help..., . |
 Nov 15 2004, 06:55 PM
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#1
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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 |
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| *tweeak* |
Nov 15 2004, 07:02 PM
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#2
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Guest |
there really should be hmwk help place. if youd posted this earlier, i could have given you the address to this live hmwk help place, but its too late for that now
 are irrational roots like imaginary numbers? i dont know, i have a c in algebra 
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Nov 15 2004, 07:03 PM
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#3
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te quiero Group: Banned Posts: 2,586 Joined: Apr 2004 Member No: 14,678 |
QUOTE(my_papaya @ Nov 15 2004, 6:02 PM) there really should be hmwk help place. if youd posted this earlier, i could have given you the address to this live hmwk help place, but its too late for that now are irrational roots like imaginary numbers? i dont know, i have a c in algebra well.. this is for precalc  but no, they're not imaginary numbers.. they're fractions. |
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| *tweeak* |
Nov 15 2004, 07:05 PM
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#4
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oh, for some reason i thought youd said you were taking alg2 (in another thread)
oh well, sorry |
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| *autumn.* |
Nov 15 2004, 08:25 PM
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#5
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ugh. i dont really remember things from precalc, but i believe you use division to figure this one out? do you already know one of the roots?
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Nov 15 2004, 08:34 PM
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#6
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 Senior Member Group: Member Posts: 110 Joined: Aug 2004 Member No: 42,702 |
whoa.
 im getting dizzy just looking at that...umm.. calculator?
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Nov 15 2004, 09:22 PM
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#7
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 RiKACHANtEL Group: Member Posts: 3,876 Joined: Sep 2004 Member No: 51,230 |
 sorry i cant help you. i'm in geometry
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Nov 15 2004, 09:25 PM
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#8
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 Saap!? Group: Member Posts: 568 Joined: Feb 2004 Member No: 5,151 |
www.ask.com is what i use for hmk
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Nov 15 2004, 09:57 PM
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#9
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 hi. call me linda. Group: Official Member Posts: 8,187 Joined: Feb 2004 Member No: 3,475 |
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. |
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Nov 15 2004, 10:14 PM
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#10
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 We are the cure. Group: Staff Alumni Posts: 4,936 Joined: Jan 2004 Member No: 1,456 |
^ OH MAN.
I'm gonna fail high school. |
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Nov 15 2004, 10:35 PM
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#11
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 hello : ) Group: Official Member Posts: 4,227 Joined: Apr 2004 Member No: 13,139 |
heh how sad. i`m in precalc and i don`t kno how to solve this. no wonder why i have a D in that class. & yes there should be a homework help forum or something
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Nov 15 2004, 11:14 PM
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#12
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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. |
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