Pearson | PTE
Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK
Class 10 Maths Chapter 2 Solutions Polynomials will discuss about the polynomials with any degree and its solutions using a graphical representation. These solutions are prepared by expert teachers keeping in mind the latest CBSE syllabus 2023. Also, Class 10 maths chapter 2 solutions polynomials gives a detailed explanation of questions given in the NCERT Book of Class 10 Maths.
You must refer to these Polynomial Class 10 notes to check the correct answers and proper explanation of each question. These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials are prepared by experts and are easy understand. You can check the NCERT solutions for Class 10 for other subjects here.
Also Read,
The general Polynomial Expression is represented as follows:
F(x) = anxn + an-1xn-1 + an-2xn-2 + ... + rx + s.
For a natural number n, an-bn = (a - b)(an-1 + an-2b + ... + bn-2a + bn-1)
When n is an even number (expressed as n = 2a),
xn + yn = (x + y)(xn-1 - xn-2y + ... + yn-2x - yn-1)
For an odd number n,
xn + yn = (x + y)(xn-1 - xn-2y + ... - yn-2x + yn-1)
Different types of polynomials encompass various essential concepts and properties, each delineated in detail in the provided table.
Algebraic polynomial identity
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
(a + b)(a - b) = a2 - b2
(x + a)(x + b) = x2 + (a + b)x + ab
(x + a)(x - b) = x2 + (a - b)x - ab
(x - a)(x + b) = x2 + (b - a)x - ab
(x - a)(x - b) = x2 - (a + b)x + ab
(a + b)3 = a3 + b3 + 3ab(a + b)
(a - b)3 = a3 - b3 - 3ab(a - b)
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz
(x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz
(x -y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz
(x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
x2 + y2 = 12[(x + y)2 + (x - y)2]
(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
x3 + y3 = (x + y)(x2 - xy + y2)
x3 - y3 = (x - y)(x2 + xy + y2)
x2 + y2 + z2 - xy - yz - zx = 1/2[(x - y)2 + (y - z)2 + (z - x)2]
The division algorithm for polynomials states that when considering any two polynomials p(x) and g(x). Where g(x) is not equal to zero, Then we can find polynomials q(x) and r(x) such that:
p(x) = q(x) g(x) + r(x)
Here, r(x) = 0 or deg r(x) ≥ deg g(x)
Free download NCERT Solutions for Class 10 Maths Chapter 2 Polynomials PDF for CBSE Exam.
NCERT Solutions for Chapter 2 Maths Class 10 Polynomials Excercise: 2.1
Class 10 Maths Chapter 2 Solutions Polynomials Excercise: 2.2
Answer:
x 2 - 2x - 8 = 0
x 2 - 4x + 2x - 8 = 0
x(x-4) +2(x-4) = 0
(x+2)(x-4) = 0
The zeroes of the given quadratic polynomial are -2 and 4
VERIFICATION
Sum of roots:
Verified
Product of roots:
Verified
The zeroes of the given quadratic polynomial are 1/2 and 1/2
VERIFICATION
Sum of roots:
Verified
Product of roots:
Verified
6x 2 - 3 - 7x = 0
6x 2 - 7x - 3 = 0
6x 2 - 9x + 2x - 3 = 0
3x(2x - 3) + 1(2x - 3) = 0
(3x + 1)(2x - 3) = 0
The zeroes of the given quadratic polynomial are -1/3 and 3/2
Sum of roots:
Verified
Product of roots:
Verified
4u(u + 2) = 0
The zeroes of the given quadratic polynomial are 0 and -2
VERIFICATION
Sum of roots:
Verified
Product of roots:
Verified
The zeroes of the given quadratic polynomial are and
VERIFICATION
Sum of roots:
Verified
Product of roots:
Verified
3x 2 + 3x - 4x - 4 = 0
3x(x + 1) - 4(x + 1) = 0
(3x - 4)(x + 1) = 0
The zeroes of the given quadratic polynomial are 4/3 and -1
VERIFICATION
Sum of roots:
Verified
Product of roots:
Verified
Answer:
The required quadratic polynomial is
The required quadratic polynomial is
The required quadratic polynomial is x 2 + .
The required quadratic polynomial is x 2 - x + 1
The required quadratic polynomial is 4x 2 + x + 1
The required quadratic polynomial is x 2 - 4x + 1
NCERT Solutions for Chapter 2 Maths Class 10 Polynomials Excercise: 2.3
The quotient is x-3 and the remainder is 7x-9
The division is carried out as follows
The quotient is
and the remainder is 8
The quotient is and the remainder is
After dividing we got the remainder as zero. So is a factor of
After division, the remainder is zero thus is a factor of
The remainder is not zero, there for the first polynomial is not a factor of the second polynomial
Q3 Obtain all other zeroes of , if two of its zeroes are
Answer:
Two of the zeroes of the given polynomial are .
Therefore two of the factors of the given polynomial are and
is a factor of the given polynomial.
To find the other factors we divide the given polynomial with
The quotient we have obtained after performing the division is
(x+1) 2 = 0
x = -1
The other two zeroes of the given polynomial are -1.
remainder =-2x+4
Carrying out the polynomial division as follows
Q5 (2) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and deg q(x) = deg r(x)
example for the polynomial which satisfies the division algorithm with r(x)=0 is given below
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Excercise: 2.4
p(x) = 2x 3 + x 2 -5x + 2
p(1) = 2 x 1 3 + 1 2 - 5 x 1 + 2
p(1) =2 + 1 - 5 + 2
p(1) = 0
p(-2) = 2 x (-2) 3 + (-2) 2 - 5 x (-2) +2
p(-2) = -16 + 4 + 10 + 2
p(-2) = 0
Therefore the numbers given alongside the polynomial are its zeroes
Verification of relationship between the zeroes and the coefficients
Comparing the given polynomial with ax 3 + bx 2 + cx + d, we have
a = 2, b = 1, c = -5, d = 2
The roots are
Verified
Verified
Verified
p(x) = x 3 - 4x 2 + 5x - 2
p(2) = 2 3 - 4 x 2 2 + 5 x 2 - 2
p(2) = 8 - 16 + 10 - 2
p(-2) = 0
p(1) = 1 3 - 4 x 1 2 + 5 x 1 - 2
p(1) = 1 - 4 + 5 - 2
p(1) = 0
Therefore the numbers given alongside the polynomial are its zeroes
Verification of relationship between the zeroes and the coefficients
Comparing the given polynomial with ax 3 + bx 2 + cx + d, we have
a = 1, b = -4, c = 5, d = -2
The roots are
Verified
Verified
Verified
Hence the required cubic polynomial is x 3 - 2x 2 - 7x + 14 = 0
Q3 If the zeroes of the polynomial are a – b, a, a + b, find a and b.
The roots of the above polynomial are a, a - b and a + b
Sum of the roots of the given plynomial = 3
a + (a - b) + (a + b) = 3
3a = 3
a = 1
The roots are therefore 1, 1 - b and 1 + b
Product of the roots of the given polynomial = -1
1 x (1 - b) x (1 + b) = - 1
1 - b 2 = -1
b 2 - 2 = 0
Therefore a = 1 and .
Q4 If two zeroes of the polynomial
are
, find other zeroes .
therefore the factors are
We have to find the remaining two factors. To find the remaining two factors we have to divide the polynomial with the product of the above factors
Now carrying out the polynomial division
Now we get
So the zeroes are
Given the remainder =x+a
The obtained remainder after division is
now equating the coefficient of x
which gives the value of
now equating the constants
Therefore k=5 and a=-5
The degree of a polynomial
Number of zeroes in a quadratic polynomial
The sum & product of quadratic polynomial
Division algorithm
The number of zeroes in a cubic polynomial
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | Polynomials |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
The chapter 2 Maths Class 10 are important for board exams as well as for competitive exams.
These NCERT solutions are prepared by experts. These solutions are very well explained and are to the point.
To make the Class 10 Maths chapter 2 solutions easy to understand, relevant diagrams or images are given with the solutions wherever required.
You can check these solutions online or can save the webpage to get these solutions offline. NCERT Solutions for Class 10 Maths Chapter 2 PDF download will also be made available soon.
Also See,
You will find yourself much improved in terms of concepts, their applications, and problem-solving. 90% of the questions in the board examinations come from the NCERT syllabus.
Now when you have gone through the NCERT solutions for Maths chapter 2 Polynomials Class 10 , you must have learnt to answer a question in the step by step manner.
Once you have studied the Class 10 Maths chapter 2 NCERT solutions, your next target should be previous year questions papers.
NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After going through these two things, you can jump to the next chapters.
At the careers360 official website, you can find polynomials class 10 ncert solutions. These solutions for ch2 maths class 10 are designed and developed by reliable experts keeping in mind the new pattern provided by the state and central boards. Also, you can download it as a polynomials class 10 pdf.
The topics covered in NCERT maths class 10 chapter 2 (Polynomials) are:
Definition and types of polynomials
Zeroes of a polynomial
Division algorithm for polynomials
Fundamental Theorem of Algebra
Relationship between zeroes and coefficients of a polynomial
Nature of roots of a polynomial
There are several reasons why you should use the ncert solutions for class 10 maths chapter 2 polynomials:
The NCERT Solutions are accurate and reliable
The NCERT Solutions cover all the concepts and exercises given in the textbook
The NCERT Solutions can help you develop problem-solving skills
The NCERT Solutions can be used as a reference guide
Overall, using the polynomials class 10 questions with answers can be a useful resource for students to improve their understanding of the concepts and perform well in exams.
There are a total of 10 exercises in Chapter 2 (Polynomials) of the NCERT Solutions for Class 10 Mathematics. These exercises consist of a variety of questions, including multiple choice questions, fill in the blanks, true or false, and short answer questions. The number of questions in each exercise may vary.
According to the NCERT chapter 2 Class 10 Maths Polynomials, a polynomial is a mathematical expression that consists of variables (such as x, y, etc.) and coefficients (numbers), which are combined using only the operations of addition, subtraction, and multiplication.
Exam Date:01 January,2025 - 14 February,2025
Exam Date:01 January,2025 - 14 February,2025
Hello
Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.
1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.
2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.
3. Counseling and Seat Allocation:
After the KCET exam, you will need to participate in online counseling.
You need to select your preferred colleges and courses.
Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.
4. Required Documents :
Domicile Certificate (proof that you are a resident of Karnataka).
Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
https://medicine.careers360.com/neet-college-predictor
Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide
Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK