If the sides of the triangle are the roots of the equation `x^(3)-2x^(2)-x-16`=0, then the product of the in-radius and circum-radius of the triangle ,is

Options

A. 3
B. 6
C. 4
D. 2

Answer

To solve the problem, we need to find the product of the in-radius (r) and circum-radius (R) of a triangle whose sides are the roots of the cubic equation \(x^3 - 2x^2 - x - 16 = 0\).

### Step-by-step Solution:

1. **Identify the roots of the cubic equation**:
The given equation is:
\[ x^3 - 2x^2 - x - 16 = 0 \]
Let the roots of this equation be \(a\), \(b\), and \(c\), which represent the sides of the triangle.

2. **Use Vieta's Formulas**:
According to Vieta's formulas, for a cubic equation \(x^3 + px^2 + qx + r = 0\):
- The sum of the roots \(a + b + c = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -(-2) = 2\).
- The product of the roots \(abc = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -(-16) = 16\).

3. **Calculate the semi-perimeter (s)**:
The semi-perimeter \(s\) of the triangle is given by:
\[ s = \frac{a + b + c}{2} = \frac{2}{2} = 1 \]

4. **Calculate the area (Δ) of the triangle**:
The area of the triangle can be calculated using Heron's formula:
\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]
However, we can also use the relation between the circum-radius \(R\) and the area:
\[ R = \frac{abc}{4\Delta} \]
Thus, we need to express \(\Delta\) in terms of \(s\) and \(abc\).

5. **Express the product of in-radius and circum-radius**:
The product of the in-radius \(r\) and circum-radius \(R\) is given by:
\[ r \cdot R = \frac{\Delta}{s} \cdot \frac{abc}{4\Delta} = \frac{abc}{4s} \]

6. **Substitute the values**:
We have \(abc = 16\) and \(s = 1\):
\[ r \cdot R = \frac{16}{4 \cdot 1} = \frac{16}{4} = 4 \]

### Final Answer:
The product of the in-radius and circum-radius of the triangle is \(4\).