Solve the equation
`3x^3 -23x^2 + 72 x -70=0` one root being ` 3+ sqrt(-5)`

Answer

To solve the equation \(3x^3 - 23x^2 + 72x - 70 = 0\) given that one root is \(3 + \sqrt{-5}\), we will follow these steps:

### Step 1: Identify the roots
Given that one root is \(3 + \sqrt{-5}\), we can express this as:
\[ 3 + \sqrt{-5} = 3 + i\sqrt{5} \]
Since the coefficients of the polynomial are real, the complex conjugate must also be a root. Therefore, the second root is:
\[ 3 - \sqrt{-5} = 3 - i\sqrt{5} \]

### Step 2: Find the quadratic factor
Let the roots be \(\alpha = 3 + i\sqrt{5}\) and \(\beta = 3 - i\sqrt{5}\). The quadratic factor corresponding to these roots can be written as:
\[ (x - \alpha)(x - \beta) = (x - (3 + i\sqrt{5}))(x - (3 - i\sqrt{5})) \]
Using the difference of squares:
\[ = (x - 3)^2 - (i\sqrt{5})^2 \]
Calculating this gives:
\[ = (x - 3)^2 - (-5) = (x - 3)^2 + 5 \]
Now expanding \((x - 3)^2\):
\[ = x^2 - 6x + 9 + 5 = x^2 - 6x + 14 \]

### Step 3: Divide the original polynomial by the quadratic factor
We will divide the original polynomial \(3x^3 - 23x^2 + 72x - 70\) by the quadratic factor \(x^2 - 6x + 14\) using polynomial long division.

1. Divide the leading term: \(3x^3 \div x^2 = 3x\).
2. Multiply \(3x\) by \(x^2 - 6x + 14\):
\[ 3x(x^2 - 6x + 14) = 3x^3 - 18x^2 + 42x \]
3. Subtract this from the original polynomial:
\[ (3x^3 - 23x^2 + 72x - 70) - (3x^3 - 18x^2 + 42x) = -5x^2 + 30x - 70 \]
4. Now divide the leading term: \(-5x^2 \div x^2 = -5\).
5. Multiply \(-5\) by \(x^2 - 6x + 14\):
\[ -5(x^2 - 6x + 14) = -5x^2 + 30x - 70 \]
6. Subtract this from \(-5x^2 + 30x - 70\):
\[ (-5x^2 + 30x - 70) - (-5x^2 + 30x - 70) = 0 \]

### Step 4: Find the remaining root
The result of the division shows that:
\[ 3x^3 - 23x^2 + 72x - 70 = (x^2 - 6x + 14)(3x - 5) \]
Now, we can find the remaining root by setting \(3x - 5 = 0\):
\[ 3x = 5 \implies x = \frac{5}{3} \]

### Conclusion
The roots of the equation \(3x^3 - 23x^2 + 72x - 70 = 0\) are:
1. \(3 + i\sqrt{5}\)
2. \(3 - i\sqrt{5}\)
3. \(\frac{5}{3}\)