ihope | What's this about scribd? |

noodle_snacks | hey anyone 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 | hehe my bad 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 any pointers? |

ihope | noodle_snacks: you mean x-1-i and x-1+i are factors? |

noodle_snacks | yeah sorry |

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. |

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 | thanks is 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. |