|ihope||What's this about scribd?|
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.
anyway, 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 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
|ihope||noodle_snacks: you mean x-1-i and x-1+i are factors?|
|ihope||Do you know synthetic division?|
|noodle_snacks||doesn't ring a bell no|
should 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.|
|ihope||If you want to remove both roots at once, divide by (x - (1+i))(x - (1-i)), whatever that is.|
is it better to simplify (x - (1+i))(x - (1+i))
|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?
|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.|