Rational Root Theorm: x^3 + 9x^2 +15x-114=0
ok rational root theorem says that the "rational roots" have to be of the form \[\frac{p}{q}\] where p divides the constant and q divides the leading coefficient.
Fundamental Theorem of Algebra, the Rational Root Theorem, Descartes' Rule of Signs, and the Factor Theorem.
since your leading coefficient is 1, that means they will have to be integers that divide 114
So, how i write it? is that's all the working?
\[114=2\times 3\times 19\] so you have a bunch of choices: \[\{\pm1,\pm2,\pm3,\pm19,\pm114,\pm6,\pm38,\pm57\]
which method is that?
that is finding the possible rational roots
fundamental theorem doesn't say anything about finding the roots, only says that there is one
what study group
you polynomial must have a root in the complex plane
at the top it says openstudy is having study groups. i know nothing about it
because your polynomial is of degree 3 it must have one REAL zero, but it doesn't have to be rational
we should ask saifoo about it he knows everything
factor theorem says if r is a root of p(x) then \[p(x)=(x-r)q(x)\] in other words if you know a zero, you can factor
here, im blank!! literally.
any myininaya will have to tell you about descartes rule of sign, because i forget
Sorry sat, im lost again. which one is: Fundamental Theorem of Algebra,? the Rational Root Theorem,? Factor therom?
says something about the possible positive or negative zeros having to do with the change of signs of the coefficients of the polynomial
1) fundamental theorem of algebra: every polynomial has a zero (in the complex plane)
im trying to figure it out from 5 hours now. :(
And where is the working for FTA?
there is no "working" for it. it is just an "existence " theorem. tells you there is one, not how to find it
tell me the exact words which i should write?
rational root theorem is as i described.
\[f(x)=x^3+9x^2+15x-114\] + + + - there is only 1 change so this means there is one possible real root \[f(-x)=-x^3+9x^2-15x-114\] - + - - 2 changes so this means there are 2 possible or 0 possible negative real roots so there could be 1+ and 2- real roots or there could be 1+ and 0- real roots which means there is 2 complex roots
"by the fundamental theorem of algebra, \[x^3 + 9x^2 +15x-114=0\] as a solution in the complex plane."
since the complex zeros come in conjugate pairs, there must be one real zero, since your polynomial has odd degree (it is 3)
ok so one change in sign of \[p(x)\] means one possible positive root?
that one line is suppose to say possible real root
positive dop
i keep on leaving out the word positive
and two changes of sign of \[p(-x)\] means 0 or 2 possible negative roots yes?
Is that correct.
yes satellite
@saifoo i listed the possible rational roots for you
we want 3 solutions so we have two possiblilities 1 positive real +2 negative real+0 complex 1 positive real +0 negative real +2 complex
it is a long list. of course they are only the "possible rational roots" no one guarantees that the roots are rational
If you guys are free, can you help me out? feel free to decline. i would not mind. can u guys write me on a doc file and send it here? we can divide the word among both.
I'm really in a terrible situation now!!
in fact you can check that the one real root is in fact positive and not rational
work*
what do you guys say?
i don't understand you want us to write what we wrote here but put it in a doc file?
why don't you just copy it down
Yes, this is b/c i can't really copy it from here, it changes it's format.
Let me do a thing, i will type all this what you wrote in doc file and shot it you you guys.. then just let me know if that's correct or not.
i cnt help with the prob but i am here so everyones day just got better!!
give me a minute
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