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BSEB Class 9 Maths Chapter 2 Polynomials Ex 2.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 2 Polynomials Ex 2.3 Book Answers

 BSEB Class 9 Maths Chapter 2 Polynomials Ex 2.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 2 Polynomials Ex 2.3 Book Answers

BSEB Class 9th Maths Chapter 2 Polynomials Ex 2.3 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 9th 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 9th Maths Chapter 2 Polynomials Ex 2.3 books are published by the Bihar Board Publishers. These Bihar Board Class 9th Maths Chapter 2 Polynomials Ex 2.3 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 9th Maths Chapter 2 Polynomials Ex 2.3 book solutions pdf online from this page.

Bihar Board Class 9th Maths Chapter 2 Polynomials Ex 2.3 Books Solutions

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BSEB Bihar Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) x – 12
(iii) x
(iv) x + π
(v) 5 + 2x
Solution:
(i) By remainder theorem, the required remainder is equal to p(- 1).
Now, p(x) = x³ + 3x² + 3x + 1
∴ p(- 1) = (- 1)³ + 3(- 1)² + 3(- 1) + 1
= – 1 + 3 – 3 + 1 = 0
Hence, required remainder = p(- 1) = 0

(ii) By remainder theorem, the required remainder is equal p(12)

(iii) By remainder theorem, the required remainder is equal to p(0).
Now, p(x) = x³ + 3x² + 3x + 1
∴ p(0) = 0 + 0+ 0 + 1 = 1
Hence, the required remainder = p(0) = 1

(iv) By remainder theorem the required remainder is p(~ n)
Now p(x) = x³ + 3x² + 3x + 1
∴ p(- π) = (- π)³ + 3(- π)² + 3(- π) + 1
= – π³ + 3π² – 3π + 1

(v) By remainder theorem, the required remainder is 5 p(- 52)

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Solution:
Let p(x) = x³ – ax² + 6x – a
By remainder theorem, when p(x) is divided by x – a. Then, remainder = p(a)
∴ p(x) = a³ – a . a² + 6a – a
= a³ – a³ + 6a – a = 5a

Question 3.
Check whether 7 + 3x is a factor of 3x² + 7x
Solution:
7 + 3x will be a factor of p{x) = 3x² + 7x if p(-73) = 0

Bihar Board Class 9th Maths Chapter 2 Polynomials Ex 2.3 Textbooks for Exam Preparations

Bihar Board Class 9th Maths Chapter 2 Polynomials Ex 2.3 Textbook Solutions can be of great help in your Bihar Board Class 9th Maths Chapter 2 Polynomials Ex 2.3 exam preparation. The BSEB STD 9th Maths Chapter 2 Polynomials Ex 2.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 2 Polynomials Ex 2.3 Books State Board syllabus with maximum efficiency.

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