`x^(3)(x^(2)-5)=-4x`
If `x gt 0`, what is one possible solutions to the equation above?

Answer

To solve the equation \( x^3(x^2 - 5) = -4x \) for \( x > 0 \), we can follow these steps:

### Step 1: Rearrange the Equation
Start by moving all terms to one side of the equation:
\[ x^3(x^2 - 5) + 4x = 0 \]

### Step 2: Expand the Left Side
Now, expand the left side:
\[ x^5 - 5x^3 + 4x = 0 \]

### Step 3: Factor Out Common Terms
Next, factor out the common term \( x \):
\[ x(x^4 - 5x^2 + 4) = 0 \]

### Step 4: Set Each Factor to Zero
Since \( x > 0 \), we discard the solution \( x = 0 \) and focus on the quadratic part:
\[ x^4 - 5x^2 + 4 = 0 \]

### Step 5: Substitute for Simplicity
Let \( y = x^2 \). Then the equation becomes:
\[ y^2 - 5y + 4 = 0 \]

### Step 6: Factor the Quadratic Equation
Now, factor the quadratic:
\[ (y - 4)(y - 1) = 0 \]

### Step 7: Solve for \( y \)
Setting each factor to zero gives us:
\[ y - 4 = 0 \quad \text{or} \quad y - 1 = 0 \]
Thus, we have:
\[ y = 4 \quad \text{or} \quad y = 1 \]

### Step 8: Substitute Back for \( x \)
Recall that \( y = x^2 \). Therefore:
\[ x^2 = 4 \quad \Rightarrow \quad x = 2 \quad (\text{since } x > 0) \]
\[ x^2 = 1 \quad \Rightarrow \quad x = 1 \quad (\text{since } x > 0) \]

### Conclusion
The possible solutions for \( x \) are \( x = 2 \) and \( x = 1 \). Since we are looking for a solution where \( x > 0 \), both values are valid.

One possible solution to the equation is:
\[ \boxed{2} \]