The equation whose roots are ` 1,2 +- 3i` is

Options

A. `x^3 -4x^2 +x+6=0`
B. `x^3 -x^2 -3x +3=0`
C. `x^3-4x^2 -8x +8=0`
D. `x^3 -5x^2 +17x-13=0`

Answer

To find the equation whose roots are \(1\), \(2 + 3i\), and \(2 - 3i\), we can follow these steps:

### Step 1: Identify the Roots
The roots of the equation are given as:
- \( \alpha = 2 + 3i \)
- \( \beta = 2 - 3i \)
- \( \gamma = 1 \)

### Step 2: Form the Polynomial
The polynomial can be formed using the roots by the formula:
\[ f(x) = (x - \alpha)(x - \beta)(x - \gamma) \]

### Step 3: Multiply the Complex Roots
First, we will multiply the complex roots \( \alpha \) and \( \beta \):
\[ (x - \alpha)(x - \beta) = (x - (2 + 3i))(x - (2 - 3i)) \]
Using the difference of squares:
\[ = (x - 2 - 3i)(x - 2 + 3i) = (x - 2)^2 - (3i)^2 \]
Calculating this gives:
\[ = (x - 2)^2 - 9(-1) = (x - 2)^2 + 9 \]
Expanding \( (x - 2)^2 \):
\[ = x^2 - 4x + 4 + 9 = x^2 - 4x + 13 \]

### Step 4: Multiply by the Real Root
Now, we multiply the result by \( (x - \gamma) = (x - 1) \):
\[ f(x) = (x^2 - 4x + 13)(x - 1) \]
Expanding this:
\[ = x^3 - x^2 - 4x^2 + 4x + 13x - 13 \]
Combining like terms:
\[ = x^3 - 5x^2 + 17x - 13 \]

### Final Equation
Thus, the required polynomial equation is:
\[ f(x) = x^3 - 5x^2 + 17x - 13 \]