BSEB Class 10 Maths Chapter 2 Polynomials Ex 2.3 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.3 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

BSEB Class 10 Maths Chapter 2 Polynomials Ex 2.3 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.3 Book Answers

BSEB Class 10th Maths Chapter 2 Polynomials Ex 2.3 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 books are published by the Bihar Board Publishers. These Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 10th Maths Chapter 2 Polynomials Ex 2.3 book solutions pdf online from this page.

Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 Textbooks Solutions PDF

Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.3 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 10th Maths Chapter 2 Polynomials Ex 2.3 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.3 Textbooks. These Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 Books Solutions

Board	BSEB
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths Chapter 2 Polynomials Ex 2.3
Chapters	All
Provider	Hsslive

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BSEB Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

p(x) = r³ – 3x² + 5x – 3, g(x) = x² – 2
p(x) = x⁴ – 3x² + 4x + 5, g(x) = x²+ 1 – x
p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Solution:
1. Here, dividend and divisor are both in standard forms. So, we have:

2. Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as
x² – x + 1.
We have:

∴ The quotient is x² + x – 3 and the remainder is 8.

3. To carry out the division, we first write divisor in the standard form.
So, divisor = – x² + 2
We have:

∴ The quotient is – x² – 2 and the remainder is – 5x + 10.

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution:
1. Let us divide 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3.
We have:

Since the remainder is 0, therefore t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

2. Let us divide 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1.
We get,

Since, the remainder is 0, therefore x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2

3. Let us divide x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1. We get,

Here, remainder is 2(≠ 0). Therefore, x³ – 3x + 1 is not a factor of
x⁵ – 4x³ + x² + 3x + 1.

Question 3.
Obtain all the zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are
53‾‾√ and – 53‾‾√.
Solution:
Since two zeroes are 53‾‾√ and – 53‾‾√, so (x – 53‾‾√) and (x + 53‾‾√) are the factors of the given polynomial.
Now, (x – 53‾‾√) (x + 53‾‾√) = x² – 53.
So, (3x² – 5) is a factor of the given polynomial.
Applying the division algorithm to the given polynomial and 3x² – 5, we have:

∴ 3x⁴ + 6x³ – 2x² – 10x – 5 = (3x² – 5)(x² + 2x + 1)
Now, x² + 2x + 1 = x² + x + x + 1
= x(x + 1) + 1(x + 1)
= (x + 1)(x + 1)
So, its other zeroes are – 1 and – 1.
Thus, all the zeroes of the given fourth degree polynomial are 53‾‾√, – 53‾‾√, – 1 and – 1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quoitent and remainder were x – 2 and – 2x + 4 respectively. Find g(x).
Solution:
Since on dividing x³ – 3x² + x + 2 by a polynomial g(x), the quoitent and remainder were (x – 2) and (- 2x + 4) respectively, therefore
Quoitent × Divisor + Remainder = Dividend
or (x – 2) × g(x) + (- 2x + 4) = x³ – 3x² + x + 2
or (x – 2) × g(x) = x³ – 3x² + x + 2 + 2x – 4
or g(x) = 𝑥3−3𝑥2+3𝑥−2𝑥−2 …………… (1)
Let us divide x³ – 3x² + 3x – 2 by x – 2. We get

∴ (1) gives g(x) = x² – x + 1.

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

deg p(x) = deg q(x)
deg q(x) = deg r(x)
deg r(x) = 0

Solution:
There can be several examples for each of (i), (ii) and (iii).
However, one example for each case may be taken as under:

p(x) = 14, g(x)= 2, q(x) = x² – x + 7, r(x) = 0
p(x) = x³ + x² + x + 1, g(x) = x² – 1, q(x) = x + 1, r(x) = 2x + 2
p(x) = x³ + 2x² – x + 2, g(x) = – 1, q(x) = x + 2, r(x) = 4

BSEB Textbook Solutions PDF for Class 10th

Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 Textbooks for Exam Preparations

Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 Textbook Solutions can be of great help in your Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.3 exam preparation. The BSEB STD 10th Maths Chapter 2 Polynomials Ex 2.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 2 Polynomials Ex 2.3 Books State Board syllabus with maximum efficiency.

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