Solve the equation
`3x^3 -4x^2 +x +88 =0` given that ` 2- sqrt(-7)` is a root .

Answer

To solve the equation \(3x^3 - 4x^2 + x + 88 = 0\) given that \(2 - \sqrt{-7}\) is a root, we can follow these steps:

### Step 1: Identify the roots
Since \(2 - \sqrt{-7}\) is a root, we can express it as \(2 - i\sqrt{7}\). By the property of complex roots, the conjugate \(2 + i\sqrt{7}\) must also be a root. Let's denote:
- \( \alpha = 2 - i\sqrt{7} \)
- \( \beta = 2 + i\sqrt{7} \)

Let the third root be \( \gamma \).

### Step 2: Form the quadratic from the known roots
We can form a quadratic polynomial from the roots \( \alpha \) and \( \beta \):
\[ (x - \alpha)(x - \beta) = (x - (2 - i\sqrt{7}))(x - (2 + i\sqrt{7})) \]
Using the identity \( (a - b)(a + b) = a^2 - b^2 \):
\[ = (x - 2)^2 - (i\sqrt{7})^2 \]
Calculating this gives:
\[ = (x - 2)^2 - (-7) = (x - 2)^2 + 7 \]
Expanding \( (x - 2)^2 \):
\[ = x^2 - 4x + 4 + 7 = x^2 - 4x + 11 \]

### Step 3: Divide the original polynomial by the quadratic
Now, we need to divide \(3x^3 - 4x^2 + x + 88\) by \(x^2 - 4x + 11\) to find the third root \( \gamma \).

#### Long Division:
1. Divide \(3x^3\) by \(x^2\) to get \(3x\).
2. Multiply \(3x\) by \(x^2 - 4x + 11\):
\[ 3x(x^2 - 4x + 11) = 3x^3 - 12x^2 + 33x \]
3. Subtract this from the original polynomial:
\[ (3x^3 - 4x^2 + x + 88) - (3x^3 - 12x^2 + 33x) = (8x^2 - 32x + 88) \]
4. Divide \(8x^2\) by \(x^2\) to get \(8\).
5. Multiply \(8\) by \(x^2 - 4x + 11\):
\[ 8(x^2 - 4x + 11) = 8x^2 - 32x + 88 \]
6. Subtract this from \(8x^2 - 32x + 88\):
\[ (8x^2 - 32x + 88) - (8x^2 - 32x + 88) = 0 \]

### Step 4: Solve for the third root
After division, we find that:
\[ 3x + 8 = 0 \]
Solving for \(x\):
\[ x = -\frac{8}{3} \]

### Final Roots
Thus, the roots of the equation \(3x^3 - 4x^2 + x + 88 = 0\) are:
1. \( \alpha = 2 - i\sqrt{7} \)
2. \( \beta = 2 + i\sqrt{7} \)
3. \( \gamma = -\frac{8}{3} \)