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NCERT Solutions for Class 9th Mathematics

 

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Chapter 2. Polynomials

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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