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Chapter 2: Problem 33

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

Short Answer

Expert verified
The possible number of positive real roots is 0 and the possible number of negative real roots is 1.

Step by step solution

01

Identify the Function and Apply Descartes's Rule of Signs

Firstly, we look at the function \(f(x) = x^3+2x^2+5x+4\). As per Descartes's rule of signs, the possible number of positive real roots for a function is determined by the number of sign changes in the function's terms. Here, all coefficients are positive, so there are no changes in signs. Hence, the number of positive real roots is 0.
02

Find Possible Negative Roots

For negative roots, we need to substitute \(x\) with \(-x\) in the given function. This leads us to the function \(f(-x) = (-x)^3+2(-x)^2+5(-x)+4\). After simplifying, the substituted equation is \(-x^3+2x^2-5x+4\). Now, count the number of sign changes in the polynomial. The sign changes once in the transition from \(+2x^2\) to \(-5x\), hence there is possibly one negative real root.
03

Result

Therefore, the given function \(f(x) = x^3+2x^2+5x+4\) might have 0 positive real roots and 1 negative real root. This means the function might have one real root overall. However, remember that Descartes's rule of signs gives us the possible number of roots, the function could still have complex roots, more investigation would be required to confirm the exact roots.

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