Find all roots of the equation `x^4+2x^3-16x^2-22x+7=0`, if one root is `2+sqrt3`

Explanation

To find all the roots of the polynomial equation \( x^4 + 2x^3 - 16x^2 - 22x + 7 = 0 \) given that one root is \( 2 + \sqrt{3} \), we can follow these steps:

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

### Step 2: Use Vieta's formulas
Let the roots of the polynomial be \( r_1 = 2 + \sqrt{3} \), \( r_2 = 2 - \sqrt{3} \), \( r_3 = C \), and \( r_4 = D \). According to Vieta's formulas:
- The sum of the roots \( r_1 + r_2 + r_3 + r_4 = -\frac{\text{coefficient of } x^3}{\text{coefficient of } x^4} = -2 \).
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