2007-05-27 10:20:08 · 5 answers · asked by franchette11 1 in Science & Mathematics ➔ Mathematics
By synthetic division to divide by c - (-5) the following layout is used:-
- 5| 1--- 7 ---6 --- -18--- 1
---|---- - 5 - -10---20- -10
______________________
----1-----2--- - 4----2----R = - 9
This shows a remainder of - 9 when divide by c - (-5)
Therefore c + 5 is not a factor.
2007-05-28 07:15:22 · answer #1 · answered by Como 7 · 0⤊ 0⤋
if (c+5) were a factor of (c^4 + 7c^3 + 6c^2 - 18c + 1),
it would have to go into it evenly (that is, with no remainder),
but this operation gives you a remainder of -9, so it cannot be a factor.
It's very hard to show you the calculations here, but it's just a division problem ....
2007-05-27 10:52:58 · answer #2 · answered by Concetta Z 2 · 0⤊ 0⤋
http://www.mathworld.wolfram.com
2007-05-27 10:22:55 · answer #3 · answered by Anonymous · 0⤊ 0⤋
umm... helloooo
that makes no sense!!!!!!!!!
2007-05-27 10:23:58 · answer #4 · answered by Rose 3 · 0⤊ 0⤋
???????????
2007-05-27 10:22:37 · answer #5 · answered by Cammy 2 · 0⤊ 0⤋