For all `x gt 5, (5x^2-x^3)/(x^2-x- 20)` = ?

Options

A. `(-x^2)/(x+4)`
B. `x^2/(x+4)`
C. `(-5+x)/(20)`
D. `(5x-x^2)/(x-19)`

Answer

To simplify the expression \(\frac{5x^2 - x^3}{x^2 - x - 20}\) for \(x > 5\), we will follow these steps:

### Step 1: Factor the numerator
The numerator is \(5x^2 - x^3\). We can factor out \(-x^2\) from the numerator:

\[ 5x^2 - x^3 = -x^2(x - 5) \]

### Step 2: Factor the denominator
Now, let's factor the denominator \(x^2 - x - 20\). We need to find two numbers that multiply to \(-20\) and add to \(-1\). These numbers are \(-5\) and \(4\). Therefore, we can factor the denominator as follows:

\[ x^2 - x - 20 = (x - 5)(x + 4) \]

### Step 3: Rewrite the expression
Now we can rewrite the original expression using the factored forms:

\[ \frac{5x^2 - x^3}{x^2 - x - 20} = \frac{-x^2(x - 5)}{(x - 5)(x + 4)} \]

### Step 4: Cancel common factors
Since \(x > 5\), we can safely cancel the common factor \((x - 5)\) from the numerator and the denominator:

\[ = \frac{-x^2}{x + 4} \]

### Final Result
Thus, the simplest form of the expression is:

\[ \frac{-x^2}{x + 4} \]