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( kindly use the remainder theorem to solve the question.)

iii) x

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Answer 1

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 (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

☆ I HOPE ITS HELP YOU ☆

Answer 2

It's..Ur answer..have refer with both....