logic and thought. It provides a powerful method for developing mental discipline and improvinglogical thinking. Additionally, understanding Mathematics is crucial in order to comprehend the concepts of other topics, like Science, Social Science, even Music and the Arts. Numerous professions and fields use Mathematics. Engineering, Science, and Economics questions are resolved using the principles and techniques of Mathematics. Mathematics is one of the most important subjects in a student’s academic career. If students wish to succeed in the subject, they must practise consistently. In order to perform well on their Class 10 board examinations, students should also have a strong understanding of mathematics because it is a conceptual subject.

Extramarks offers NCERT Solutions for Class 10 Maths, Chapter 2, Exercise 2.4. Polynomials are one of the most important topics in Class 10 Mathematics. It contains a comprehensive set of questions along with a summary. The NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 is an optional exercise, so it covers all topics and allows students to review the entire chapter. The NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 is designed to test students’ skills and cognitive thinking. It is used to see if a student can apply concepts learned throughout the chapter to complex problems in order to understand their progress. There are 5 problems in this section, each containing several sub-problems. With the help of this exercise, students will master the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 and Polynomial Topics, Polynomial Division Algorithms, and Polynomial Zeros.

**Topics Covered**

The NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 are detailed notes for Chapter 2 of the Mathematics textbook for Class 10 in the CBSE. Here, students will get an in-depth overview of Polynomials, including their types, algebraic expressions, degree, a graphical depiction of polynomial equations, factorization, the link between a polynomial’s zeros and coefficient, and more—all with an ample number of examples.

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**What is a Polynomial?**

Students need to have a strong conceptual understanding of Mathematics, and if learning is done simply, it will be easier for students to understand the material. The Extramarks website is aware of the challenges and obstacles that many students confront, therefore, the teachers at Extramarks create the lesson plans and the course materials in a way that best meets their requirements. Class 10th Math Exercise 2.4 consists of important topics, so students should clearly understand every topic with the help of the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4.

A polynomial is an expression in Mathematics made up of coefficients and variables, commonly referred to as Indeterminates. It is a mathematical term for numerous terms since, as its name implies, “poly” means “many” and “nominal” means “phrases”. Typically, Polynomials are the product or difference of exponents and variables. Each term in the Polynomial is referred to as a “part.” Class 10 is considered a board class and students are under a lot of pressure to score high marks in their examinations. The Extramarks website helps them by providing them with the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4. With the help of the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 students can easily understand the concepts.

**Types of Polynomial**

Polynomials are grouped into three categories based on the number of terms they contain. There are three different kinds of polynomials that are, Monomial, Binomial, and Trinomial.

These polynomials can be combined using addition, subtraction, multiplication, and division, but never division by a variable.

An expression with only one term is called a monomial. The single term in an expression needs to be non-zero in order for it to be a monomial.

A polynomial expression with exactly two terms is referred to as a binomial. One way to think of a binomial is as the difference or sum of two or more monomials.

An expression with precisely three terms is called a trinomial.

**What Are Zero Polynomials?**

A Zero Polynomial is a kind of polynomial in which all the coefficients of the variables are equal to zero. In other words, it indicates that the powers of all the variables are equal to zero. A polynomial is an expression made up of variables and coefficients. The only polynomial with an uncertain degree is a zero polynomial. A polynomial’s degree is the highest degree of its non-zero terms; a polynomial with a value of zero has no non-zero terms. As a result, we cannot determine the degree of a polynomial using terms with degrees.

**Exercise – 2.4**

The NCERT Solutions Of Class 10 Maths Chapter 2 Exercise 2.4 are available on the Extramarks website to help students with their academics. Extramarks subject matter experts created the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.4 to assist students in their preparation for the Class 10 board examination.For students to quickly solve the NCERT questions, these specialists produce the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4. Along with ensuring that students can learn quickly from this, they also consider their ability to understand the NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.4.

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**NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4 **

The NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 provide a thorough understanding of Polynomials. This chapter will teach students about polynomials, their relevance, and their significance.Their comprehension of the idea of Polynomials will improve with the aid of the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4. To do well in mathematics, students must understand numerous key ideas in the Polynomials NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.4.Students can download the PDF of NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4. Additionally, they may get the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 to help students review the entire syllabus and do better on tests.

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**Practice Problems**

Extramarks provides the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 to assist students in preparing for and doing well in the CBSE Class 10 Maths examination.The NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 is based on the board’s most recent syllabus. Students can prepare effectively for the board examination by practising the key questions for each chapter of the 10th standard Mathematics course. Students will get thorough answers. They can consult the solutions if they face trouble when attempting to answer NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4.

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**Knowing About Polynomials Is Inevitable**

Polynomials are an important aspect of Mathematics and Algebra “language.” They are used to express numbers that are the result of mathematical operations in almost every area of Mathematics. Other types of mathematical expressions, such as rational expressions, also use Polynomials as “building blocks.” Polynomials are used frequently by people. They are used by people to simulate a variety of structures and things, as well as in industries and construction. Even marketing, finance, and stocks employ them. The form of polynomial models is straightforward.

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**Facts**

Polynomial equations are those created using exponents, coefficients, and variables. The greater of its possible exponents is referred to as the degree of the polynomial. By factoring Polynomials according to their degree and the variables in the equation, students can find their solutions. The degree of a polynomial is determined by the highest degree monomial in the polynomial.Polynomials can be prime or factorable into products of primes, just like whole numbers. As long as each variable’s power is a nonnegative integer, they can include any number of variables. They serve as the foundation for solving algebraic equations.

**Method of Division Polynomials**

The step-by-step method of solving polynomial equations can be found on the Extramarks website and mobile application. Students can make use of these solutions to score well in their examinations.

**Applications of Division Algorithm**

An algorithm for dividing two integers, N and D, into their quotient and remainder, or the product of Euclidean division, is known as a Division Algorithm. Some are done manually, while software and digital circuit designs are used for others. Students can refer to the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 for help in clearly understanding the concept.

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**NCERT Solutions Class 10 Maths All Chapters**

One of the most important subjects in a student’s academic career is Mathematics as it is a conceptual subject. If students wish to succeed in the subject, they must practise consistently. Students should also have a strong understanding of Mathematics in order to do well on their Class 10 board examination. NCERT Solutions for Class 10 Mathematics for all exercises from Chapters 1 to 15 are available on the Extramarks’ website. Extramarks’ knowledgeable faculty has selected these NCERT solutions to aid students in their test preparation. Students looking for the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 can download each PDF to find a more effective way to approach the problems.

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**NCERT Solutions Class 10 Maths Chapter 2 Exercises**

Mathematics is an important subject for students in Class 10. The curriculum is structured so that students can learn new information and build a conceptual base before moving on to more complex topics. Extramarks’ NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.4 created by teachers would be of particular assistance with Class 10 Mathematical Chapter 2.Students may view the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 PDF file offline by downloading it.

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**Q.1 **

$\begin{array}{l}\text{Verify that the numbers given alongside of the cubic}\\ \text{polynomials below are their zeroes. Also verify the}\\ \text{relationship between the zeroes and the coefficients}\\ \text{in each case:}\\ {\text{(i) 2x}}^{3}+{\mathrm{x}}^{2}-5\mathrm{x}+2;\text{}\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-2\text{\hspace{0.17em}}\\ {\text{(ii) x}}^{3}-4{\mathrm{x}}^{2}+5\mathrm{x}-2;\text{2, 1, 1}\end{array}$

**Ans.**

$\begin{array}{l}{\text{(i) 2x}}^{3}+{\mathrm{x}}^{2}-5\mathrm{x}+2;\text{}\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-2\text{\hspace{0.17em}}\\ \text{Let}\mathrm{p}\left(\mathrm{x}\right)={\text{2x}}^{3}+{\mathrm{x}}^{2}-5\mathrm{x}+2\\ \text{Then}\\ \mathrm{p}\left(\frac{1}{2}\right)=\text{2}{\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{2}\right)}^{2}-5\left(\frac{1}{2}\right)+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=\frac{1}{2}+2-\frac{5}{2}=0\\ \\ \mathrm{p}\left(1\right)=\text{2}{\left(1\right)}^{3}+{\left(1\right)}^{2}-5\left(1\right)+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2+1-5+2=0\\ \mathrm{p}(-2)=\text{2}{(-2)}^{3}+{(-2)}^{2}-5(-2)+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-16+4+10+2=0\\ \\ \text{Therefore,}\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\text{and\hspace{0.17em}\hspace{0.17em}}-2\text{are zeroes of the given poynomial.}\\ \text{Comparing the given poynomial with}{\mathrm{ax}}^{3}+{\mathrm{bx}}^{2}+\mathrm{cx}+\mathrm{d},\text{we get}\\ \mathrm{a}=2,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=-5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=2\\ \text{Let the given roots are}\mathrm{\alpha},\text{\hspace{0.17em}}\mathrm{\beta}\text{and}\mathrm{\gamma}\text{. Then}\\ \mathrm{\alpha}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{1}{2},\text{}\mathrm{\beta}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\gamma}=-2\\ \text{Now,}\\ \mathrm{\alpha}+\mathrm{\beta}+\mathrm{\gamma}=\frac{1}{2}+1-2=-\frac{1}{2}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \mathrm{\alpha \beta}+\mathrm{\beta \gamma}+\mathrm{\gamma \alpha}=\frac{1}{2}\times 1+1\times (-2)+(-2)\times \frac{1}{2}=\frac{-5}{2}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{\alpha \beta \gamma}=\frac{1}{2}\times 1\times (-2)=-1=\frac{-2}{2}=\frac{-\mathrm{d}}{\mathrm{a}}\\ \\ \text{The relationship between roots and coefficients is verified.}\\ {\text{(ii) x}}^{3}-4{\mathrm{x}}^{2}+5\mathrm{x}-2;\text{2, 1, 1}\\ \text{Let}\mathrm{p}\left(\mathrm{x}\right)={\text{x}}^{3}-4{\mathrm{x}}^{2}+5\mathrm{x}-2\\ \text{Then}\\ \mathrm{p}\left(\text{2}\right)={\left(2\right)}^{3}-4{\left(2\right)}^{2}+5\left(2\right)-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8-16+10-2=0\\ \\ \mathrm{p}\left(1\right)={\left(1\right)}^{3}-4{\left(1\right)}^{2}+5\left(1\right)-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-4+5-2=0\\ \\ \text{Therefore,}2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\text{and\hspace{0.17em}\hspace{0.17em}}1\text{are zeroes of the given poynomial.}\\ \text{Comparing the given poynomial with}{\mathrm{ax}}^{3}+{\mathrm{bx}}^{2}+\mathrm{cx}+\mathrm{d},\text{we get}\\ \mathrm{a}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=-4,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=-2\\ \text{Let the given roots are}\mathrm{\alpha},\text{\hspace{0.17em}}\mathrm{\beta}\text{and}\mathrm{\gamma}\text{. Then}\\ \mathrm{\alpha}\text{\hspace{0.17em}=\hspace{0.17em}}2,\text{}\mathrm{\beta}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\gamma}=1\\ \text{Now,}\\ \mathrm{\alpha}+\mathrm{\beta}+\mathrm{\gamma}=2+1+1=4=-\frac{-4}{1}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \mathrm{\alpha \beta}+\mathrm{\beta \gamma}+\mathrm{\gamma \alpha}=2\times 1+1\times 1+1\times 2=5=\frac{5}{1}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{\alpha \beta \gamma}=2\times 1\times 1=2=\frac{-(-2)}{1}=\frac{-\mathrm{d}}{\mathrm{a}}\\ \text{The relationship between roots and coefficients is verified.}\end{array}$

**Q.2 **

$\begin{array}{l}\text{Find a cubic polynomial with the sum of zeroes , sum of the}\\ \text{product of its zeroes taken two at a time, and the}\\ \text{product of its zeroes as 2,}-\text{7,}-\text{14 respectively.}\end{array}$

**Ans.**

$\begin{array}{l}\text{Let the polynomial be}{\mathrm{ax}}^{3}+{\mathrm{bx}}^{2}+\mathrm{cx}+\mathrm{d}\text{and its zeroes}\\ \text{be}\mathrm{\alpha}\text{,\hspace{0.17em}\hspace{0.17em}}\mathrm{\beta}\text{and}\mathrm{\gamma}.\\ \text{Given that}\\ \mathrm{\alpha}+\mathrm{\beta}+\mathrm{\gamma}=2=\frac{-(-2)}{1}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \mathrm{\alpha \beta}+\mathrm{\beta \gamma}+\mathrm{\gamma \alpha}=-7=\frac{-7}{1}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{\alpha \beta \gamma}=-14=\frac{-14}{1}=\frac{-\mathrm{d}}{\mathrm{a}}\\ \text{If we take}\mathrm{a}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=-2,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=-7\text{and}\mathrm{d}=14\text{\hspace{0.17em}\hspace{0.17em}then the}\\ \text{required polynomial is}{\mathrm{x}}^{3}-2{\mathrm{x}}^{2}-7\mathrm{x}+14.\end{array}$

**Q.3 **

$\begin{array}{l}\text{If the zeroes of the polynomial\hspace{0.33em}}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+1\\ \text{are a}-\text{b, a,\hspace{0.33em}}\mathrm{a}+\mathrm{b}\text{, find a and b.}\end{array}$

**Ans.**

$\begin{array}{l}\text{The given polynomial is}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+1\text{\hspace{0.17em}\hspace{0.17em}and its zeroes}\\ \text{are a}-\text{b, a,}\mathrm{a}+\mathrm{b}.\\ \text{Sum of the roots = (a}-\text{b})+\mathrm{a}+(\mathrm{a}+\mathrm{b})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{a}=-\frac{\text{Coefficient of}{\mathrm{x}}^{2}}{\text{Coefficient of}{\mathrm{x}}^{3}}=3\\ \Rightarrow \mathrm{a}=1\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sum of the products of zero taken two at a time}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{a}(\mathrm{a}-\mathrm{b})+\mathrm{a}(\mathrm{a}+\mathrm{b})+(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{2}-\mathrm{ab}+{\mathrm{a}}^{2}+\mathrm{ab}+{\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3{\mathrm{a}}^{2}-{\mathrm{b}}^{2}=\frac{\text{Coefficient of}\mathrm{x}}{\text{Coefficient of}{\mathrm{x}}^{3}}=1\\ \Rightarrow 3{\mathrm{a}}^{2}-{\mathrm{b}}^{2}=1\\ \Rightarrow 3.1-{\mathrm{b}}^{2}=1\\ \Rightarrow {\mathrm{b}}^{2}=2\\ \Rightarrow \mathrm{b}=\pm \sqrt{2}\\ \text{Hence,}\mathrm{a}=1\text{and}\mathrm{b}=\sqrt{2}\text{\hspace{0.17em}\hspace{0.17em}or}-\sqrt{2}\end{array}$

**Q.4 **

\begin{array}{l}\text{If two zeroes of the polynomial}\text{\hspace{0.33em}}{x}^{4}-6{x}^{3}-26{x}^{2}+138x-35\\ \text{are}\text{\hspace{0.33em}}2\pm \sqrt{3}\text{,}\text{\hspace{0.33em}}\text{find other zeroes}\text{.}\end{array}

**Ans.**

$\begin{array}{l}\text{Two zeroes of the polynomial}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35\\ \text{are}2\pm \sqrt{3}.\\ \text{Therefore,}\left[\mathrm{x}-\left(2+\sqrt{3}\right)\right]\text{and}\left[\mathrm{x}-\left(2-\sqrt{3}\right)\right]\text{are factors of the}\\ \text{given polynomial.}\\ \text{Now},\\ \left[\mathrm{x}-\left(2+\sqrt{3}\right)\right]\left[\mathrm{x}-\left(2-\sqrt{3}\right)\right]=\left[\left(\mathrm{x}-2\right)-\sqrt{3})\right]\left[\left(\mathrm{x}-2\right)+\sqrt{3})\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\mathrm{x}-2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}-4\mathrm{x}+4-3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}-4\mathrm{x}+1\\ {\mathrm{x}}^{2}-4\mathrm{x}+1\text{is also a factor of the given polynomial.}\\ \text{Therefore, we divide}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35\\ \text{by}{\mathrm{x}}^{2}-4\mathrm{x}+1\text{to find other factors.}\\ \\ {\mathrm{x}}^{2}-4\mathrm{x}+1\overline{){\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{4}-4{\mathrm{x}}^{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+{\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\overline{\begin{array}{l}-2{\mathrm{x}}^{3}-27{\mathrm{x}}^{2}+138\mathrm{x}-35\\ -2{\mathrm{x}}^{3}+8{\mathrm{x}}^{2}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}\\ +\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\overline{\begin{array}{l}-35{\mathrm{x}}^{2}+140\mathrm{x}-35\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ -35{\mathrm{x}}^{2}+140\mathrm{x}-35\\ +\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\overline{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\\ {\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left({\mathrm{x}}^{2}-4\mathrm{x}+1\right)\left({\mathrm{x}}^{2}-2\mathrm{x}-35\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left[\mathrm{x}-\left(2+\sqrt{3}\right)\right]\left[\mathrm{x}-\left(2-\sqrt{3}\right)\right]\left({\mathrm{x}}^{2}-7\mathrm{x}+5\mathrm{x}-35\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left[\mathrm{x}-\left(2+\sqrt{3}\right)\right]\left[\mathrm{x}-\left(2-\sqrt{3}\right)\right]\left(\mathrm{x}-7\right)\left(\mathrm{x}+5\right)\\ \left(\mathrm{x}-7\right)\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\left(\mathrm{x}+5\right)\text{\hspace{0.17em}}\mathrm{are}\text{\hspace{0.17em}}\mathrm{factors}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35.\\ \mathrm{Hence},\text{\hspace{0.17em}}7\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}-5\text{\hspace{0.17em}}\mathrm{are}\text{\hspace{0.17em}}\mathrm{zeroes}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35.\end{array}$

**Q.5 **

$\begin{array}{l}\text{If the polynomial\hspace{0.33em}}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-25\mathrm{x}+10\text{\hspace{0.33em}is divided}\\ \text{by another polynomial\hspace{0.33em}}{\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k}\text{,\hspace{0.33em}the remainder comes}\\ \text{out to be\hspace{0.33em}}\mathrm{x}+\mathrm{a}\text{,\hspace{0.33em}find\hspace{0.33em}}\mathrm{k}\text{\hspace{0.33em}and\hspace{0.33em}}\mathrm{a}\text{.}\end{array}$

**Ans.**

$\begin{array}{l}\text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividend = Divisor}\times \text{Quotient + Remainder}\\ \text{or Dividend}-\text{Remainder = Divisor}\times \text{Quotient}\\ \text{or}({\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-25\mathrm{x}+10)-(\mathrm{x}+\mathrm{a})=({\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k})\times \text{Quotient}\\ \text{or}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-26\mathrm{x}+10-\mathrm{a}=({\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k})\times \text{Quotient}\\ \text{Now},\\ {\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k}\begin{array}{c}{\mathrm{x}}^{2}-4\mathrm{x}+(8-\mathrm{k})\\ \overline{)\begin{array}{l}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-26\mathrm{x}+10-\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{\xaf}{\begin{array}{l}{\mathrm{x}}^{4}-2{\mathrm{x}}^{3}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{kx}}^{2}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{\xaf}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-4{\mathrm{x}}^{3}+(16-\mathrm{k}){\mathrm{x}}^{2}-26\mathrm{x}+10-\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-4{\mathrm{x}}^{3}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8{\mathrm{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}}4\mathrm{kx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{\xaf}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-(8-\mathrm{k}){\mathrm{x}}^{2}+(4\mathrm{k}-26)\mathrm{x}+10-\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-(8-\mathrm{k}){\mathrm{x}}^{2}-(-2\mathrm{k}+16)\mathrm{x}+8\mathrm{k}-{\mathrm{k}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}(2\mathrm{k}-10)\mathrm{x}+{\mathrm{k}}^{2}-8\mathrm{k}+10-\mathrm{a}\\ \text{Remainder\hspace{0.17em}\hspace{0.17em}}(2\mathrm{k}-10)\mathrm{x}+{\mathrm{k}}^{2}-8\mathrm{k}+10-\mathrm{a}\text{must be zero.}\\ \mathrm{So},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}(2\mathrm{k}-10)\mathrm{x}+{\mathrm{k}}^{2}-8\mathrm{k}+10-\mathrm{a}\text{}=0\\ \Rightarrow 2\mathrm{k}-10=0\text{or}{\mathrm{k}}^{2}-8\mathrm{k}+10-\mathrm{a}=0\\ \Rightarrow \mathrm{k}=5\text{or}{5}^{2}-8\times 5+10-\mathrm{a}=0\\ \Rightarrow \mathrm{k}=5\text{and}\mathrm{a}=-5\end{array}$

## FAQs (Frequently Asked Questions)

### 1. Are the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 complex for students to understand?

Students will not find the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 to be difficult. Students can simply study the solutions of NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 from the Extramarks website and excel in their examination with consistent practice and the appropriate assistance from Extramarks. Extramarks provides thorough answers to the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 questions to help students comprehend the material. Students can also take part in doubt sessions offered on the Extramarks website if they are reluctant to ask questions in front of their peers at school.

The solutions to the Class 10 Maths Chapter 2 Exercise 2.4 should be familiar to students in order to give them an idea of the kinds of questions that can appear in the examination. Therefore, Extramarks advises students to study a number of sample papers and past years’ papers before taking their examination. Understanding the concepts presented in the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 will help students advance in their preparation. Although the chapter’s material might occasionally be complex, students can easily understand it with the help of Extramarks’ NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4.

### 2. Are the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 sufficient for students to study for the chapter?

For students to adequately prepare for the chapter, the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 are sufficient. The NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 are enough for helping students strengthen their fundamental concepts and get ready to manage any chapter-related questions. Students can also practice past papers and sample papers for better preparation. According to the Extramarks, students who thoroughly grasp the NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.4 perform better on examination.

### 3. Why are students required to use Extramarks for examination preparation?

Students are required to use Extramarks for examination preparation to ensure that all of the chapter’s concepts and formulas are thoroughly understood. Extramarks provides chapter-based worksheets, interactive activities, an infinite number of practice questions, and more. To monitor the progress of students who have registered on the Extramarks website, the website offers adaptive quizzes, MCQs, and mock tests. In-depth reports and analyses are produced by the Extramarks experts to assist students to make the most of their learning experience on the learning platform. This performance analysis shows the students’ strong and weak areas, allowing them to work on the concepts they struggle with in order to help them perform well on the examination.