Question

f(x) = 9x^4 − 82x^2 + 9
1-Use Descartes's Rule of Signs to determine the possible numbers of positive real zeros of f.
2-Use Descartes's Rule of Signs to determine the possible numbers of negative real zeros of f.
3- Use the Rational Zero Test to create a list of all the possible rational zeros of f

Answer 1

Do you know how to use DRoS?

First, we write the polynomial in standard form, with decreasing powers of the variable. Got that already. Now we look at the highest power to get the total number of roots. Here we have an \(x^4\) term, so there will be 4 roots (aka zeros).

Next, we scan from left to right, counting sign changes in the coefficients. I find it easiest to just copy them first: +-+

The number of sign changes gives the maximum possible number of positive, real zeros. This number may be decreased by a multiple of 2 to reflect pairs of complex conjugate roots. In a polynomial with only real coefficients, complex roots will always come in conjugate pairs \(a\pm bi\)

So, if our count of sign changes turned out to be 7, for example, the possible positive real root count could be 7, 5, 3 or 1, with 0, 2, 4, or 6 complex roots respectively.

Next we find the number of possible negative real roots. To do this, we rewrite the polynomial as f(-x) and repeat the counting of the sign changes. The easiest way to do this is to simply rewrite f(x), changing the sign of any term which has an odd exponent (1,3,5,7, etc).

Again, this number is the maximum number of negative real roots, and may be decreased by a multiple of 2 to account for pairs of complex conjugate roots.

Finally, any difference between the total number of roots (gotten from the highest exponent) and the sum of the counts of positive real roots and negative real roots is made up by additional complex roots. For example, consider the polynomial \[x^4+10x^2+9\]
There are no sign changes, so 0 positive real roots. f(-x) is identical to f(x) because there are no terms with odd exponents, so there are also 0 negative real roots. However, we need 4 roots in total, so there must be 4 - 0 - 0 = 4 complex roots and indeed, there are: \[\{x = -i\},\{x= i\},\{x= -3 i\},\{x= 3 i\}\]

I stress again that this is an example and NOT the answer to your problem.

Answer 2

The rational root theorem says that the possible zeros will be the set of fractions whose numerators are factors of the constant term and denominators are factors of the coefficient of the leading term, with either sign possible. It's much more pleasant when the leading term has a coefficient of 1 :-)

Answer 3

still don't understand

Answer 4

It would be helpful if you could be a bit more specific about what you don't understand. Or tell me what you do understand...