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mathmateBest ResponseYou've already chosen the best response.1
Are you familiar with Descarte's rule of signs?
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Using Descartes rule of signs, we know that there is one positive real root, and either two negative real roots, or two complex roots.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
What have you been using to calculate the root of a polynomial equation that does not have rational roots? There is one more thing you could do, doublecheck the question. This polynomial does not have rational roots.
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
maybe a numerical method to find the roots?
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
I agree,... if the question had been correctly posted.
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
or graphing calculator?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
I would go out on a limb, and say that this isn't the correct equation. Some slight sign changes result in an easily factored equation.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
True, that is for getting an initial estimate.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
@evy15 we're waiting anxiously your confirmation of the equation!! :)
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
agree... check that last term... should it really be 15 ???
 one year ago

evy15Best ResponseYou've already chosen the best response.0
yes the equation is correct
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
As @dpalnc said, if the last term is +15, then there are three rational roots (i.e. the equation can be factorized and solved with 3 real rational roots).
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
In that case, you have 2 complex roots and a real irrational root. What methods have you used so far to solve for irrational roots?
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Cubics can be solved using Cardano's formula, which is overly complex. We can usually find the real root using a numerical method. Does all this sound familiar to you?
 one year ago

wioBest ResponseYou've already chosen the best response.0
@evy15 Is this homework? Are there certain methods you are supposed to be learning?
 one year ago

wioBest ResponseYou've already chosen the best response.0
Once you find the 1st root, you can use synthetic division an the quadratic equation to find the other two.
 one year ago

wioBest ResponseYou've already chosen the best response.0
The rational root theorem says that it's got to be \[ \pm\frac{1,3,5,15}{1} \]So \(3\) is a possibility. Try plugging it in.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Are you absolutely sure that you didn't type it incorrectly? With a small sign change, one of the roots is indeed 3. However, as written, it has one irrational root, and two complex roots.
 one year ago

evy15Best ResponseYou've already chosen the best response.0
the thing is im completely confused because I missed a day in class and now idk how to do it
 one year ago

joemath314159Best ResponseYou've already chosen the best response.0
I checked the solutions to the equation in wolfram, they are pretty intense. If you didnt type the problem incorrectly, then the teacher/professor typed it incorrectly, because there is no way a teacher should expect a student to find those roots by hand =/
 one year ago

joemath314159Best ResponseYou've already chosen the best response.0
and like others have mentioned, if only one of the signs is changed, it becomes an easy regular standard problem.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Unless, of course, you're learning about methods to approximate roots.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.0
oh yeah. that could be the case. Newtons Method :)
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
the only way i can think of if the equation is correct as you say is to approximate the root(s) using newton's method...
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
or graphing calc. or wolfram.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.0
If the problem as typed is correct, there is nothing we (or anyone) can do. Not without a calc or comupter, or something.
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
try to see if x=15 is a root by plugging that back into the original equation... i don't think that's right.
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
wait... do you mean to find the yintercept of \(\large y=x^33x^25x15 \) ?? because 15 is the yintercept...
 one year ago

joemath314159Best ResponseYou've already chosen the best response.0
if thats the actual problem....then lol.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Isn't the question: "Find the roots of the polynomial eq. x^33x^25x15=0"
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
So you need the roots of the equation, not just the yintercept. As I said in the other post, from the type of question you have, it seems likely that either you or your prof had a typo in this question. To make sure it'd better be your prof, you want to triple check for typos in your post.
 one year ago
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