NCERT Solutions Class 9 Mathematics Polynomials Exercise 2.3

Class 9 - Mathematics
Polynomials - Exercise 2.3

NCERT Solutions Class 9 Mathematics Textbook
Top Block 1

Exercise 2.3


Question : Find the remainder when x3 + 3x2 + 3x + 1 is divided by

(i) x + 1           (ii) x – 1/2              (iii) x                (iv) x + π                  (v) 5 + 2x

Answer :

Let p(x) = x3 + 3x2 + 3x + 1

(i) put x + 1 = 0, we get

x = -1

Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x + 1, remainder is given

by p(-1).

Now, p(-1) = (-1)3 + 3 * (-1)2 + 3 * (-1) + 1

                    = -1 + 3 – 3 + 1

                    = -4 + 4

                    = 0

Hence, the remainder is 0.    

(ii) put x – 1/2 = 0, we get

x = 1/2

Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x – 1/2, remainder is given

by p(1/2).

Now, p(1/2) = (1/2)3 + 3 * (1/2)2 + 3 * (1/2) + 1

                    = 1/8 + 3 * 1/4 + 3/2 + 1

                    = 1/8 + 3/4 + 3/2 + 1

                    = (1 * 1 + 3 * 2 + 3 * 4 + 1 * 8)/8                            [LCM(8, 4, 2, 1) = 8]

                  = (1 + 6 + 12 + 8)/8

                  = 27/8

Hence, the remainder is 27/8              

(iii) put x = 0

Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x, remainder is given

by p(0).

Now, p(-1) = (0)3 + 3 * (0)2 + 3 * (0) + 1

                    = 0 + 0 + 0 + 1

                    = 1

Hence, the remainder is 1.               

(iv) put x + π = 0

Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x + π, remainder is given

by p(-π).

Now, p(-π) = (-π)3 + 3 * (-π)2 + 3 * (-π) + 1

                    = -π3 + 3 π3 – 3 π + 1

Hence, the remainder is -π3 + 3 π3 – 3 π + 1.                  

(v) put 5 + 2x = 0, we get

     2x = -5

⇒ x = -5/2

Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by 5 + 2x, remainder is given

by p(-5/2).

Now, p(-5/2) = (-5/2)3 + 3 * (-5/2)2 + 3 * (-5/2) + 1

                    = -125/8 + 3 * 25/4 – 15/2 + 1

                    = -125/8 + 75/4 – 15/2 + 1

                    = (-1 * 125 + 75 * 2 – 15 * 4 + 1 * 8)/8             [LCM(8, 4, 2, 1) = 8]

                    = (-125 + 150 – 60 + 8)/8

                    = -27/8

Hence, the remainder is 27/8

Mddle block 1

Question : 2: Find the remainder when x3 – ax2 + 6x – a is divided by x – a.

Answer :

Let p(x) = x3 – ax2 + 6x – a

Put x – a = 0, we get

x = a

Using remainder theorem, when p(x) = x3 – ax2 + 6x – a is divided by x – a, remainder is given

by p(a).

Now, p(a) = a3 – a * a2 + 6 * a – a

                   = a3 – a3 + 6a – a

                   = 5a

Hence, the remainder is 5a.  


Question : 3: Check whether 7 + 3x is a factor of 3x2 + 7x

Answer :

Given, 3x2 + 7x = 3 * x * x + 7 * x

                            = x(3 * x + 7)

                            = x(3x + 7)

                            = x(7 + 3x)

Hence, 7 + 3x is a factor of 3x2 + 7x

Bottom Block 3
Share with your friends

Leave a Reply