Algebra 2 roots

Question

4. P(x) = -9x3 - x2 +5x -8
Possible Roots:  

 

i have to use Descartes' Rule

 

please explain

Answer 1

Take a survey of the polynomial, from left to right, observing successive sign changes.  This count is TWO.

M. Descartes says that therefore there are either two Real positive roots, or none (two less because complex roots occur in conjugate pairs when all coefficients & constant are Real).

 

Now form P(-x) = 9x³ -x² -5x -8. This has exactly one sign change, so M. Descartes says that there is exactly one Real negative zero.

 

Since one negative root is guaranteed, one would seek that first. Using the rational roots theorem, this root could only be one of these: -1, -2, -4, -8, or any of these integers divided by 3 or 9.

 

Personally, I like to graph the function to find the zero, then divide it out synthetically, and then work with the remaining quadratic factor.