Determine which of the following polynomials has (x + 1) a factor :

Question 1. Determine which of the following polynomials has (x + 1) a factor :
(i) x³ + x³ + x + 1
(ii) x⁴ + x³ + x³ + x + 1
(iii) x⁴ + 3x³ + 3x³ + x + 1
(iv) x³ – x³ – (2+√2)x + √2

 Determine  x³ + x³ + x + 1  polynomials has (x + 1) a factor

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

 Determine x⁴ + x³ + x³ + x + 1 polynomials has (x + 1) a factor

(ii)x⁴ + x³ + x³ + x + 1
Apply remainder theorem
=>x + 1 =0
=> x = - 1
Replace x by – 1 we get
=>x⁴ + x³ + x² + x + 1
=>  (-1)⁴+ (-1)³ + (-1)² + (-1) + 1
=> 1 -1 + 1 - 1 + 1
=> 1
Remainder is not equal to 0 so that x+1 is not a factor of x⁴ + x³ + x³ + x + 1
 

 Determine   x⁴ + 3x³ + 3x³ + x + 1  polynomials has (x + 1) a factor

(iii)x⁴ + 3x³ + 3x³ + x + 1
Apply remainder theorem
=>x + 1 =0
=> x = - 1
Replace x by – 1 we get
=>x⁴ + 3x³ + 3x³ + x + 1
=>  (-1)⁴+ 3(-1)³ + 3(-1)² + (-1) + 1
=> 1 -3 + 3 - 1 + 1
=> 1
Remainder is not equal to 0 so that x+1 is not a factor of x⁴ + 3x³ + 3x³ + x + 1

 Determine x³ – x³ – (2+√2)x + √2 polynomials has (x + 1) a factor

(iv) x³ – x³ – (2+√2)x + √2
Apply remainder theorem
=>x + 1 =0
=> x = - 1
Replace x by – 1 we get
=>x³ – x³ – (2+√2)x + √2
=>  (-1)³ – (-1)² – (2 + √2)(-1) + √2
=> 1 - 1 + 2 + √2 + √2
=> 2 + 2√2
Remainder is not equal to 0 so that x+1 is not a factor of x³ – x³ – (2+√2)x + √2

remainder theorem How Find remainder when

Question 1  Find 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

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

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

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

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

(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