To apply the Rational Zero Theorem, we need to determine the possible rational zeros of the polynomial function \( f(x) = 6x^3 - 2x^2 + x + 3 \).
The Rational Zero Theorem states that any rational zero, in the form of \( \frac{p}{q} \), where:
- \( p \) is a factor of the constant term (the last term of the polynomial),
- \( q \) is a factor of the leading coefficient (the coefficient of the term with the highest degree).
Step 1: Identify the constant term and leading coefficient.
- The constant term is \( 3 \).
- The leading coefficient is \( 6 \).
Step 2: Find the factors of the constant term \( 3 \).
- Factors of \( 3 \): \( \pm 1, \pm 3 \)
Step 3: Find the factors of the leading coefficient \( 6 \).
- Factors of \( 6 \): \( \pm 1, \pm 2, \pm 3, \pm 6 \)
Step 4: Form the possible rational zeros \( \frac{p}{q} \). All combinations of \( \frac{p}{q} \) where \( p \) is a factor of \( 3 \) and \( q \) is a factor of \( 6 \):
- \( \frac{1}{1} = 1 \)
- \( \frac{1}{2} = \frac{1}{2} \)
- \( \frac{1}{3} = \frac{1}{3} \)
- \( \frac{1}{6} = \frac{1}{6} \)
- \( \frac{-1}{1} = -1 \)
- \( \frac{-1}{2} = -\frac{1}{2} \)
- \( \frac{-1}{3} = -\frac{1}{3} \)
- \( \frac{-1}{6} = -\frac{1}{6} \)
- \( \frac{3}{1} = 3 \)
- \( \frac{3}{2} = \frac{3}{2} \)
- \( \frac{3}{3} = 1 \)
- \( \frac{3}{6} = \frac{1}{2} \)
- \( \frac{-3}{1} = -3 \)
- \( \frac{-3}{2} = -\frac{3}{2} \)
- \( \frac{-3}{3} = -1 \)
- \( \frac{-3}{6} = -\frac{1}{2} \)
Step 5: Gather the possible rational zeros. From the calculations, the possible rational zeros are:
- \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{3}{2}, \pm \frac{1}{6}, \pm 6 \)
Now, let's analyze your provided options:
- negative 3: Yes, this is a possible zero.
- negative 2 thirds: No, this is not a possible zero.
- 3 halves: Yes, this is a possible zero.
- 6: Yes, this is not directly listed, but it corresponds since the ratio of \( \frac{3}{1} \) can be \( 6 \) if we treat the factors correctly.
Thus, the possible rational zeros from the provided options are:
- Negative 3
- 3 halves
- 6
Final Answer:
- negative 3
- 3 halves
- 6