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2. Polynomials Mathematics class 10 in English Medium ncert book solutions Exercise 2.3
2. Polynomials Exercise 2.3 – Complete NCERT Book Solutions for Class 10 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 2. Polynomials Exercise 2.3 to help you master concepts and score higher.
2. Polynomials Mathematics class 10 in English Medium ncert book solutions Exercise 2.3
NCERT Solutions for Class 10 Mathematics play an important role in helping students understand the concepts of the chapter 2. Polynomials clearly. This chapter includes the topic Exercise 2.3, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 10 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 2. Polynomials is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 2.3. By studying these updated NCERT Solutions for Class 10 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
2. Polynomials
Exercise 2.3
Exercise 2.3 class 10 maths chapter 2. Polynomials
Q1. 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
Solution: (i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Quotients q(x) = x – 3 and Remainder = 7x – 9
Solution: (ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
Quotients q(x) = x2 + x – 3 and Remainder = 8
Solution: (iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Quotients q(x) = – x2 – 2 and Remainder = – 5x + 10
Q2. Check whether the first polynomial is a factor of the second polynomial 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
Solution: (i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
Hence Remainder r(x) is 0
Therefore, t2 – 3 is the factor of 2t4 + 3t3 – 2t2 – 9t – 12
Solution: (ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
Hence Remainder r(x) is 0
Therefore, x2 + 3x + 1 is the factor of 3x4 + 5x3 – 7x2 + 2x + 2
Solution: (iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Hence Remainder r(x) = 2
Therefore, x3 – 3x + 1, is not a factor of x5 – 4x3 + x2 + 3x + 1
Q3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are
Solution:
Given that : p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Or 3x2 - 5 = 0
Therefore, 3x2 - 5 is the factor of p(x)
Now Dividing 3x4 + 6x3 - 2x2 - 10x - 5 by 3x2 - 5
Therefore, p(x) = (3x2 – 5) (x2 + 2x + 1)
Now, factorizing and getting zeroes x2 + 2x + 1 -
= x2 + x + x + 1 = 0
= x(x + 1) + 1(x + 1) = 0
= (x + 1) (x + 1) = 0
Or x + 1 = 0, x + 1 = 0
Or x = – 1, x = – 1
Therefore, two zeroes are – 1 and – 1.
Q4. 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).
Solution:
Given that: Dividend p(x) = x3 – 3x2 + x + 2
Quotient q(x) = x – 2,
Remainder r(x) = – 2x + 4
Divisor g(x) =?
Dividend = divisor × quotient + remainder
p(x) = g(x) × q(x) + r(x)
x3 – 3x2 + x + 2 = g(x) (x – 2) + (– 2x + 4)
x3 – 3x2 + x + 2 + 2x – 4 = g(x) (x – 2)
g(x) (x – 2) = x3 – 3x2 + 3x – 2
Dividing x3 – 3x2 + 3x – 2 by x - 2 we obtain g(x)-
Therefore, Divisor g(x) = x2 – x + 1
Q5. 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
Solution:
Using Euclid’s Division algorithm:
p(x) = g(x) × q(x) + r(x) where q(x) – 0
(i) deg p(x) = deg q(x)
The deg of dividend p(x) and quotient q(x) can be equal when deg of divisor is 0 or any number.
Example : Let p(x) = 2x2 - 6x + 3
And let g(x) = 2
On dividing
p(x) = 2x2 - 6x + 2 + 1
= 2(x2 - 3x + 1) + 1
Now comparing 2(x2 - 3x + 1) + 1 by p(x) = g(x) × q(x) + r(x) we get:
So, q(x) = x2 - 3x + 1and r(x) = 1
By which we obtain deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
Solution: This situation comes when deg p(x) and deg g(x) is equal-
Let p(x) = 2x2 + 6x + 7 and g(x) = x2 + 3x + 2
On dividing: q(x) = 2 and r(x) = 3
Therefore, deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
r(x) = 0 is obtained when p(x) is completely divisible by g(x):
Let p(x) = x2 – 1 and g(x) = x + 1
On dividing we obtain:
q(x) = x – 1 and r(x) = 0
See other sub-topics of this chapter:
1. Exercise 2.1 2. Exercise 2.2 3. Exercise 2.3 4. Exercise 2.4