Question 113799
4 a


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 {{{12x^4-67x^3+108x^2-47x+6}}}, simply count the sign changes


Here is the list of sign changes:

<ol><li>{{{12x^4}}} to {{{-67x^3}}} (positive to negative)</li><li>{{{-67x^3}}} to {{{108x^2}}} (negative to positive)</li><li>{{{108x^2}}} to {{{-47x}}} (positive to negative)</li><li>{{{-47x}}} to {{{6}}} (negative to positive)</li></ol>



So there are 4 sign changes, this means there are a maximum of 4 positive roots


So the number of positive real roots is 4,2, or 0




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Now lets find the number of possible negative real roots


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


{{{f(-x)=12(-x)^4-67(-x)^3+108(-x)^2-47(-x)+6}}} Plug in -x (just replace every x with -x)


{{{f(-x)=12x^4+67x^3+108x^2+47x+6}}} 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)=12x^4+67x^3+108x^2+47x+6}}}



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Now lets count the sign changes for {{{12x^4+67x^3+108x^2+47x+6}}}:

By looking at {{{12x^4+67x^3+108x^2+47x+6}}} we can see that there are no sign changes (all the terms are positive).

So there are no negative roots


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Summary:



So there are 4,2, or 0 positive roots  and  0 negative roots