SOLUTION: Could anyone please help me with this problem? Thank You! Use Descartes's rule of signs to discuss the possibilities for the roots of the equation. DO NOT solve the equation.

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Question 135276: Could anyone please help me with this problem? Thank You!
Use Descartes's rule of signs to discuss the possibilities for the roots of the equation. DO NOT solve the equation.
4x^3-9x^2+6x+4=0
Thanks again:)
Found 2 solutions by vleith, stanbon:
Answer by vleith(2983) (Show Source):
You can put this solution on YOUR website!
Here is a great description of the rule --> http://www.purplemath.com/modules/drofsign.htm
Using that information. we see that if x is positive, has 2 changes in sign. So there are either 2 or 0 real roots.
If x is negative, becomes , which has one change in sign. So there is exactly one negative real root.
does this help ?

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Use Descartes's rule of signs to discuss the possibilities for the roots of the equation. DO NOT solve the equation.
4x^3-9x^2+6x+4=0
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The number of changes of sign of f(x) is 2 so there are 2 or 0 positive Real roots.
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f(-x) = -4x^3-9x^2-6x+4
The number of changes of sign of f(-x) is one, so there are no negative
Real roots.
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You can find DesCartes's Rule of signs on-line by using Google or some other
search engine.
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Cheers,
Stan H.