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\( \leftarrow \) Use the Rational Zero Theorem to list all possible rational zero: \[ f(x)=-3 x^{4}+16 x^{3}+17 x^{2}-17 x+6 \]

Ask by Ingram Hampton. in the United States
Mar 01,2025

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Answer

The possible rational zeros are: ±1, ±2, ±3, ±6, ±1/3, ±2/3.

Solution

To use the Rational Zero Theorem, we start by noting that any possible rational zero for the polynomial   f(x) = -3x⁴ + 16x³ + 17x² - 17x + 6 must be of the form p/q, where:  • p is a factor of the constant term, 6,  • q is a factor of the leading coefficient, -3. Step 1. List the factors:  • Factors of the constant term 6 are: ±1, ±2, ±3, ±6.  • Factors of the leading coefficient -3 are: ±1, ±3. Step 2. Write down all possible fractions p/q:  Using q = 1:   p/1 gives: ±1, ±2, ±3, ±6.  Using q = 3:   p/3 gives: ±1/3, ±2/3, ±3/3, ±6/3.   Note that ±3/3 simplifies to ±1 and ±6/3 simplifies to ±2, which are already listed. Step 3. Combine the unique values:  The possible rational zeros are:   ±1, ±2, ±3, ±6, ±1/3, ±2/3. Thus, by the Rational Zero Theorem, the list of all possible rational zeros for f(x) is:   1, -1, 2, -2, 3, -3, 6, -6, 1/3, -1/3, 2/3, -2/3.

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To apply the Rational Zero Theorem, we first identify the possible rational roots using the factors of the constant term and the leading coefficient. In this case, the constant term is \(6\) and the leading coefficient is \(-3\). The factors of \(6\) (constant term) are \( \pm 1, \pm 2, \pm 3, \pm 6\), and the factors of \(-3\) (leading coefficient) are \( \pm 1, \pm 3\). The possible rational zeros are formed by taking the ratio of these factors. Thus, we get the following possibilities: \[ \frac{\text{Factors of } 6}{\text{Factors of } -3} = \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{3}{3} (which is \pm 1, already listed) \] So the complete list of possible rational zeros is: \[ \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3} \]

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