Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
What are the basic concepts of polynomials Class 10?
Polynomials are algebraic expressions that consist of variables and coefficients.
What are the important points of polynomials?
For any expression to become a polynomial, the power of the variable should be a whole number. The addition and subtraction of a polynomial are possible between like terms only. All the numbers in the universe are called constant polynomials.
What are the facts about polynomials Class 10?
A polynomial is made up of terms that are only added, subtracted or multiplied.
What is the zero of a polynomial?
The zeros of a polynomial p(x) are all the x-values that make the polynomial equal to zero.
What is the degree of zero in a polynomial?
The degree of the zero-degree polynomial (0) is not defined.
RD Sharma Solutions Class 10 Maths Chapter 2 Exercise 2.3 Polynomials has been provided here. Students can refer to these questions before their examinations for better preparation.
Neha Tanna29 Oct, 2024
Share
RD Sharma Solutions Class 10 Maths Chapter 2 Exercise 2.3:
Chapter 2, Exercise 2.3 in RD Sharma's Class 10 Maths textbook focuses on polynomials, particularly on solving equations that involve finding the zeros of a polynomial.
This exercise emphasizes understanding the relationship between a polynomial's zeros and its coefficients, specifically for quadratic polynomials. Problems include calculating zeros and verifying that they satisfy the given polynomial. It also covers the concepts of factorization and the Factor Theorem, which helps students break down polynomials into simpler linear or quadratic factors, deepening their foundational understanding of algebraic expressions and polynomial functions.
Students can better grasp basic polynomial principles by working through the problems in Exercise 2.3 of RD Sharma's Class 10 Maths Chapter 2. By working through these problems, students gain vital algebraic skills such as determining a polynomial's zeros and comprehending the connection between the zeros and the polynomial's coefficients.
Their factorisation abilities, which are essential for decomposing complicated expressions and resolving equations, are strengthened by this practice. These skills are extremely significant in advanced mathematics since solving these issues provides a strong basis for more complex algebraic topics including quadratic equations, calculus, and graphing polynomial functions.
RD Sharma Solutions Class 10 Maths Chapter 2 Exercise 2.3 PDF
Below, we have provided the PDF containing the RD Sharma Solutions for Class 10 Maths, specifically for Chapter 2, Exercise 2.3 on Polynomials. This exercise covers important concepts and problems related to polynomials, helping students enhance their understanding and problem-solving skills.
By referring to this comprehensive solution guide, students can effectively tackle their assignments and prepare for exams with confidence. Feel free to download the PDF for detailed explanations and solutions.
Given,
g(t) = t
2
– 3; f(t) =2t
4
+ 3t
3
– 2t
2
– 9t – 12
Since, the remainder r(t) = 0 we can say that
the first polynomial is a factor of the second polynomial.
(ii) g(x) = x
3
– 3x + 1; f(x) = x
5
– 4x
3
+ x
2
+ 3x + 1
Solution:
Given,
g(x) = x
3
– 3x + 1; f(x) = x
5
– 4x
3
+ x
2
+ 3x + 1
Since, the remainder r(x) = 2 and not equal to zero we can say that
the first polynomial is not a factor of the second polynomial.
(iii) g(x) = 2x
2
– x + 3; f(x) = 6x
5
− x
4
+ 4x
3
– 5x
2
– x –15
Solution:
Given,
g(x) = 2x
2
– x + 3; f(x)=6x
5
− x
4
+ 4x
3
– 5x
2
– x –15
Since, the remainder r(x) = 0 we can say that
the first polynomial is not a factor of the second polynomial.
3.
Obtain all zeroes of the polynomial f(x)= 2x
4
+ x
3
– 14x
2
– 19x–6, if two of its zeroes are -2 and -1.
Solution:
Given,
f(x)= 2x
4
+ x
3
– 14x
2
– 19x – 6
If the two zeros of the polynomial are -2 and -1, then its factors are (x + 2) and (x + 1)
⇒ (x+2)(x+1) = x
2
+ x + 2x + 2 = x
2
+ 3x +2 …… (i)
This means that (i) is a factor of f(x). So, performing division algorithm we get,
The quotient is 2x
2
– 5x – 3.
⇒ f(x)= (2x
2
– 5x – 3)( x
2
+ 3x +2)
For obtaining the other 2 zeros of the polynomial
We put,
2x
2
– 5x – 3 = 0
⇒ (2x + 1)(x – 3) = 0
∴ x = -1/2 or 3
Hence, all the zeros of the polynomial are -2, -1, -1/2 and 3.
4.
Obtain all zeroes of f(x) = x
3
+ 13x
2
+ 32x + 20, if one of its zeros is -2.
Solution:
Given,
f(x)=
x
3
+ 13x
2
+ 32x + 20
And, -2 is one of the zeros. So, (x + 2) is a factor of f(x),
Performing division algorithm, we get
⇒ f(x)= (x
2
+ 11x + 10)( x + 2)
So, putting x
2
+ 11x + 10 = 0 we can get the other 2 zeros.
⇒ (x + 10)(x + 1) = 0
∴ x = -10 or -1
Hence, all the zeros of the polynomial are -10, -2 and -1.
5.
Obtain all zeroes of the polynomial f(x) = x
4
– 3x
3
– x
2
+ 9x – 6, if the two of its zeroes are −√3 and √3.
Solution:
Given,
f(x) = x
4
– 3x
3
– x
2
+ 9x – 6
Since, two of the zeroes of polynomial are −√3 and √3 so, (x + √3) and (x–√3) are factors of f(x).
⇒ x
2
– 3 is a factor of f(x). Hence, performing division algorithm, we get
⇒ f(x)= (x
2
– 3x + 2)( x
2
– 3)
So, putting x
2
– 3x + 2 = 0 we can get the other 2 zeros.
⇒ (x – 2)(x – 1) = 0
∴ x = 2 or 1
Hence, all the zeros of the polynomial are −√3, 1, √3 and 2.
6.
Obtain all zeroes of the polynomial f(x)= 2x
4
– 2x
3
– 7x
2
+ 3x + 6, if the two of its zeroes are −√(3/2) and √(3/2).
Solution:
Given,
f(x)= 2x
4
– 2x
3
– 7x
2
+ 3x + 6
Since, two of the zeroes of polynomial are −√(3/2) and √(3/2) so, (x + √(3/2)) and (x –√(3/2)) are factors of f(x).
⇒ x
2
– (3/2) is a factor of f(x). Hence, performing division algorithm, we get
⇒ f(x)= (2x
2
– 2x – 4)( x
2
– 3/2)= 2(x
2
– x – 2)( x
2
– 3/2)
So, putting x
2
– x – 2 = 0 we can get the other 2 zeros.
⇒ (x – 2)(x + 1) = 0
∴ x = 2 or -1
Hence, all the zeros of the polynomial are −√(3/2), -1, √(3/2) and 2.
7.
Find all the zeroes of the polynomial x
4
+ x
3
– 34x
2
– 4x + 120, if the two of its zeros are 2 and -2.
Solution:
Let,
f(x) = x
4
+ x
3
– 34x
2
– 4x + 120
Since, two of the zeroes of polynomial are −2 and 2 so, (x + 2) and (x – 2) are factors of f(x).
⇒ x
2
– 4 is a factor of f(x). Hence, performing division algorithm, we get
⇒ f(x)= (x
2
+ x – 30)( x
2
– 4)
So, putting x
2
+ x – 30 = 0, we can get the other 2 zeros.
⇒ (x + 6)(x – 5) = 0
∴ x = -6 or 5
Hence, all the zeros of the polynomial are 5, -2, 2 and -6.
Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 2 Exercise 2.3
Solving the RD Sharma Solutions for Class 10 Maths, specifically Chapter 2 Exercise 2.3 on Polynomials, offers several benefits for students:
Concept Clarity
: It reinforces the fundamental concepts of polynomials, ensuring students understand their properties, operations, and applications.
Step-by-Step Guidance
: The solutions provide a systematic approach to problem-solving, helping students grasp the methods used to tackle various polynomial problems.
Varied Problem Types
: The exercise includes a range of problems, from basic to complex, allowing students to build confidence and proficiency in handling different scenarios.
Self-Assessment
: Working through the solutions enables students to assess their understanding and identify areas needing improvement.
Exam Preparation
: The comprehensive practice prepares students for exams, enhancing their ability to solve questions under time constraints.
Enhanced Problem-Solving Skills
: Regular practice with these solutions helps develop critical thinking and analytical skills, which are essential for higher-level mathematics.
Talk to a counsellorHave doubts? Our support team will be happy to assist you!
Thus,
q(x) = x – 7 and r(x) = 17x +1
Thus,
q(x) = 15x + 10 and r(x) = 3x – 32
Since, the remainder r(t) = 0 we can say that
the first polynomial is a factor of the second polynomial.
Since, the remainder r(x) = 2 and not equal to zero we can say that
the first polynomial is not a factor of the second polynomial.
Since, the remainder r(x) = 0 we can say that
the first polynomial is not a factor of the second polynomial.
The quotient is 2x
2
– 5x – 3.
⇒ f(x)= (2x
2
– 5x – 3)( x
2
+ 3x +2)
For obtaining the other 2 zeros of the polynomial
We put,
2x
2
– 5x – 3 = 0
⇒ (2x + 1)(x – 3) = 0
∴ x = -1/2 or 3
Hence, all the zeros of the polynomial are -2, -1, -1/2 and 3.
⇒ f(x)= (x
2
+ 11x + 10)( x + 2)
So, putting x
2
+ 11x + 10 = 0 we can get the other 2 zeros.
⇒ (x + 10)(x + 1) = 0
∴ x = -10 or -1
Hence, all the zeros of the polynomial are -10, -2 and -1.
⇒ f(x)= (x
2
– 3x + 2)( x
2
– 3)
So, putting x
2
– 3x + 2 = 0 we can get the other 2 zeros.
⇒ (x – 2)(x – 1) = 0
∴ x = 2 or 1
Hence, all the zeros of the polynomial are −√3, 1, √3 and 2.
⇒ f(x)= (2x
2
– 2x – 4)( x
2
– 3/2)= 2(x
2
– x – 2)( x
2
– 3/2)
So, putting x
2
– x – 2 = 0 we can get the other 2 zeros.
⇒ (x – 2)(x + 1) = 0
∴ x = 2 or -1
Hence, all the zeros of the polynomial are −√(3/2), -1, √(3/2) and 2.
⇒ f(x)= (x
2
+ x – 30)( x
2
– 4)
So, putting x
2
+ x – 30 = 0, we can get the other 2 zeros.
⇒ (x + 6)(x – 5) = 0
∴ x = -6 or 5
Hence, all the zeros of the polynomial are 5, -2, 2 and -6.
