Part 2 of 3 Number of possible negative real zeros:

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1. Write the polynomial in standard form. For a polynomial
   $$
   f(x)=a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0,
   $$
   ensure that the terms are ordered from highest degree to lowest degree.

2. Substitute $$-x$$ for $$x$$ in the polynomial to obtain
   $$
   f(-x)=a_n(-x)^n+a_{n-1}(-x)^{n-1}+\cdots+a_1(-x)+a_0.
   $$
   
3. Simplify $$f(-x)$$, remembering that $$(-x)^k$$ is:
   - $$x^k$$ if $$k$$ is even,
   - $$-x^k$$ if $$k$$ is odd.

4. Apply Descartes’ Rule of Signs: Count the number of sign changes in the sequence of coefficients of $$f(-x)$$. Each sign change corresponds to a possible negative real zero. (The actual number of negative real zeros is either equal to that count or less than it by an even number.)

5. Conclude that the number of possible negative real zeros is equal to the number of sign changes in $$f(-x)$$, or that number minus a nonnegative even integer.

Thus, without further specifics for $$f(x)$$, the answer is:

If $$f(-x)$$ shows $$S$$ sign changes, then the polynomial has at most $$S$$ negative real zeros (or it may have $$S-2k$$ negative real zeros for some nonnegative integer $$k$$).
To find the number of possible negative real zeros of a polynomial: 1. Write the polynomial in standard form. 2. Substitute $$-x$$ for $$x$$ in the polynomial. 3. Simplify the new polynomial by changing the sign of the terms with odd exponents. 4. Count the number of times the signs of the coefficients change in the simplified polynomial. 5. The number of possible negative real zeros is equal to the number of sign changes, or it may be less by an even number. So, if there are $$S$$ sign changes in $$f(-x)$$, the polynomial has at most $$S$$ negative real zeros, or possibly $$S-2k$$ zeros for some nonnegative integer $$k$$.

Part 2 of 3 Number of possible negative real zeros:

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