Log In · Register

 
help..., .
expoised
post Nov 15 2004, 06:55 PM
Post #1


te quiero
******

Group: Banned
Posts: 2,586
Joined: Apr 2004
Member No: 14,678



yknow there should be a homework help forum...

anyways... i'm pretty sure this isn't allowed, but i'm in trouble and this is my last resort. i've googled it, but i really dont understand how some sites explain it, but how would you find irrational roots to a polynomial equation without using a graphing calculator?

more specifically...

can you find all the irrational roots to this equation:
1x^8 - 5x^7 + 8x^5 + 6x^4 - 1x^3 - 9x^2 + 9x + 4

thanks.

and if you're gonna close this... can you PLEASE PLEASE PLEASE wait until 12 midnight tonight?? PLEASE PLEASE PLEASE.

much love,
melissa
 
 
Start new topic
Replies
expoised
post Nov 15 2004, 11:14 PM
Post #2


te quiero
******

Group: Banned
Posts: 2,586
Joined: Apr 2004
Member No: 14,678



QUOTE(linke3 @ Nov 15 2004, 8:57 PM)
yeah, there really should be a homework help. anyways, i know this- we just did this in class like a week ago. anyways, the really long way is to first find the factors of the last number (p), in this case, it's 4. so the factors are +/- 1, +/- 2, +/- 4. Then, you find the factors of the first coefficient (q), 1, which are +/- 1. Then, you divide p/q for every possible choice. So, you'll get that the possible roots are +/- 1, +/- 2, +/- 4. Then, you have to use synthetic division and divide the equation by each possible root. When you find one real root, then factor that out and continue. Or, you can find the upper and lower bound by picking one root and dividing the equation by that root. If the numbers from the division are all positive, then that root is the upper bound, and you can eliminate any number greater than that root. If the numbers from the division alternates (it goes +, -, +, -, etc) then that is the lower bound, and you can eliminated any number less than that.
Eh, sorry if that sounds confusing or doesn't work. that's how i did it though.

yes, thank you... that's how you find RATIONAL roots.

in this case, there aren't any.

i'm trying to find IRRATIONAL roots. =/

but thanks anyways.
 

Posts in this topic


Reply to this topicStart new topic
1 User(s) are reading this topic (1 Guests and 0 Anonymous Users)
0 Members: