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Chapter 2: Problem 88

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$h(x)=4 x^{2}-8 x+3$$

Short Answer

Expert verified
The function \(h(x)=4x^2-8x+3\) potentially has 2 or 0 positive real zeros and 0 negative real zeros.

Step by step solution

01

Identify the Coefficients

Look at the function \(h(x)=4x^2-8x+3\), the coefficients of the polynomial are 4, -8, and 3.
02

Apply Descartes's Rule of Signs for Positive Zeros

For positive zeros count the number of sign changes in the coefficients of the polynomial as read from left to right: 4 to -8 is a sign change, but -8 to 3 is also a sign change. So, there are 2 sign changes, meaning there are 2 or 0 positive real zeros.
03

Apply Descartes's Rule of Signs for Negative Zeros

For negative zeros count the sign changes in the coefficients of the polynomial \(h(-x)\). We get \(h(-x)= 4x^2+8x+3\). The polynomial has no sign changes, which means there are 0 negative real zeros.

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