Descartes rule of sign question?

Question

Use Descartes' Rule of Signs to analyze the zeros of the following. List all possibilities.

P(x) = x^5 - 4x^4 + 3x^3 + 2x - 6

P(x) = -5x^4 + x^3 + 2x^2 - 1

P(x) = 2x^5 + 7x^3 + 6x^2 - 2

Could someone explain how I would do this? Maybe show an example? Thank You! :)

But how do you figure out the complex solutions?

Answer 1

Descarte's rule of signs simply states that the number of negative or positive roots by analyzing the equation for example:
For a positive root, the rule states that the positive roots of the ploynomial are equal to the number of sign differences between nonzero coefficients

P(x) = x^5 - 4x^4 + 3x^3 + 2x - 6
there is two sign changes between the first and second term , and second and third, so at least two positive roots exist (the six is raised to the zero power so it does not count as a sign change to find the positive root)

The number of sign changes after negating the coefficients of odd-power terms gives the negative roots:

P(x) = -x^5 + 4x^4 - 3x^3 - 2x + 6
There are three sign changes between the first and second, second and third, and fourth and fifth so there are three negative roots (in this case the six raised to the zero power is counted as essential to finding a negative root).

Answer 2

Descartes' Rule of Signs is that the maximum number of positive roots is the number of sign changes in P(x). The first one therefore has at most 2 positive roots, and can also have 0. It also states that the number of negative roots is the number of sign changes in P(-x). P(-x) = -x^5-4x^4-3x^3-2x-6, so the first one has 0 negative.
If the number of positive or negative roots is diffrent, it will differ by multiples of two.

Answer 3

there are two tests done with the Descartes rule of signs...the first is to look at the function as it is written, making sure you write the terms in order of decreasing exponent, and count the number of sign changes in coefficients...

this test gives you the maximum number of POSSIBLE possible roots; in your first example, there are three sign changes (between the first and second terms, the second and third terms, and the fourth and last terms).

the second test is to change x to -x, and count the sign changes, this tells you the maximum number of POSSIBLE negative roots

in your first example, your equation becomes:

-x^5-4x^4-3x^3-2x-6, so there are no sign changes and there are no possible negative roots

so, for your first example, you can say there are at most 3 positive solutions (there might be only 1) and at least two complex solutions. Using only the Descartes Rule, you can't determine exactly how many positive roots there are, just the max number. (I worked out the solutions to this case, and there is 1 pos root and 4 complex roots).

In your second example, P(x) has two sign changes, so there are either 2 or 0 positive roots; P(-x) has two sign changes, so there are at most 2 neg roots. (In this particular case, all four roots are complex)

In the final example, P(x) has one sign change, so there is exactly one pos root, P(-x) has two sign changes, so there are at max 2 neg roots, and you know there are at least two complex roots

without drawing graphs of the function, the Descartes rule alone will not allow determination of how many complex roots there are

Answer 4

have a look here :
http://www.purplemath.com/modules/drofsign.htm

Answer 5

roots_p1 :

3.0000
-1.0000
1.0000 + 1.0000i
1.0000 - 1.0000i
--------------------
roots_p2 :

0.6218 + 0.3187i
0.6218 - 0.3187i
-0.5218 + 0.3707i
-0.5218 - 0.3707i

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

roots_p3 :

-2.2585
-0.8489 + 0.5000i
-0.8489 - 0.5000i
0.4562