Find the remainder when `f(x)=x^3-6x^2+2x-4` is divided by `g(x)=3x-1.`

Options

A. `-107/27`
B. `-190/56`
C. `-179/79`
D. `907/25`

Explanation

To find the remainder when \( f(x) = x^3 - 6x^2 + 2x - 4 \) is divided by \( g(x) = 3x - 1 \), we can use the Remainder Theorem. According to the theorem, the remainder of the division of a polynomial \( f(x) \) by a linear polynomial \( g(x) = ax + b \) can be found by evaluating \( f \) at \( x = -\frac{b}{a} \).

### Step-by-Step Solution:

1. **Identify the coefficients of \( g(x) \)**:
\[ g(x) = 3x - 1 \implies a = 3, \, b = -1 \]

2. **Find the value of \( x \) where \( g(x) = 0 \)**:
\[ 3x - 1 = 0 \implies 3x = 1 \implies x = \frac{1}{3} \]

3. **Evaluate \( f(x) \) at \( x = \frac{1}{3} \)**:
\[ f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 6\left(\frac{1}{3}\right)^2 + 2\left(\frac{1}{3}\right) - 4 \]

4. **Calculate each term**:
- First term:
\[ \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]
- Second term:
\[ -6\left(\frac{1}{3}\right)^2 = -6 \cdot \frac{1}{9} = -\frac{6}{9} = -\frac{2}{3} \]
- Third term:
\[ 2\left(\frac{1}{3}\right) = \frac{2}{3} \]
- Fourth term:
\[ -4 \]

5. **Combine all the terms**:
\[ f\left(\frac{1}{3}\right) = \frac{1}{27} - \frac{2}{3} + \frac{2}{3} - 4 \]
The \( -\frac{2}{3} \) and \( +\frac{2}{3} \) cancel each other out:
\[ = \frac{1}{27} - 4 \]

6. **Convert \( -4 \) to a fraction with a denominator of 27**:
\[ -4 = -\frac{108}{27} \]

7. **Final calculation**:
\[ f\left(\frac{1}{3}\right) = \frac{1}{27} - \frac{108}{27} = \frac{1 - 108}{27} = \frac{-107}{27} \]

### Conclusion:
The remainder when \( f(x) \) is divided by \( g(x) \) is:
\[ \boxed{\frac{-107}{27}} \]