SOLUTION: Use Descartes rule of signs to determine the number of positive and negative roots {{{x^4+2x^3+3x^2-4x-2}}}

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Question 138663: Use Descartes rule of signs to determine the number of positive and negative roots

x%5E4%2B2x%5E3%2B3x%5E2-4x-2
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Using Descartes' Rule of Signs, 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%5E4%2B2x%5E3%2B3x%5E2-4x-2, simply count the sign changes

Here is the list of sign changes:
  1. 3x%5E2 to -4x (positive to negative)

(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



Now lets find the number of possible negative real roots

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First we need to find f%28-x%29:

f%28-x%29=%28-x%29%5E4%2B2%28-x%29%5E3%2B3%28-x%29%5E2-4%28-x%29-2 Plug in -x (just replace every x with -x)

f%28-x%29=x%5E4-2x%5E3%2B3x%5E2%2B4x-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%28-x%29=x%5E4-2x%5E3%2B3x%5E2%2B4x-2


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Now lets count the sign changes for x%5E4-2x%5E3%2B3x%5E2%2B4x-2:
Here is the list of sign changes:
  1. x%5E4 to -2x%5E3 (positive to negative)
  2. -2x%5E3 to 3x%5E2 (negative to positive)
  3. 4x to -2 (positive to negative)

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

So for x%5E4-2x%5E3%2B3x%5E2%2B4x-2 there are a maximum of 3 negative roots
So the number of negative real roots is 1



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Summary:


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