SOLUTION: Use Descartes' Rule of Signs to determine the possible number of positive and negative zeros for the function P(x) = -3x^5 -7x^3 -4x

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Question 92096: Use Descartes' Rule of Signs to determine the possible number of positive and negative zeros for the function P(x) = -3x^5 -7x^3 -4x - 5.
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
P(x) = -3x^5 -7x^3 -4x - 5
P(x) has 0 changes of sign so it has no positive rational roots.
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P(-x) = 3x^5 +7x^3 +4x-5
P(-x) has 1 change of sign; P(x) has one negative rational root.
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Cheers,
Stan H.

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!

First lets find the number of possible positive real zeros:

For , simply count the sign changes

By looking at we can see that there are no sign changes (all the terms are negative).
So there are no positive zeros

Now lets find the number of possible negative real zeros

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First we need to find :

Plug in -x (just replace every x with -x)

Simplify (note: if the exponent of the given term is odd, simply negate the sign of the term)

So

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Now lets count the sign changes for :
Here is the list of sign changes:

to (positive to negative)

So for there are a maximum of 1 negative zero
So there is exactly one negative zero

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Summary:

So there are 0 positive zeros and 1 negative zero

SOLUTION: Use Descartes' Rule of Signs to determine the possible number of positive and negative zeros for the function P(x) = -3x^5 -7x^3 -4x - 5.