Question

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

Answer 1

how many sign changes are there for h(x)
how many sign changes are there for h(-x)

Answer 2

one sign change for h(x) and im not sure about h(-x)

Answer 3

h(-x)=2(-x)^3+5(-x)^2-31(-x)-15

Answer 4

is it also one?

Answer 5

h(-x)=2(-x)^3+5(-x)^2-31(-x)-15
= -2x^3 + 5x^2 + 31x - 15

Answer 6

i see two sign changes

Answer 7

oh, oops! now what?

Answer 8

the first part of the question find the possible zeros

Answer 9

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.

Answer 10

By descartes law of signs there must be 1 positive root and two or zero negative roots.

Answer 11

how do i find the combinations after knowing that?

Answer 12

the possible rational roots are of the form
p/q where p divides the last coefficient, and q divides the first coefficient

Answer 13

so would it be -15/2

Answer 14

$$ \Large \frac{p}{q} = \frac{\pm 1, \pm 3, \pm 5, \pm 15}{\pm 1,\pm 2} $$

Answer 15

descartes was not a complete fool just because he did not discover calculus.

Answer 16

thanks @jayzdd :) i get it haha

Answer 17

It turns out all the zeroes are irrational.

Answer 18

but the number of real roots did obey descartes law of signs