12) List the potential rational zeros of the polynomial function. Do not find the zeros. f (x) = -3x2 – 2x + 8

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### Problem 12: Potential Rational Zeros of a Polynomial Function

**Instruction:**
List the potential rational zeros of the polynomial function. Do not find the zeros.

**Polynomial Function:**
\[ f(x) = -3x^2 - 2x + 8 \]

To list the potential rational zeros for the polynomial \( f(x) = -3x^2 - 2x + 8 \), we use the Rational Root Theorem. This theorem states that any potential rational zero, expressed as a fraction \(\frac{p}{q}\), is such that \(p\) is a factor of the constant term (8 in this case), and \(q\) is a factor of the leading coefficient (-3 in this case).

**Factors of 8 (constant term):** \( \pm 1, \pm 2, \pm 4, \pm 8 \)

**Factors of -3 (leading coefficient):** \( \pm 1, \pm 3 \)

**Potential Rational Zeros:**
The potential rational zeros are all possible fractions \(\frac{p}{q}\), where \(p\) is a factor of 8, and \(q\) is a factor of -3. These include:

\[
\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{8}{3}
\]
Transcribed Image Text:### Problem 12: Potential Rational Zeros of a Polynomial Function **Instruction:** List the potential rational zeros of the polynomial function. Do not find the zeros. **Polynomial Function:** \[ f(x) = -3x^2 - 2x + 8 \] To list the potential rational zeros for the polynomial \( f(x) = -3x^2 - 2x + 8 \), we use the Rational Root Theorem. This theorem states that any potential rational zero, expressed as a fraction \(\frac{p}{q}\), is such that \(p\) is a factor of the constant term (8 in this case), and \(q\) is a factor of the leading coefficient (-3 in this case). **Factors of 8 (constant term):** \( \pm 1, \pm 2, \pm 4, \pm 8 \) **Factors of -3 (leading coefficient):** \( \pm 1, \pm 3 \) **Potential Rational Zeros:** The potential rational zeros are all possible fractions \(\frac{p}{q}\), where \(p\) is a factor of 8, and \(q\) is a factor of -3. These include: \[ \pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{8}{3} \]
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