Solve the equation
`x^3 +6x +20=0` one root being `l +3i`

Answer

To solve the equation \( x^3 + 6x + 20 = 0 \) given that one root is \( 1 + 3i \), we can follow these steps:

### Step 1: Identify the roots
Since the coefficients of the polynomial are real numbers, if \( 1 + 3i \) is a root, then its complex conjugate \( 1 - 3i \) must also be a root.

### Step 2: Let the roots be defined
Let the roots of the polynomial be:
- \( r_1 = 1 + 3i \)
- \( r_2 = 1 - 3i \)
- \( r_3 = a \) (where \( a \) is the real root we need to find)

### Step 3: Use the sum of the roots
The sum of the roots for a cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ r_1 + r_2 + r_3 = -\frac{b}{a} \]
In our case, \( a = 1 \) and \( b = 0 \). Therefore:
\[ (1 + 3i) + (1 - 3i) + a = -0 \]
This simplifies to:
\[ 2 + a = 0 \]
From this, we can solve for \( a \):
\[ a = -2 \]

### Step 4: Use the product of the roots
The product of the roots is given by:
\[ r_1 \cdot r_2 \cdot r_3 = -\frac{d}{a} \]
Here, \( d = 20 \) and \( a = 1 \). Thus:
\[ (1 + 3i)(1 - 3i)(-2) = -20 \]
Calculating \( (1 + 3i)(1 - 3i) \):
\[ (1 + 3i)(1 - 3i) = 1^2 - (3i)^2 = 1 - 9(-1) = 1 + 9 = 10 \]
So we have:
\[ 10 \cdot (-2) = -20 \]
This confirms that our roots are consistent.

### Step 5: List the roots
The roots of the polynomial \( x^3 + 6x + 20 = 0 \) are:
- \( r_1 = 1 + 3i \)
- \( r_2 = 1 - 3i \)
- \( r_3 = -2 \)

### Final Solution
Thus, the roots of the equation \( x^3 + 6x + 20 = 0 \) are \( 1 + 3i, 1 - 3i, \) and \( -2 \).

---