NCRT math class 9th polynomial exersice 2.3

                       EXERSICE 2.3

Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x – 
(iii) x
(iv) x + π
(v) 5 + 2x
     Solution:
     Let p(x) = x3 + 3x2 + 3x +1
(i) The zero of x + 1 is -1.
∴  p(x=-1) = (-1)^3 + 3(-1)^2 + 3(-1) +1
    = -1 + 3- 3 + 1 = 0
   Thus, the  remainder = 0

(ii) The zero of  is 

     Thus, the  remainder = 

iii) The zero of x is 0.
∴ p(0) = (0)^3 + 3(0)^2 + 3(0) + 1
= 0 + 0 + 0 + 1 = 1
Thus, the  remainder = 1.

(iv) The zero of x + π is -π.
       p(-π) = (-π)^3 + 3(- π)^2  + 3(- π) +1
       = -π3 + 3π2 + (-3π) + 1
       = – π3 + 3π2 – 3π +1
       Thus, the required remainder is                        -π3 + 3π2 – 3π+1.

(v) The zero of 5 + 2x is  .


     Thus, the remainder is  .

Question 2.
Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Solution:
We have, p(x) = x3 – ax2 + 6x – a and zero       of x – a is a.
  ∴ p(a) = (a)3 – a(a)2 + 6(a) – a
  = a3 – a3 + 6a – a = 5a
 Thus, the  remainder is 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x3+7x.
Solution:
We have, p(x) = 3x3+7x. and zero of 7 + 3x is .

Since,( ) ≠ 0
i.e. the remainder is not 0.
∴ 3x3 + 7x is not divisib1e by 7 + 3x.
Thus, 7 + 3x is not a factor of 3x3 + 7x.