Question

find the remainder when x cube + 3 X square + 3 X + 1 is divided by 5 + 2 x using remainder theorem​

Answer 1

Answer:

Apply remainder theroem

Apply remainder theroem5+2x=0

Apply remainder theroem5+2x=02x=-5

Apply remainder theroem5+2x=02x=-5x=5/2

Apply remainder theroem5+2x=02x=-5x=5/2replace x by -5/2 , we get 

Apply remainder theroem5+2x=02x=-5x=5/2replace x by -5/2 , we get x3+3x2+3x+1

Apply remainder theroem5+2x=02x=-5x=5/2replace x by -5/2 , we get x3+3x2+3x+1= (-5/2)3+3(-5/2)2+3(-5/2)+1

Apply remainder theroem5+2x=02x=-5x=5/2replace 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

Apply remainder theroem5+2x=02x=-5x=5/2replace 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= (-125+150-60+8)/125

Apply remainder theroem5+2x=02x=-5x=5/2replace 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= (-125+150-60+8)/125= -27/8

Apply remainder theroem5+2x=02x=-5x=5/2replace 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= (-125+150-60+8)/125= -27/8therefore the remainder is -27/8