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Chapter 2: Problem 30

Use Descartes' Rule of Signs to state the number of possible positive and negative real zeros of each polynomial function. $$P(x)=3 x^{3}+11 x^{2}-6 x-8$$

Short Answer

Expert verified
The function \(P(x)=3 x^{3}+11 x^{2}-6 x-8\) has possibly two positive real zeros and one negative real zero.

Step by step solution

01

Recognise the coefficients of the function

Given \(P(x)=3 x^{3}+11 x^{2}-6 x-8\). The coefficients of the terms are 3, 11, -6, and -8.
02

Determine potential positive real zeros

To find the number of positive real zeros, count the number of sign changes of the coefficients from the function as it is given. The coefficients are '3, 11, -6, -8'. There are two changes in signs (+ to - and - to +). By Descartes' Rule of Signs, the number of positive real zeros is the number of sign changes (2) or that number less an even integer.
03

Determine potential negative real zeros

To find the number of negative real zeros, replace the variable \(x\) with \(-x\) in the original function, which will be represented as \(P(-x)=3(-x)^{3}+11(-x)^{2}-6(-x)-8\). After simplifying, this results in \(-3 x^{3}+11 x^{2}+6 x-8\). By Descartes' Rule of Signs, the number of negative real zeros corresponds to the number of sign changes in this function. In this case, the sign of the coefficients change one time. So there is one potential negative real zero or that number less an even integer.

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