anonymous
  • anonymous
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^2-31x-15
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
how many sign changes are there for h(x) how many sign changes are there for h(-x)
anonymous
  • anonymous
one sign change for h(x) and im not sure about h(-x)
anonymous
  • anonymous
h(-x)=2(-x)^3+5(-x)^2-31(-x)-15

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anonymous
  • anonymous
is it also one?
anonymous
  • anonymous
h(-x)=2(-x)^3+5(-x)^2-31(-x)-15 = -2x^3 + 5x^2 + 31x - 15
anonymous
  • anonymous
i see two sign changes
anonymous
  • anonymous
oh, oops! now what?
triciaal
  • triciaal
the first part of the question find the possible zeros
anonymous
  • anonymous
The rule states that if the terms of a single-variable 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
  • anonymous
By descartes law of signs there must be 1 positive root and two or zero negative roots.
anonymous
  • anonymous
how do i find the combinations after knowing that?
anonymous
  • anonymous
the possible rational roots are of the form p/q where p divides the last coefficient, and q divides the first coefficient
anonymous
  • anonymous
so would it be -15/2
anonymous
  • anonymous
$$ \Large \frac{p}{q} = \frac{\pm 1, \pm 3, \pm 5, \pm 15}{\pm 1,\pm 2} $$
nincompoop
  • nincompoop
descartes was not a complete fool just because he did not discover calculus.
triciaal
  • triciaal
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anonymous
  • anonymous
thanks @jayzdd :) i get it haha
anonymous
  • anonymous
It turns out all the zeroes are irrational.
anonymous
  • anonymous
but the number of real roots did obey descartes law of signs

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