Find all the values of p for which the equation `x^(4)-4x^(3)-8x^(2)+p=0` has
two real solutions

Answer

To find all the values of \( p \) for which the equation

\[ x^4 - 4x^3 - 8x^2 + p = 0 \]

has exactly two real solutions, we can analyze the function \( f(x) = x^4 - 4x^3 - 8x^2 + p \).

### Step 1: Differentiate the function
First, we differentiate the function to find its critical points:

\[ f'(x) = \frac{d}{dx}(x^4 - 4x^3 - 8x^2 + p) = 4x^3 - 12x^2 - 16x \]

### Step 2: Factor the derivative
Next, we can factor out the common term from the derivative:

\[ f'(x) = 4(x^3 - 3x^2 - 4x) = 4x(x^2 - 3x - 4) \]

Now, we can further factor the quadratic:

\[ x^2 - 3x - 4 = (x - 4)(x + 1) \]

Thus, the derivative can be expressed as:

\[ f'(x) = 4x(x - 4)(x + 1) \]

### Step 3: Find critical points
Setting the derivative equal to zero gives us the critical points:

\[ 4x(x - 4)(x + 1) = 0 \]

This yields the critical points:

\[ x = 0, \quad x = 4, \quad x = -1 \]

### Step 4: Analyze the intervals
We will analyze the behavior of \( f(x) \) in the intervals determined by the critical points: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 4) \), and \( (4, \infty) \).

- For \( x < -1 \), \( f'(x) < 0 \) (decreasing).
- For \( -1 < x < 0 \), \( f'(x) > 0 \) (increasing).
- For \( 0 < x < 4 \), \( f'(x) < 0 \) (decreasing).
- For \( x > 4 \), \( f'(x) > 0 \) (increasing).

### Step 5: Determine conditions for two real solutions
To have exactly two real solutions, the function must touch the x-axis at two points, which means it must have a double root at one of the critical points and cross the x-axis at another.

1. **At \( x = -1 \)**:
\[ f(-1) = (-1)^4 - 4(-1)^3 - 8(-1)^2 + p = 1 + 4 - 8 + p = p - 3 \]
For \( f(-1) < 0 \):
\[ p - 3 < 0 \implies p < 3 \]

2. **At \( x = 0 \)**:
\[ f(0) = 0^4 - 4(0)^3 - 8(0)^2 + p = p \]
For \( f(0) < 0 \):
\[ p < 0 \]

3. **At \( x = 4 \)**:
\[ f(4) = 4^4 - 4(4)^3 - 8(4)^2 + p = 256 - 256 - 128 + p = p - 128 \]
For \( f(4) < 0 \):
\[ p - 128 < 0 \implies p < 128 \]

### Step 6: Combine conditions
The conditions we derived are:
- \( p < 3 \)
- \( p < 0 \)
- \( p < 128 \)

The most restrictive condition is \( p < 0 \).

### Final Answer
Thus, the values of \( p \) for which the equation \( x^4 - 4x^3 - 8x^2 + p = 0 \) has exactly two real solutions are:

\[ \boxed{(-\infty, 0)} \]