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2. Polynomials Mathematics Exercise - 2.3 class 9 Maths in English - CBSE Study

2. Polynomials Mathematics Class 9 exercise - 2.3 class 9 Maths cbse board school study materials like cbse notes in English medium, all chapters and exercises are covered the ncert latest syllabus 2026 - 27.

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2. Polynomials Mathematics Exercise - 2.3 class 9 Maths in English - CBSE Study

NCERT Solutions for Class 9 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 2. Polynomials with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 2.3 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.

Class 9 English Medium Mathematics All Chapters:

2. Polynomials

3. Exercise 2.3

Chapter 2. Polynomials 


Exercise 2.3

Solution: 

(i)By remainder theorem, the required remainder is equal to p(x) = (-1)

              P (-1) = x3+ 3x2+ 3x+1

                        = (-1)3 + 3 (-1)2 + 3(-1) + 1

                        = -1 + 3 – 3 + 1 = 0 

              Required remainder is p (-1) = 0

          

          Required remainder is p (1/2) = 0

(iv) By remainder theorem, the required remainder is equal to p(x) = - π

                   P(x) = x3+ 3x2+ 3x+1

                   P(π) = (- π)3 + 3(-π)2 + 3(-π) + 1

                           = (-π)3 + 3π2 + 3π + 1

    Required remainder is p (π) = 0 

 Required remainder is p (- 5/2) = 0 

Q.2. find the remainder  when x3- ax+ 6x - a is divided by x- a.

Solution: Let P(x) = x3- ax2+ 6x- a .   P(x) = a.

                   By remainder theorem 

              P (a) = (a)3- a(a)2+ 6(a) - a

                       =a3-a3 + 6a - a

     Remainder = 5a

Q3. Check whether 7 + 3x is a factor of 3x3 + 7x. 

Solution: 

g(x) = 7 + 3x = 0 

       ⇒ 3x = - 7

       

P(x)  0

Therefore,  7 + 3x is not a factor of P(x). 

 

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