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Chapter 2: Problem 33

Use Descartes' Rule of Signs to state the number of possible positive and negative real zeros of each polynomial function. $$P(x)=4 x^{4}-12 x^{3}-3 x^{2}+12 x-7$$

Short Answer

Expert verified
In the polynomial function \(P(x)\), there are either 3 or 1 positive real zeros and 1 negative real zero.

Step by step solution

01

Counting Sign Changes for Positive Real Zeros

The given polynomial is \(P(x)=4x^4-12x^3-3x^2+12x-7\). To find the number of positive real zeros, count the number of sign changes in \(P(x)\). The signs in order from highest degree to lowest are 'positive', 'negative', 'negative', 'positive', 'negative' which gives 3 sign changes. Therefore, there are either 3 or 3-2=1 positive real zeros, as per Descartes' rule.
02

Substituting and Counting Sign Changes for Negative Real Zeros

To find the number of negative real zeros, substitute \(x\) with \(-x\) in \(P(x)\) and then count the number of sign changes in \(P(-x)\). Doing this yields \(P(-x)=4x^4+12x^3-3x^2-12x-7\). The signs in order from highest degree to lowest are 'positive', 'positive', 'negative', 'negative', 'negative', yielding only 1 sign change. Therefore, there is either 1 or 1-2=-1 negative real zero. However, the number of real zeros can't be negative, so there is 1 negative real zero.
03

Writing Conclusion

Based on the previous steps using Descartes' Rule of Signs, there are either 3 or 1 positive real zeros and 1 negative real zero in the polynomial function \(P(x)\).

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