Answer :
We start with the function
$$
g(x)=3x^5-2x^4+9x^3-x^2+12.
$$
According to the Rational Root Theorem, any potential rational root of a polynomial of the form
$$
a_nx^n + \cdots + a_0
$$
is given by
$$
\pm \frac{p}{q},
$$
where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$.
For $g(x)$ we have:
- The leading coefficient is $3$, so its factors are $\pm 1$ and $\pm 3$.
- The constant term is $12$, so its factors are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, and $\pm 12$.
Thus, the potential rational roots for $g(x)$ are given by
$$
\pm \frac{p}{q} = \pm\frac{\text{factor of } 12}{\text{factor of } 3}.
$$
This leads to the set of candidates
$$
\left\{ \pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 12,\ \pm \frac{1}{3},\ \pm \frac{2}{3},\ \pm \frac{4}{3} \right\}.
$$
Now, let’s analyze the first option:
$$
f(x)=3x^5-2x^4-9x^3+x^2-12.
$$
For this function:
- The leading coefficient is also $3$, so its factors remain $\pm 1$ and $\pm 3$.
- The constant term is $-12$. When applying the theorem, we consider the absolute value, which is $12$, and its factors are the same: $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, and $\pm 12$.
Thus, the potential rational roots for $f(x)$ are also
$$
\pm \frac{\text{factor of } 12}{\text{factor of } 3} = \left\{ \pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 12,\ \pm \frac{1}{3},\ \pm \frac{2}{3},\ \pm \frac{4}{3} \right\}.
$$
Since the set of potential rational roots of $g(x)$ and this $f(x)$ are identical, we conclude that the function
$$
f(x)=3x^5-2x^4-9x^3+x^2-12
$$
has the same set of potential rational roots as $g(x)$.
Therefore, the answer is that the first option is correct.
$$
g(x)=3x^5-2x^4+9x^3-x^2+12.
$$
According to the Rational Root Theorem, any potential rational root of a polynomial of the form
$$
a_nx^n + \cdots + a_0
$$
is given by
$$
\pm \frac{p}{q},
$$
where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$.
For $g(x)$ we have:
- The leading coefficient is $3$, so its factors are $\pm 1$ and $\pm 3$.
- The constant term is $12$, so its factors are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, and $\pm 12$.
Thus, the potential rational roots for $g(x)$ are given by
$$
\pm \frac{p}{q} = \pm\frac{\text{factor of } 12}{\text{factor of } 3}.
$$
This leads to the set of candidates
$$
\left\{ \pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 12,\ \pm \frac{1}{3},\ \pm \frac{2}{3},\ \pm \frac{4}{3} \right\}.
$$
Now, let’s analyze the first option:
$$
f(x)=3x^5-2x^4-9x^3+x^2-12.
$$
For this function:
- The leading coefficient is also $3$, so its factors remain $\pm 1$ and $\pm 3$.
- The constant term is $-12$. When applying the theorem, we consider the absolute value, which is $12$, and its factors are the same: $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, and $\pm 12$.
Thus, the potential rational roots for $f(x)$ are also
$$
\pm \frac{\text{factor of } 12}{\text{factor of } 3} = \left\{ \pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 12,\ \pm \frac{1}{3},\ \pm \frac{2}{3},\ \pm \frac{4}{3} \right\}.
$$
Since the set of potential rational roots of $g(x)$ and this $f(x)$ are identical, we conclude that the function
$$
f(x)=3x^5-2x^4-9x^3+x^2-12
$$
has the same set of potential rational roots as $g(x)$.
Therefore, the answer is that the first option is correct.
