Question 1. Find the Remainder When X3+3x2 + 3x + 1 is Divided B Exercise 2.3 Chapter 2 Polynomial Maths Class 9

Question 1. 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

Solution: (i) x + 1         
Apply remainder theorem
=>x + 1 =0
=> x = - 1
Replace x by – 1 we get
=>x³+3x² + 3x + 1
=>(-1)³ + 3(-1)² + 3(-1) + 1
=> -1 + 3 - 3 + 1
=> 0
Remainder is 0

(ii) x –1/2       
Apply remainder theorem
=>x – 1/2 =0
=> x =  1/2
Replace x by  1/2 we get
=>x³+3x² + 3x + 1
=>(1/2)³ + 3(1/2)² + 3(1/2) + 1
=> 1/8 + 3/4 + 3/2 + 1
Add the fraction taking LCM of denominator we get
=>(1 + 6 + 12 + 8)/8
=>27/8
Remainder is 27/8

(iii) x              
Apply remainder theorem
=>x =0
Replace x by 0 we get
=>x³+3x² + 3x + 1
=>(0)³ + 3(0)² + 3(0) + 1
=> 0+0 +0 + 1
=> 1
Remainder is 1

(iv) x + π        
Apply remainder theorem
=>x + π =0
=> x = - π
Replace x by – π we get
=>x³+3x² + 3x + 1
=>(- π)³ + 3(-π)² + 3(-π) + 1
=> - π3 + 3π2 - 3π + 1

Remainder is - π³ + 3π² - 3π + 1

(v) 5 + 2x
Apply remainder theorem
=>5+2x =0
=> 2x = - 5
=> x = - 5/2
Replace x by – 5/2 we get
=>x³+3x² + 3x + 1
=>(-5/2)³ + 3(-5/2)² + 3(-5/2) + 1
=> -125/8  + 75/4 – 15/2 + 1
Add the fraction taking LCM of denominator
=>(-125 + 150  - 60 + 8 )/125
=> -27/8
Remainder is -27/8