`(x^(2)-4x-5)/((x-1)^(2)-4(x-1))`
The expression above is equivalent to

Options

A. `x+1`
B. `(x+5)/(x-1)`
C. `(x+1)/(x-1)`
D. `(x+1)/(x-4)`

Answer

To solve the expression \((x^2 - 4x - 5) / ((x - 1)^2 - 4(x - 1))\), we will follow these steps:

### Step 1: Factor the numerator
The numerator is \(x^2 - 4x - 5\). We need to factor this quadratic expression.

To factor \(x^2 - 4x - 5\), we look for two numbers that multiply to \(-5\) (the constant term) and add to \(-4\) (the coefficient of \(x\)). The numbers \(-5\) and \(1\) satisfy this condition.

Thus, we can factor the numerator as:
\[ x^2 - 4x - 5 = (x - 5)(x + 1) \]

### Step 2: Simplify the denominator
The denominator is \((x - 1)^2 - 4(x - 1)\). We can simplify this expression.

First, expand \((x - 1)^2\):
\[ (x - 1)^2 = x^2 - 2x + 1 \]
Now substitute this back into the denominator:
\[ x^2 - 2x + 1 - 4(x - 1) = x^2 - 2x + 1 - 4x + 4 = x^2 - 6x + 5 \]

Next, we factor \(x^2 - 6x + 5\). We look for two numbers that multiply to \(5\) and add to \(-6\). The numbers \(-5\) and \(-1\) work.

Thus, we can factor the denominator as:
\[ x^2 - 6x + 5 = (x - 5)(x - 1) \]

### Step 3: Rewrite the expression
Now we can rewrite the original expression:
\[ \frac{(x - 5)(x + 1)}{(x - 5)(x - 1)} \]

### Step 4: Cancel common factors
We see that \((x - 5)\) is a common factor in both the numerator and denominator. We can cancel this out (assuming \(x \neq 5\)):
\[ \frac{x + 1}{x - 1} \]

### Conclusion
The expression \((x^2 - 4x - 5) / ((x - 1)^2 - 4(x - 1))\) simplifies to:
\[ \frac{x + 1}{x - 1} \]

Thus, the equivalent expression is option **C: \(\frac{x + 1}{x - 1}\)**.

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