SOLUTION: (2u^3-13u^2-8u+7)divided by (u-7) I have the quotient as 2u^2+u-15 with a remainder of 105 Useing long division This seems wrong because the reamainder was 0-105. Please he

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: (2u^3-13u^2-8u+7)divided by (u-7) I have the quotient as 2u^2+u-15 with a remainder of 105 Useing long division This seems wrong because the reamainder was 0-105. Please he      Log On


   

Question 23672: (2u^3-13u^2-8u+7)divided by (u-7)
I have the quotient as 2u^2+u-15 with a
remainder of 105
Useing long division
This seems wrong because the reamainder
was 0-105. Please help me out.
Thank you
Found 2 solutions by venugopalramana, Earlsdon:
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
(2u^3-13u^2-8u+7)divided by (u-7)...YOU CAN USE SHORT DIVISION FOR THIS ...
LET U-7=0...U=7
7....|.......2........-13.......-8..........7
.....|.......0........14........7..........-7
-------------------------------------------------
.....|.......2.........1.........-1.........0
HENCE REMINDER IS ZERO
QUOTIENT IS 2U^2+U-1....
IF YOU SHOW YOUR WORKING , WE CAN TELL WHERE YOU HAVE ERRED.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Hmmm...I get the following when I do the long division:
%282u%5E3-13u%5E2-8u%2B7%29%2F%28u-7%29 = 2u%5E2%2Bu-1 with no remainder.
Check: Multiply your divisor u-7by the quotient 2u%5E2%2Bu-1 to see that you get the dividend 2u%5E3-13u%5E2-8u%2B7 back.
%282u%5E2%2Bu-1%29%28u-7%29+=+2u%5E3-14u%5E2%2Bu%5E2-7u%2B-u%2B7 Simplify the right side.
2u%5E3-13u%5E2-8u%2B7 Which is your original dividend.