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anonymous
 one year ago
Help please :)
List the possible zeros for the polynomial. Also, use Descartes' rule of signs to describe possible combinations of roots for the polynomials.
h(x)=2x^3+5x^231x15
anonymous
 one year ago
Help please :) List the possible zeros for the polynomial. Also, use Descartes' rule of signs to describe possible combinations of roots for the polynomials. h(x)=2x^3+5x^231x15

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how many sign changes are there for h(x) how many sign changes are there for h(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one sign change for h(x) and im not sure about h(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0h(x)=2(x)^3+5(x)^231(x)15

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0h(x)=2(x)^3+5(x)^231(x)15 = 2x^3 + 5x^2 + 31x  15

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i see two sign changes

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1the first part of the question find the possible zeros

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The rule states that if the terms of a singlevariable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number. Multiple roots of the same value are counted separately.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By descartes law of signs there must be 1 positive root and two or zero negative roots.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i find the combinations after knowing that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the possible rational roots are of the form p/q where p divides the last coefficient, and q divides the first coefficient

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would it be 15/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$ \Large \frac{p}{q} = \frac{\pm 1, \pm 3, \pm 5, \pm 15}{\pm 1,\pm 2} $$

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0descartes was not a complete fool just because he did not discover calculus.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks @jayzdd :) i get it haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It turns out all the zeroes are irrational.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the number of real roots did obey descartes law of signs
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