x^4+16x^2+7x-11=0 Descartes’ rule of signs to find the nature of the roots of the equation

I would suggest you print out and study this example

http://www.purplemath.com/modules/drofsign.htm

To apply Descartes' Rule of Signs to find the nature of the roots of the given equation:

1. Count the number of sign changes in the coefficients of the polynomial when the terms are written in descending order of degree.

In the given equation, rearranging terms to the highest degree first, we have:
x^4 + 16x^2 + 7x - 11 = 0

The sign changes occur between the second and third term (from positive to negative) and between the third and fourth term (from negative to positive). Hence, there are 2 sign changes.

2. Determine the possible number of positive roots.

The possible number of positive roots is either equal to the number of sign changes (2) or less than it by an even number (2 - 2 = 0).

Therefore, according to Descartes' Rule of Signs, the equation x^4 + 16x^2 + 7x - 11 = 0 can have either 2 positive roots or 0 positive roots.

3. Replace x with -x and repeat steps 1 and 2 to determine the possible number of negative roots.

When we substitute -x for x in the equation, we get:
(-x)^4 + 16(-x)^2 - 7x - 11 = 0
x^4 + 16x^2 - 7x - 11 = 0

Again, there are 2 sign changes: between the second and third term (from positive to negative) and between the third and fourth term (from negative to positive).

The possible number of negative roots is also either equal to the number of sign changes (2) or less than it by an even number (2 - 2 = 0).

According to Descartes' Rule of Signs, the equation x^4 + 16x^2 + 7x - 11 = 0 can have either 2 negative roots or 0 negative roots.

In summary, based on Descartes' Rule of Signs, the equation x^4 + 16x^2 + 7x - 11 = 0 can have either 2 positive roots, 2 negative roots, or 0 positive and 0 negative roots.