Solve the equation
` x^4- 9x^3 + 27 x^2 - 29 x + 6 =0` given that ` 2 - sqrt(3)` is a root.

Answer

To solve the equation \( x^4 - 9x^3 + 27x^2 - 29x + 6 = 0 \) given that \( 2 - \sqrt{3} \) is a root, we will follow these steps:

### Step 1: Identify the conjugate root
Since \( 2 - \sqrt{3} \) is a root, its conjugate \( 2 + \sqrt{3} \) must also be a root of the polynomial.

### Step 2: Form a quadratic equation from the known roots
Let \( \alpha = 2 - \sqrt{3} \) and \( \beta = 2 + \sqrt{3} \). The sum and product of these roots can be calculated as follows:
- Sum: \( \alpha + \beta = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4 \)
- Product: \( \alpha \beta = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1 \)

Using the sum and product of the roots, we can form the quadratic equation:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \implies x^2 - 4x + 1 = 0 \]

### Step 3: Divide the original polynomial by the quadratic
Now, we will divide the original polynomial \( x^4 - 9x^3 + 27x^2 - 29x + 6 \) by \( x^2 - 4x + 1 \) using polynomial long division.

1. Divide the leading term: \( x^4 \div x^2 = x^2 \)
2. Multiply \( x^2 \) by \( x^2 - 4x + 1 \):
\[ x^2(x^2 - 4x + 1) = x^4 - 4x^3 + x^2 \]
3. Subtract from the original polynomial:
\[ (x^4 - 9x^3 + 27x^2 - 29x + 6) - (x^4 - 4x^3 + x^2) = -5x^3 + 26x^2 - 29x + 6 \]
4. Repeat the process:
- Divide \( -5x^3 \div x^2 = -5x \)
- Multiply and subtract:
\[ -5x(x^2 - 4x + 1) = -5x^3 + 20x^2 - 5x \]
\[ (-5x^3 + 26x^2 - 29x + 6) - (-5x^3 + 20x^2 - 5x) = 6x^2 - 24x + 6 \]
5. Divide \( 6x^2 \div x^2 = 6 \)
6. Multiply and subtract:
\[ 6(x^2 - 4x + 1) = 6x^2 - 24x + 6 \]
\[ (6x^2 - 24x + 6) - (6x^2 - 24x + 6) = 0 \]

### Step 4: Result of the division
The result of the division is:
\[ x^2 - 5x + 6 = 0 \]

### Step 5: Solve the quadratic equation
Now, we can solve the quadratic equation \( x^2 - 5x + 6 = 0 \) using the factorization method:
\[ (x - 2)(x - 3) = 0 \]
Thus, the roots are:
\[ x = 2 \quad \text{and} \quad x = 3 \]

### Step 6: List all roots
Now we have found all the roots of the original polynomial:
1. \( 2 - \sqrt{3} \)
2. \( 2 + \sqrt{3} \)
3. \( 2 \)
4. \( 3 \)

### Final Answer
The roots of the equation \( x^4 - 9x^3 + 27x^2 - 29x + 6 = 0 \) are:
\[ 2 - \sqrt{3}, \quad 2 + \sqrt{3}, \quad 2, \quad 3 \]