#math - Sat 5 May 2007 between 00:11 and 00:41

NY Lost Funds



ihopeWhat's this about scribd?
noodle_snackshey
anyone here
ihopenoodle_snacks: yep.
By the way, on IRC, the general protocol is to simply ask a question, then wait at least, say, ten minutes to get an answer.
noodle_snackshehe
my bad
anyway, no idea if i am in the right place, but:
ihopeSo you're fine, but it's very annoying to see that someone's asked a question and then left when you could have answered it.
noodle_snacksI have the equation x^4 - 3x^3 + 7x^2 - 8x + 6 = 0, given that 1+i is a factor, from that i know that 1-i is also a factor
but now i am getting confused as to the method to take out those two factors so i can use the quadratic formula to solve for the last two roots
any pointers?
ihopenoodle_snacks: you mean x-1-i and x-1+i are factors?
noodle_snacksyeah
sorry
ihopeDo you know synthetic division?
noodle_snacksdoesn't ring a bell no
should i look it up?
ihopeThat's not a bad idea.
The general way to remove a root from a polynomial is to divide by the corresponding factor: (x^4 - 3x^3 + 7x^2 - 8x + 6)/(x - (1+i)) is the polynomial with the root 1+i removed.
Steve|OfficeSynthetic division was one of those things I learned in junior high school and then never remembered or used since.
echelon--lol.
ihopeIf you want to remove both roots at once, divide by (x - (1+i))(x - (1-i)), whatever that is.
noodle_snacksthanks
is it better to simplify (x - (1+i))(x - (1+i))
first?
ihopex^2 - 2x + 2.
If you use synthetic division, though, you'll have to do it one at a time.
Steve|Office: what do you use when you want to solve a polynomial? Or do you not solve them?
Steve|Officembot, usually.
noodle_snackswith synthetic division, it seems likely you end up with a remainder, do you just add that to the x^0 term?
ihopenoodle_snacks: if the thing you use to divide with is a root, you don't end up with a remainder.
If you do end up with a remainder, it wasn't a root.
noodle_snacksoh ok, must have been the example i was looking at
rod-calis not i an imaginary number?
ihoperod-cal: i is an imaginary number, yes.
I am not an imaginary number, but i is. :-)
rod-calIs not the science of imaginary numbers called 'complex numbers'?
noodle_snacksthats what it says in my text book
ihoperod-cal: a complex number is anything that's either a real number or an imaginary number.

Page: 2 9 16 23 30 37 44 

IrcArchive

NY Lost Funds