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

### NY Lost Funds

 ihope What's this about scribd? noodle_snacks heyanyone here ihope noodle_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_snacks hehemy badanyway, no idea if i am in the right place, but: ihope So 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_snacks I 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 factorbut 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 rootsany pointers? ihope noodle_snacks: you mean x-1-i and x-1+i are factors? noodle_snacks yeahsorry ihope Do you know synthetic division? noodle_snacks doesn't ring a bell noshould i look it up? ihope That'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|Office Synthetic division was one of those things I learned in junior high school and then never remembered or used since. echelon-- lol. ihope If you want to remove both roots at once, divide by (x - (1+i))(x - (1-i)), whatever that is. noodle_snacks thanksis it better to simplify (x - (1+i))(x - (1+i))first? ihope x^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|Office mbot, usually. noodle_snacks with synthetic division, it seems likely you end up with a remainder, do you just add that to the x^0 term? ihope noodle_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_snacks oh ok, must have been the example i was looking at rod-cal is not i an imaginary number? ihope rod-cal: i is an imaginary number, yes.I am not an imaginary number, but i is. :-) rod-cal Is not the science of imaginary numbers called 'complex numbers'? noodle_snacks thats what it says in my text book ihope rod-cal: a complex number is anything that's either a real number or an imaginary number.