State the number of possible positive and negative real zeros of the function f(x)=9x^6-3x^5+33x^4-11x^3+18x^2-6x

To determine the number of possible positive and negative real zeros of the function f(x) = 9x^6 - 3x^5 + 33x^4 - 11x^3 + 18x^2 - 6x, we can use the Descartes' Rule of Signs.

First, we count the sign changes in the sequence of coefficients. Then we can additionally consider the effect of any multiple roots.

For the number of possible positive real zeros:

There are 3 sign changes in the original function as we consider the coefficients of x^6, x^5, x^4, x^3, x^2, and x to determine the sign changes. Note that if there are any multiple roots, these do not contribute to the count of sign changes, as they don't represent distinct zeros.

By Descartes' Rule of Signs, the number of positive real zeros of the function is either 3 or 1 less than a multiple of 2, which means it's 3 or 1.

For the number of possible negative real zeros:

We consider the sign changes in f(-x), which means we replace x with -x in the original function. The sign changes in the coefficients of the resulting polynomial will give us the number of possible negative real zeros.

In the modified function f(-x) = 9x^6 + 3x^5 + 33x^4 + 11x^3 + 18x^2 + 6x, there are no sign changes. Therefore, the modified function f(-x) has 0 sign changes and, according to Descartes' Rule of Signs, the number of negative real zeros is 0 or a difference of an even number.

So, the function f(x) = 9x^6 - 3x^5 + 33x^4 - 11x^3 + 18x^2 - 6x has either 3 or 1 possible positive real zeros, and 0 possible negative real zeros.

semsee45 semsee45 · Answer 2 · 2024-01-03T13:38:57+01:00

Answer:

Number of possible positive real zeros: 5 or 3 or 1

Number of possible negative real zeros: 0

Step-by-step explanation:

Given polynomial function:

[tex]f(x)=9x^6-3x^5+33x^4-11x^3+18x^2-6x[/tex]

To find the number of possible positive and negative real zeros of the given function, we can use Descartes' Rule of Signs.

Positive Real Zeros

To determine the number of possible positive real zeros, we need to examine the non-zero coefficients of the polynomial in descending order, and count the number of sign changes.

In the given polynomial, there are 5 sign changes between the terms, so the number of possible positive real zeros is 5.

Since some of the roots may be complex, we need to count down by 2's to find the complete list of the possible number of zeroes. So, the number of possible positive real zeros is 5 or 3 or 1.

Negative Real Zeros

To find the possible number of negative real zeros, begin by replacing each x with -x in the polynomial:

[tex]f(-x)=9(-x)^6-3(-x)^5+33(-x)^4-11(-x)^3+18(-x)^2-6(-x)[/tex]

When (-x) is raised to an odd power, the resulting coefficient is negative, and when (-x) is raised to an even power, the resulting coefficient is positive. Therefore, we negate the sign in front of the terms with odd powers, and leave the signs in front of the terms with even powers unchanged:

[tex]f(-x)=9x^6+3x^5+33x^4+11x^3+18x^2+6x[/tex]

As all the signs are now the same, there are no sign changes, which means that the number of possible negative zeros is 0.

Therefore:

Number of possible positive real zeros: 5 or 3 or 1
Number of possible negative real zeros: 0