SOLUTION: Hello, I am havign soem issues with understanding the Descartes rule of signs and am unsure of how to solve the problem. Use Descartes rule of signs to discuss the possibilities

Question 88391: Hello, I am havign soem issues with understanding the Descartes rule of signs and am unsure of how to solve the problem.
Use Descartes rule of signs to discuss the possibilities for the roots to each equation. Do not solve it.
2x^3-3x^2-5x-9=0
Here is what I have tried to do
2(-x)^3 - 3(-x)^2-5(-x)-9=-2x^3-3x^2-5x-9
How do I tell how many roots there are to this problem?
Thank you for your help.
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
2x^3 - 3x^2 - 5x - 9 = f(x)
The sign of the value of determines the sign of the roots.
2x^3 - 3x^2 - 5x - 9 = f(+x)
(+) (-) (-) (-) from a positive 2, a negative 3, a negative 5, and a negative 9
The sign changes once.
Root Types:
Pos: 1
Neg: ~
Img: ~
-2x^3 - 3x^2 + 5x - 9 = f(-x)
(-) (-) (+) (-)
The sign changes twice.
Root Types:
Pos: 1
Neg: 2
Img: ~
There are only three roots because of the type of equation, so the imaginery choice is zero. A rule is that two can be taken from a type and added for imaginery.
Root Types:
Pos: 1 .. 1
Neg: 2 .. 0
Img: 0 .. 2
~ How to Determine Maximum Amount of Roots ~
1 Root

2 Roots

3 Roots ~ Max.

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