## 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^2-31x-15

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

is it also one?

5. anonymous

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

6. anonymous

i see two sign changes

7. anonymous

oh, oops! now what?

8. triciaal

the first part of the question find the possible zeros

9. 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.

10. anonymous

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

11. anonymous

how do i find the combinations after knowing that?

12. anonymous

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

13. anonymous

so would it be -15/2

14. anonymous

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

15. nincompoop

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

16. triciaal

17. anonymous

thanks @jayzdd :) i get it haha

18. anonymous

It turns out all the zeroes are irrational.

19. anonymous

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