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To solve the polynomial equation \(2x^3 - x^2 - 8x - 3 = 0\), given that \(1 + \sqrt{2}\) is a root, we can follow these steps:

### Step 1: Use the Conjugate Root Theorem
Since \(1 + \sqrt{2}\) is a root and the polynomial has real coefficients, \(1 - \sqrt{2}\) must also be a root.

### Step 2: Form a Quadratic Factor
The roots \(1 + \sqrt{2}\) and \(1 - \sqrt{2}\) correspond to the quadratic factor:
\[
(x - (1 + \sqrt{2}))(x - (1 - \sqrt{2}))
\]
Expanding this factor, we get:
\[
(x - 1 - \sqrt{2})(x - 1 + \sqrt{2}) = (x - 1)^2 - (\sqrt{2})^2 = x^2 - 2x - 1
\]
So, \(x^2 - 2x - 1\) is a factor of \(2x^3 - x^2 - 8x - 3\).

### Step 3: Divide the Polynomial by the Quadratic Factor
We now divide \(2x^3 - x^2 - 8x - 3\) by \(x^2 - 2x - 1\) to find the remaining factor.

Using synthetic or long division, we find that:
\[
2x^3 - x^2 - 8x - 3 = (x^2 - 2x - 1)(2x + 3)
\]

### Step 4: Solve Each Factor
1. For the quadratic factor \(x^2 - 2x - 1 = 0\), the roots are:
   \[
   x = 1 \pm \sqrt{2}
   \]
2. For the linear factor \(2x + 3 = 0\), solving for \(x\) gives:
   \[
   x = -\frac{3}{2}
   \]

### Final Answer
The roots of the equation \(2x^3 - x^2 - 8x - 3 = 0\) are:
\[
x = 1 + \sqrt{2}, \quad x = 1 - \sqrt{2}, \quad x = -\frac{3}{2}
\]

Would you like further details on any specific step?

---

1. How is the Conjugate Root Theorem applied in polynomial equations?
2. How do you perform synthetic division with polynomials?
3. What are the steps to expand the product of binomials like \((x - (1 + \sqrt{2}))(x - (1 - \sqrt{2}))\)?
4. How can you identify possible factors of a polynomial?
5. How would you verify if a given solution is a root of the polynomial?

**Tip:** Always check if known roots simplify the polynomial division process.

Solve the equation 2x^3 - x^2 - 8x - 3 = 0, given that 1 + √2 is a root.

Learn how to solve the polynomial equation 2x^3 - x^2 - 8x - 3 = 0, given that 1 + √2 is a root. This solution demonstrates the application of the Conjugate Root Theorem and polynomial division, making it ideal for students in Grades 10-12.

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