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Chapter 2: Problem 87

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$g(x)=2 x^{3}-3 x^{2}-3$$

Short Answer

Expert verified
The function \(g(x)=2x^{3}-3x^{2}-3\) may have 1 positive real zero and 1 negative real zero based on Descartes's Rule of Signs.

Step by step solution

01

Identify coefficients of polynomial

Write out the coefficients of the polynomial in order, including zeros for any missing powers of x. For \(g(x)=2x^{3}-3x^{2}-3\), the coefficients are 2, -3, and -3.
02

Apply Descartes's Rule of Signs for positive zeroes

Count the number of sign changes in the list of coefficients to determine the possible number of positive real zeros. Going from 2 to -3 there is one sign change, and from -3 to -3 there is no sign change. So, there is 1 sign change, which suggests that there may be 1 positive real zero. There may also be an even number of additional positive real zeros, so in this case the possible numbers of positive real zeros is 1.
03

Apply Descartes's Rule of Signs for negative zeroes

Replace every instance of x with -x in \(g(x)=2x^{3}-3x^{2}-3\) to get \(g(-x)=-2x^{3}-3x^{2}-3\). Next count the number of sign changes. There is one sign change between -2 and -3, and none between -3 and -3. Hence, there may be 1 negative real zero.

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