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Lynienicole
 one year ago
Hello everyone, is anyone interested in helping me solve a polynomial equation?
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http://prntscr.com/88wws6
Lynienicole
 one year ago
Hello everyone, is anyone interested in helping me solve a polynomial equation? (Link below is an screenshot of the question) http://prntscr.com/88wws6

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Lynienicole
 one year ago
Best ResponseYou've already chosen the best response.0Your help would be much appreciated! :) I'm stuck on question 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u know whats important

Lynienicole
 one year ago
Best ResponseYou've already chosen the best response.0Important regarding to the question? Would be the polynomial, correct?

Lynienicole
 one year ago
Best ResponseYou've already chosen the best response.0Are you still available to help?

Lynienicole
 one year ago
Best ResponseYou've already chosen the best response.0I am still here, do you understand question 2 ? I'm totally confused on how to use the fundamental theorem and Descartes' rules.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1define the thrm and the rule

Lynienicole
 one year ago
Best ResponseYou've already chosen the best response.0The fundamental theorem of algebra states that every polynomial equation over the field of complex numbers of degree higher than one has a complex solution. Polynomials of the form , with a, b,... coefficients real or complex, can be factored completely into where the r, s,... are complex numbers. In mathematics, Descartes' rule of signs, first described by René Descartes in his work La Géométrie, is a technique for determining an upper bound on the number of positive or negative real roots of a polynomial

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1hmm, is that the thrm from your material, or from the internet? just curious

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the form i am used to is that the degree of a polynomial defines the number of possible real zeros. an nth degree polynomial therefore has at most n real roots.

Lynienicole
 one year ago
Best ResponseYou've already chosen the best response.0Google definition. My Teacher/Professor never explained what it was..

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ah, google definitions are not always the simplest constructions

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1https://www.mathsisfun.com/algebra/fundamentaltheoremalgebra.html this seems like a much simpler and easier to read description of it

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1an nth degree poly has n complex roots. but real roots are complex roots of the form: r + 0i

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1so accounting for complex roots; the degree of a polynomial tells us that a poly has at most, n real roots for some nth degree. does this make sense?

Lynienicole
 one year ago
Best ResponseYou've already chosen the best response.0I read over the link, and it makes my thinking a lot clearer aha. So basically, for question 2. I create a graph using those two functions, and explain how it matches the "construction foreman"?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1use the thrm and the sign rule to show to the foreman that your graph meets the requirements yes

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1you have an x^3 poly ... 3rd degree, and we have 3 real roots right? so the thrm holds

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the sign rule just tells us the possible number of positive and negative roots and its a little complicated depending on your abilities. let x be some positive number; a then count the number of times the 'operations' change from + to  and back again that gives us the number of possible positive roots  then let x be some negative number; a and do the same process to determine the number of possible negative roots

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1for simplicity, i just use 1 and 1 for x instead of a and a

Lynienicole
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much! Your help was greatly appreciated. :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1your welcome, and good luck
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