Use synthetic division to show that x is a solution of the equation, and use the result to factor the polynomial completely. List all real solutions of the equation. (Enter your answers as a comma-separated list.)
x^3 + 4x^2 − 2x − 8 = 0,    x =*root* 2


To determine if \( x = 2 \) is a solution of the equation \( x^3 + 4x^2 - 2x - 8 = 0 \), we will use synthetic division. Then, we will factor the polynomial completely and find all real solutions.

### Step 1: Synthetic Division

We'll divide the polynomial \( x^3 + 4x^2 - 2x - 8 \) by \( x - 2 \) using synthetic division.

1. Write down the coefficients of the polynomial \( x^3 + 4x^2 - 2x - 8 \):
   - Coefficients: \( 1, 4, -2, -8 \)
   
2. Set up the synthetic division with \( x = 2 \) as follows:

\[
\begin{array}{r|rrrr}
2 & 1 & 4 & -2 & -8 \\
  &   & 2 & 12 & 20 \\
\hline
  & 1 & 6 & 10 & 0 \\
\end{array}
\]

### Step 2: Explanation of the Synthetic Division Process

1. Bring down the first coefficient, \( 1 \), and place it under the line.
2. Multiply \( 2 \) (the value of \( x \)) by \( 1 \), which gives \( 2 \), and write it under the next coefficient.
3. Add \( 4 + 2 = 6 \).
4. Multiply \( 2 \times 6 = 12 \) and write it under the next coefficient.
5. Add \( -2 + 12 = 10 \).
6. Multiply \( 2 \times 10 = 20 \), and write it under the last coefficient.
7. Add \( -8 + 20 = 0 \), which is the remainder.

Since the remainder is zero, \( x = 2 \) is a solution, and the quotient from the division is \( x^2 + 6x + 10 \).

### Step 3: Factor the Polynomial

The result from the synthetic division gives us:

\[
x^3 + 4x^2 - 2x - 8 = (x - 2)(x^2 + 6x + 10)
\]

Next, we attempt to factor \( x^2 + 6x + 10 \). However, this quadratic does not factor nicely using real numbers, so we apply the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 6 \), and \( c = 10 \).

\[
x = \frac{-6 \pm \sqrt{6^2 - 4(1)(10)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 40}}{2} = \frac{-6 \pm \sqrt{-4}}{2} = \frac{-6 \pm 2i}{2}
\]

\[
x = -3 \pm i
\]

So, the quadratic \( x^2 + 6x + 10 \) has complex roots \( x = -3 + i \) and \( x = -3 - i \).

### Step 4: List All Real Solutions

The only real solution is \( x = 2 \), and the complete factorization of the polynomial is:

\[
(x - 2)(x^2 + 6x + 10)
\]

### Final Answer:
The real solution is:
\[
x = 2
\]

Do you want more details or have any questions? Here are some related questions for further understanding:

1. How do you interpret complex roots geometrically?
2. What is the significance of the remainder being zero in synthetic division?
3. How do you determine whether a quadratic has real or complex solutions?
4. Can synthetic division be used to divide by any polynomial, or just linear ones?
5. How would you verify the factorization of a cubic polynomial?

**Tip:** When using the quadratic formula, always check the discriminant \( \Delta = b^2 - 4ac \) to determine the nature of the roots (real or complex).

Use synthetic division to show that x is a solution of the equation, and use the result to factor the polynomial completely. List all real solutions of the equation. (Enter your answers as a comma-separated list.) x^3 + 4x^2 − 2x − 8 = 0, x = 2

Explore how to solve the cubic polynomial x^3 + 4x^2 − 2x − 8 = 0 using synthetic division with x = 2. Learn about factoring, quadratic equations, and complex roots, with a step-by-step solution suitable for Grades 10-12.

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