Assertion : The remainder when `p(x)=x^(3)-6x^(2)+2x-4` is divided by `(3x-1)"is"(-107)/(27)`
Reason : If a polynomial `p(x)` is divided by `ax-b` , the remainder is the value of `p(x)"at"(x=(b)/(a))` .

Options

A. If both assertion and reason are true and reason is the correct explanation of assertion .
B. If both assertion and reason are true but reason is not the correct explanation of assertion .
C. If assertion is true but reason is false .
D. If assertion is false but reason is true .

Answer

To solve the problem, we need to determine the remainder when the polynomial \( p(x) = x^3 - 6x^2 + 2x - 4 \) is divided by \( 3x - 1 \). We will use the Remainder Theorem, which states that if a polynomial \( p(x) \) is divided by \( ax - b \), the remainder is the value of \( p(x) \) evaluated at \( x = \frac{b}{a} \).

### Step-by-Step Solution:

1. **Identify the Polynomial and the Divisor**:
- The polynomial is \( p(x) = x^3 - 6x^2 + 2x - 4 \).
- The divisor is \( 3x - 1 \).

2. **Set the Divisor to Zero to Find the Value of \( x \)**:
- To find the value of \( x \) at which we will evaluate \( p(x) \), set \( 3x - 1 = 0 \).
- Solving for \( x \):
\[ 3x = 1 \implies x = \frac{1}{3} \]

3. **Evaluate the Polynomial at \( x = \frac{1}{3} \)**:
- Substitute \( x = \frac{1}{3} \) into the polynomial \( p(x) \):
\[ p\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 6\left(\frac{1}{3}\right)^2 + 2\left(\frac{1}{3}\right) - 4 \]

4. **Calculate Each Term**:
- Calculate \( \left(\frac{1}{3}\right)^3 = \frac{1}{27} \).
- Calculate \( 6\left(\frac{1}{3}\right)^2 = 6 \cdot \frac{1}{9} = \frac{6}{9} = \frac{2}{3} \).
- Calculate \( 2\left(\frac{1}{3}\right) = \frac{2}{3} \).
- Now substitute these values back into the polynomial:
\[ p\left(\frac{1}{3}\right) = \frac{1}{27} - \frac{2}{3} + \frac{2}{3} - 4 \]

5. **Combine the Terms**:
- The \( -\frac{2}{3} \) and \( +\frac{2}{3} \) cancel each other out:
\[ p\left(\frac{1}{3}\right) = \frac{1}{27} - 4 \]
- To combine \( \frac{1}{27} \) and \( -4 \), convert \( -4 \) to a fraction with a denominator of 27:
\[ -4 = -\frac{108}{27} \]
- Now combine:
\[ p\left(\frac{1}{3}\right) = \frac{1}{27} - \frac{108}{27} = \frac{1 - 108}{27} = \frac{-107}{27} \]

6. **Conclusion**:
- The remainder when \( p(x) \) is divided by \( 3x - 1 \) is \( \frac{-107}{27} \).

### Final Answer:
The assertion is true: the remainder when \( p(x) = x^3 - 6x^2 + 2x - 4 \) is divided by \( 3x - 1 \) is \( \frac{-107}{27} \).