Question 92096

First lets find the number of possible positive real zeros:


For {{{-3x^5-7x^3-4x-5}}}, simply count the sign changes


By looking at {{{-3x^5-7x^3-4x-5}}} we can see that there are no sign changes (all the terms are negative).

So there are no positive zeros


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Now lets find the number of possible negative real zeros


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First we need to find {{{f(-x)}}}:


{{{f(-x)=-3(-x)^5-7(-x)^3-4(-x)-5}}} Plug in -x (just replace every x with -x)


{{{f(-x)=3x^5+7x^3+4x-5}}} Simplify (note: if the exponent of the given term is odd, simply negate the sign of the term)


So {{{f(-x)=3x^5+7x^3+4x-5}}}



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Now lets count the sign changes for {{{3x^5+7x^3+4x-5}}}:

Here is the list of sign changes:

<ol><li>{{{4x}}} to {{{-5}}} (positive to negative)</li></ol>




So for {{{3x^5+7x^3+4x-5}}} 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