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Chapter 2: Problem 37

Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. $$f(x)=2 x^{4}-5 x^{3}-x^{2}-6 x+4$$

Short Answer

Expert verified
The function \( f(x) = 2x^4 - 5x^3 - x^2 - 6x + 4 \) could have 2 or 0 positive real zeros, and 2 or 0 negative real zeros.

Step by step solution

01

Identify the Function

We are given the polynomial function \(f(x) = 2x^4 - 5x^3 - x^2 - 6x + 4\). Observe that the terms are already arranged in descending order of degree.
02

Count Sign Changes for Positive Zeroes

The number sign changes for the function \(f(x) = 2x^4 - 5x^3 - x^2 - 6x + 4\) are: From \(2x^4\) to \(-5x^3\) (positive to negative), \(-5x^3\) to \(-x^2\) (negative to negative - no change), \(-x^2\) to \(-6x\) (negative to negative - no change), \(-6x\) to \(+4\) (negative to positive). Therefore, there is a total of 2 sign changes, this means according to Descartes's rule of signs the function could have 2 or 2 - 2 * 1 = 0 positive real zeros.
03

Count Sign Changes for Negative Zeroes

Before counting the sign changes apply the rule to \( f(-x) \) Substitute \(-x\) for \(x\) in the function. We get \( f(-x) = 2(-x)^4 - 5(-x)^3 - (-x)^2 - 6(-x) + 4 = 2x^4 + 5x^3 - x^2 + 6x + 4\). Now count the sign changes: From \(2x^4\) to \(5x^3\) (both positive - no change), from \(5x^3\) to \(-x^2\) (positive to negative), from \(-x^2\) to \(6x\) (negative to positive), and from \(6x\) to \(4\) (both positive - no change). Therefore, there are 2 sign changes. According to Descartes's rule of signs, it means that the function could have 2 or 2 - 2 * 1 = 0 negative real zeros.

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Most popular questions from this chapter

Use long division to rewrite the equation for \(g\) in the form $$\text {quotient }+\frac{\text {remainder}}{\text {divisor}}$$ Then use this form of the function's equation and transformations of \(f(x)=\frac{1}{x}\) to graph \(g.\) $$g(x)=\frac{3 x+7}{x+2}$$

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