Question 138663
Using <a href=http://www.purplemath.com/modules/drofsign.htm>Descartes' Rule of Signs</a>, we can find the possible number of positive roots (x-intercepts that are positive) and negative roots (x-intercepts that are negative)


First lets find the number of possible positive real roots:


For {{{x^4+2x^3+3x^2-4x-2}}}, simply count the sign changes


Here is the list of sign changes:

<ol><li>{{{3x^2}}} to {{{-4x}}} (positive to negative)</li></ol>

 (note: the rest of the terms have the same sign, so no extra sign changes occur) 


So there is 1 sign change, this means there is a maximum of 1 positive root


So there is exactly one positive root


<hr>


Now lets find the number of possible negative real roots


---------------

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


{{{f(-x)=(-x)^4+2(-x)^3+3(-x)^2-4(-x)-2}}} Plug in -x (just replace every x with -x)


{{{f(-x)=x^4-2x^3+3x^2+4x-2}}} Simplify (note: if the exponent of the given term is odd, simply negate the sign of the term. If the term has an even exponent, then the sign of the term stays the same)


So {{{f(-x)=x^4-2x^3+3x^2+4x-2}}}



------------------------

Now lets count the sign changes for {{{x^4-2x^3+3x^2+4x-2}}}:

Here is the list of sign changes:

<ol><li>{{{x^4}}} to {{{-2x^3}}} (positive to negative)</li><li>{{{-2x^3}}} to {{{3x^2}}} (negative to positive)</li><li>{{{4x}}} to {{{-2}}} (positive to negative)</li></ol>

 (note: the rest of the terms have the same sign, so no extra sign changes occur) 


So for {{{x^4-2x^3+3x^2+4x-2}}} there are a maximum of 3 negative roots

So the number of negative real roots is 1




==================================================


Summary:



So the possible roots are: 1 positive root ,  3 or 1 negative roots