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

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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

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how many sign changes are there for h(x) how many sign changes are there for h(-x)
one sign change for h(x) and im not sure about h(-x)
h(-x)=2(-x)^3+5(-x)^2-31(-x)-15

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

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