Precalc: rational zeros?

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Emerson
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Sun Oct 28, 2007 5:49 pm
Emerson says... · Sun Oct 28, 2007 5:49 pm
I have NO idea how to do this... Directions: List all potential rational zeros of (equation attached/below).

I used Descartes and found that there are 5,3, or 1 positive zeros, and one negative....But to synthetically divide the first coefficient has to be one. #_# I don't know what to do!

Help please? >_>
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Snoink
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Sun Oct 28, 2007 8:35 pm
Snoink says... · Sun Oct 28, 2007 8:35 pm
...are you saying that 0=12x^8-x^7+6x^4-x^3+x-3?
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Incandescence
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Thu Nov 08, 2007 7:29 am
Incandescence says... · Thu Nov 08, 2007 7:29 am
Suzanne--


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edit: er....not sure why that looks so bad; anyway, f(x) = x^8 - 1/12 x^7 + 1/2 x^4 - 1/12 x^3 + x - 3...

Synthetic division is obvious from here.
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Emerson
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Thu Nov 08, 2007 11:30 am
Emerson says... · Thu Nov 08, 2007 11:30 am
eeeer yeah I think that was what I was saying. XD But that was last chapter. ^.^ Thank you anyhow!
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xhalcyonx128
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Sat Nov 10, 2007 11:05 pm
xhalcyonx128 says... · Sat Nov 10, 2007 11:05 pm
you can still divide synthetically, just do this:

2/ 12 -1 0 0 6 -1 0 1 -3

then continue on like you would normally synthetically divide, just dont forget to include those 0's, theyre important.
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xhalcyonx128
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Sat Nov 10, 2007 11:06 pm
xhalcyonx128 says... · Sat Nov 10, 2007 11:06 pm
o and you dont HAVE to divide by the 2, that would just be any number that you are guessing is a 0. sythetic division is basically guess and check.
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xhalcyonx128
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Sat Nov 10, 2007 11:07 pm
xhalcyonx128 says... · Sat Nov 10, 2007 11:07 pm
if all else fails, graph it on your calculator.
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