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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 \(g(x) = 2 x^{3}-3 x^{2}-3\) has exactly one positive real root and either two or zero negative real roots.

Step by step solution

01

Identify the Coefficients of the Polynomial

The given polynomial is \(g(x) = 2 x^{3}-3 x^{2}-3\). Therefore, the coefficients are 2, -3, and -3.
02

Apply Descartes' Rule of Signs

Descartes' rule of signs states that the number of positive real zeros in a polynomial function is equal to the number of variations in the signs of its coefficients, or less than that by a multiple of 2. If we look at the given polynomial, there is one sign change (from positive 2 to negative 3) for \(x\), hence there is exactly one positive real root.
03

Find the Number of Negative Real Zeros

To find out the number of negative real zeros, replace \(x\) by \(-x\). Therefore the polynomial becomes \( g(-x) = 2(-x)^3 -3(-x)^2 - 3 = -2x^3 - 3x^2 -3 \). In this case, there are two sign changes (from negative -2\(x^3\) to negative -3\(x^2\) then to negative -3). This means that the function has either two or zero negative real roots. According to the Rule of Descartes, this is because the number of sign changes or less that by a multiple of 2 gives the number of negative real roots.

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