Polynomials | Exercise 2.3 | NCERT Solution Maths 9 | Chapter 2

Exercise 2.3 Page: 40

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

(i) x+1

Solution:

x+1= 0

⇒x = −1

∴Remainder:

p(−1) = (−1)3+3(−1)2+3(−1)+1

= −1+3−3+1

= 0

(ii) x−1/2

Solution:

x-1/2 = 0

⇒ x = 1/2

∴Remainder:

p(1/2) = (1/2)3+3(1/2)2+3(1/2)+1

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

= 27/8

(iii) x

Solution:

x = 0

∴Remainder:

p(0) = (0)3+3(0)2+3(0)+1

= 1

(iv) x+π

Solution:

x+π = 0

⇒ x = −π

∴Remainder:

p(0) = (−π)3 +3(−π)2+3(−π)+1

= −π3+3π2−3π+1

(v) 5+2x

Solution:

5+2x=0

⇒ 2x = −5

⇒ x = -5/2

∴Remainder:

(-5/2)3+3(-5/2)2+3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1

= -27/8

2. Find the remainder when x3−ax2+6x−a is divided by x-a.

Solution:

Let p(x) = x3−ax2+6x−a

x−a = 0

∴x = a

Remainder:

p(a) = (a)3−a(a2)+6(a)−a

= a3−a3+6a−a = 5a

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

Solution:

7+3x = 0

⇒ 3x = −7

⇒ x = -7/3

∴Remainder:

3(-7/3)3+7(-7/3) = -(343/9)+(-49/3)

= (-343-(49)3)/9

= (-343-147)/9

= -490/9 ≠ 0

∴7+3x is not a factor of 3x3+7x