i have posted this problem 5 times and every time people say its really hard im sorry i cant figure out. help!

Question

im supposed to analyze the zeroes and use decartes rule of signs to find all the possibilities??

P(x) = -5x^4 + x^3 + 2x^2 - 1

P(x) = 2x^5 + 7x^3 + 6x^2 - 2

Answer 1

Descartes' Rule of Signs will not tell you where the polynomial's zeroes are, but the Rule will tell you how many roots you can expect. First, look at the polynomial as it stands:

P(x)= -5x^4 + x^3 + 2x^2 -1

Ignoring the actual values of the coefficients, look at the signs:

P(x)= (-) (+) (+) (-)
[[[I have eliminated the terms leaving only the signs of each in order to clarify]]]

Now note where the signs change from positive to negative or from negative to positive. There are changes of sign from (-)5x^4 to (+)x^3, and also from (+)2x^2 to (-)1

Count the number of changes. = 2
This is the maximum number of possible zeroes.

Answer 2

First clearly and exactly define the problem, that needs to be done first before a solution can be found.
What does 'analyse the zeroes' mean?
The possibilities of what?
You have defined P(x) as two different polynomials. Please explain. Are we dealing with two different polynomials or two different problems or what?
Please use proper punctuation and capitalization when writing, especially when asking a question. It helps to be in a precise frame of mind when doing math.

From the link:
"the Rule will tell you how many roots you can expect"
"the maximum possible number of positive zeroes (x-intercepts) for the polynomial"
"some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex. Then it may be that certain pairs of roots are not real, and therefore are not graphable as x-intercepts. Because of this possibility, you have to count down by two's."
(Is this is what you're looking for, the possible number of real zeroes?)
More links:
http://en.wikipedia.org/wiki/Descartes'_rule_of_signs
http://www.mathematicshelpcentral.com/lecture_notes/precalculus_algebra_folder/descartes_rule_of_signs.htm
http://sepwww.stanford.edu/oldsep/stew_save/descartes.pdf

Answer 3

Graph the function. Use this site to help you with Descartes' Rule of Signs.

Answer 4

The first has at most two positive roots and two negative roots
The second has at most one positive root and and two negative roots Solving:
1) roots(-5,1,2 and-1)
2)0.460753543558 only real root according to Descartes´s rule

Answer 5

Dude, have a look at some of the other posts. They are asking what to buy their boyfriend for christmas and what colour underwear would a lesbian god wear. No one here wants to use their brain to help people.

I would love to help you out but its 4AM and I cant even count my fingers at this hour.

I will take it to work with me tomorrow and ask some of the math nerds there. stay tuned.

Answer 6

meanwhile you could try to solve it yourself

Answer 7

what class is this for?

Answer 8

try http://www.solvemymath.com/