Apply Descartes' Rule of Signs to x^4 - x^3 - 3x^2 - x + 2

[NEHA]

Apply Descartes' Rule of Signs to determine the possible numbers of positive roots of:

x^4 - x^3 - 3x^2 - x + 2 = 0

p(x) = x^4 - x^3 - 3^2 - x + 2
+ c - c - c - c +
5 postive real roots
or 3 postive real roots
or 1 postive real roots

 

[stapel]

Since there are only two sign changes (from positive to negative, and then back to positive), how are you getting that there are as many as five positive roots? The Rule says otherwise.

Eliz.

 

[NEHA]

i checked out the rules/site
but not sure what i did is right....correct me if i did it wrong
f(x) = +x^4 - x^3 - 3x^2 - x + 2
f(-x) = -x^4 - (-x)^3 - 3(-x)^2 - (-x) + 2
= -x^4 + x^3 + 3x^2 + x + 2

 

[galactus]

Descartes says the number of positive real roots is equal to the number of sign changes or less than that by an even integer.

Since there are 2 sign changes, it either has 2 positive real roots or 0 positive real roots.

Now, the trick. Find the roots. :wink:

 

[NEHA]

Descartes says the number of positive real roots is equal to the number of sign changes or less than that by an even integer.

Since there are 2 sign changes, it either has 2 positive real roots or 0 positive real roots.

Now, the trick. Find the roots. :wink:

umm i dont know about that....can you help me on finding the roots?

 

[galactus]

You aren't required to find the roots, are you?. Just determine how many real positive roots there may be?. Finding the actual roots may prove rather daunting. I would find them with technology of some sort.

Having ran it through Maple:

Real roots:

\(\displaystyle \L\\2.31915909373 \;\ and \;\ 0.647309872176\)

Non-real roots:

\(\displaystyle \L\\-0.983234482951+0.60456971044i \;\ and \;\ -0.983234482951-0.60456971044i\)

If you wanted to find these without technology, you could try Newton's method or the Intermediate Value theorem.